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NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Edited By Ramraj Saini | Updated on Sep 14, 2023 08:18 PM IST | #CBSE Class 12th
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NCERT Integrals Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 7 Integrals are discussed here. This chapter deals with definite and indefinite integrals. Integration class 12 also includes elementary properties of integration including basic techniques of integration. NCERT Class 12 maths chapter 7 solutions will be very helpful when you are solving the questions from NCERT books for Class 12 Maths. These NCERT Class 12 Maths solutions chapter 7 are prepared by subject matter experts that are very easy to understand. students can practice integrals class 12 ncert solutions to get good hold on the concepts.

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  1. NCERT Integrals Class 12 Questions And Answers
  2. NCERT Integrals Class 12 Questions And Answers PDF Free Download
  3. Class 12 maths chapter 7 NCERT Solutions - Important Formulae
  4. NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)
  5. NCERT Solutions for Class 12 Maths Chapter 7 Integrals - Topics
  6. NCERT solutions for class 12 maths - Chapter wise
  7. Key Features of NCERT Solutions for Class 12 Maths Chapter 7 Integrals
  8. NCERT solutions for class 12 - subject wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 12 Maths Chapter 7 Integrals
NCERT Solutions for Class 12 Maths Chapter 7 Integrals
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NCERT solutions for class 12 math chapter 7 integrals are important for board exams as well as for competitive examinations like JEE Main, VITEEE, BITSAT, etc but without command, on the concepts of integrals ncert solutions meritorious marks cant be scoured. Therefore chapter 7 maths class 12 is recommended to students. Also, you can check the NCERT solutions for other Classes here.

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NCERT Integrals Class 12 Questions And Answers PDF Free Download

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Class 12 maths chapter 7 NCERT Solutions - Important Formulae

>> Integration as Inverse of Differentiation: Integration is the inverse process of differentiation.

In differential calculus, we find the derivative of a given function, while in integral calculus, we find a function whose derivative is given.

Indefinite Integrals:

∫f(x) dx = F(x) + C

These integrals are called indefinite integrals or general integrals.

C is an arbitrary constant that leads to different anti-derivatives of the given function.

Multiple Anti-Derivatives:

A derivative of a function is unique, but a function can have infinite anti-derivatives or integrals.

Properties of Indefinite Integral:

∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx

For any real number k, ∫k f(x) dx = k∫f(x)dx.

In general, if f1, f2, …, fn are functions and k1, k2, …, kn are real numbers, then ∫[k1f1(x) + k2f2(x) + … + knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫f2(x) dx + … + kn ∫fn(x) dx

First Fundamental Theorem of Integral Calculus:

Define the area function A(x) = ∫[a, x]f(t)dt for x ≥ a, where f is continuous on [a, b].

Then A'(x) = f(x) for every x ∈ [a, b].

Second Fundamental Theorem of Integral Calculus:

If f is a continuous function on [a, b], then ∫[a, b]f(x)dx = F(b) - F(a), where F(x) is an antiderivative of f(x).

Standard Integral Formulas:

∫xn dx = xn+1/(n+1) + C (n ≠ -1)

∫cos x dx = sin x + C

∫sin x dx = -cos x + C

∫sec2 x dx = tan x + C

∫cosec2 x dx = -cot x + C

∫sec x tan x dx = sec x + C

∫cosec x cot x dx = -cosec x + C

∫ex dx = ex + C

∫ax dx = (ax)/ln(a) + C

∫(1/x) dx = ln|x| + C

Other Integral Formulas:

∫tan x dx = ln|sec x| + C

∫cot x dx = ln|sin x| + C

∫sec x dx = ln|sec x + tan x| + C

∫cosec x dx = ln|cosec x - cot x| + C

Free download Class 12 Maths Chapter 7 Question Answer for CBSE Exam.

NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)

Class 12 Integrals NCERT solutions Exercise: 7.1

Question:1 Find an anti derivative (or integral) of the following functions by the method of inspection. sin2x

Answer:

GIven sin2x ;

So, the anti derivative of sin2x is a function of x whose derivative is sin2x .

ddx(cos2x)=2sin2x

sin2x=12ddx(cos2x)

Therefore, we have sin2x=ddx(12cos2x)

Or, antiderivative of sin2x is (12cos2x) .

Question:2 Find an anti derivative (or integral) of the following functions by the method of inspection. cos3x

Answer:

GIven cos3x ;

So, the antiderivative of cos3x is a function of x whose derivative is cos3x .

ddx(sin3x)=3cos3x

cos3x=13ddx(sin3x)

Therefore, we have the anti derivative of cos3x is 13sin3x .

Question:3 Find an anti derivative (or integral) of the following functions by the method of inspection. e2x

Answer:

GIven e2x ;

So, the anti derivative of e2x is a function of x whose derivative is e2x .

ddx(e2x)=2e2x

e2x=12ddx(e2x)

e2x=ddx(12e2x)

Therefore, we have the anti derivative of e2x is 12e2x .

Question:4 Find an anti derivative (or integral) of the following functions by the method of inspection. (ax+b)2

Answer:

GIven (ax+b)2 ;

So, the anti derivative of (ax+b)2 is a function of x whose derivative is (ax+b)2 .

ddx(ax+b)3=3a(ax+b)2

(ax+b)2=13addx(ax+b)3

(ax+b)2=ddx[13a(ax+b)3]

Therefore, we have the anti derivative of (ax+b)2 is [13a(ax+b)3] .

Question:5 Find an anti derivative (or integral) of the following functions by the method of inspection. sin2x4e3x

Answer:

GIven sin2x4e3x ;

So, the anti derivative of sin2x4e3x is a function of x whose derivative is sin2x4e3x .

ddx(12cos2x43e3x)=sin2x4e3x

Therefore, we have the anti derivative of sin2x4e3x is (12cos2x43e3x) .

Question:6 Find the following integrals

(4e3x+1)dx

Answer:

Given intergral (4e3x+1)dx ;

4e3xdx+1dx=4(e3x3)+x+C

or 43e3x+x+C , where C is any constant value.

Question:7 Find the following integrals x2(11x2)dx

Answer:

Given intergral x2(11x2)dx ;

x2dx1dx

or x33x+C , where C is any constant value.

Question:8 Find the following integrals (ax2+bx+c)dx

Answer:

Given intergral (ax2+bx+c)dx ;

ax2 dx+bx dx+c dx

=ax2 dx+bx dx+cdx

=ax33+bx22+cx+C

or ax33+bx22+cx+C , where C is any constant value.

Question:9 Find the following integrals intergration of (2x2+ex)dx

Answer:

Given intergral (2x2+ex)dx ;

2x2 dx+ex dx

=2x2 dx+ex dx

=2x33+ex+C

or 2x33+ex+C , where C is any constant value.

Question:10 Find the following integrals (x1x)2dx

Answer:

Given integral (x1x)2dx ;

or (x+1x2) dx

=x dx+1x dx2dx

=x22+ln|x|2x+C , where C is any constant value.

Question:11 Find the following integrals intergration of x3+5x24x2dx

Answer:

Given intergral x3+5x24x2dx ;

or x3x2 dx+5x2x2 dx41x2 dx

x dx+51.dx4x2 dx

=x22+5x4(x11)+C

Or, x22+5x+4x+C , where C is any constant value.

Question:12 Find the following integrals x3+3x+4xdx

Answer:

Given intergral x3+3x+4xdx ;

or x3x12 dx+3xx12 dx+41x12 dx

=x52 dx+3x12 dx+4x12 dx

=x7272+3(x32)32+4(x12)12+C

Or, =27x72+2x32+8x+C , where C is any constant value.

Question:13 Find the following integrals intergration of x3x2+x1x1dx

Answer:

Given integral x3x2+x1x1dx

It can be written as

=x2(x1)+(x+1)(x1)dx

Taking (x1) common out

=(x1)(x2+1)(x1)dx

Now, cancelling out the term (x1) from both numerator and denominator.

=(x2+1)dx

Splitting the terms inside the brackets

=x2dx+1dx

=x33+x+c

Question:14 Find the following integrals (1x)xdx

Answer:

Given intergral (1x)xdx ;

x dxxx dx or

x12 dxx32 dx

=x3232x5252+C

=23x3225x52+C , where C is any constant value.

Question:15 Find the following integrals x(3x2+2x+3)dx

Answer:

Given intergral x(3x2+2x+3)dx ;

=3x2x dx+2xx dx+3x dx or =3x52 dx+2x32 dx+3x12 dx

=3x7272+2x5252+3x3232+C

=67x72+45x52+2x32+C , where C is any constant value.

Question:16 Find the following integrals (2x3cosx+ex)dx

Answer:

Given integral (2x3cosx+ex)dx ;

splitting the integral as the sum of three integrals

2x dx3cosx dx+ex dx

=2x223sinx+ex+C

=x23sinx+ex+C , where C is any constant value.

Question:17 Find the following integrals (2x23sinx+5x)dx

Answer:

Given integral (2x23sinx+5x)dx ;

2x2 dx3sinx dx+5x dx

=2x333(cosx)+5(x3232)+C

=2x33+3cosx+103x32+C , where C is any constant value.

Question:18 Find the following integrals secx(secx+tanx)dx

Answer:

Given integral secx(secx+tanx)dx ;

(sec2x+secxtanx) dx

Using the integral of trigonometric functions

=(sec2x) dx+secxtanx dx

=tanx+secx+C , where C is any constant value.

Question:19 Find the following integrals intergration of sec2xcosec2xdx

Answer:

Given integral sec2xcosec2xdx ;

1cos2x1sin2x dx

=sin2xcos2x dx

=(sec2x1) dx

=sec2x dx1 dx

=tanxx+C , where C is any constant value.

Question:20 Find the following integrals 23sinxcos2xdx

Answer:

Given integral 23sinxcos2xdx ;

(2cos2x3sinxcos2x) dx

Using antiderivative of trigonometric functions

=2tanx3secx+C , where C is any constant value.

Question:21 Choose the correct answer
The anti derivative of (x+1/x) equals

A)13x1/3+2x1/2+CB)23x2/3+12x2+CC)23x3/2+2x1/2+CD)32x3/2+12x1/2+C

Answer:

Given to find the anti derivative or integral of (x+1/x) ;

(x+1/x) dx

x12 dx+x12 dx

=x3232+x1212+C

=23x32+2x12+C , where C is any constant value.

Hence the correct option is (C).

Question:22 Choose the correct answer The anti derivative of

If ddxf(x)=4x33x4 such that f (2) = 0. Then f (x) is

A)x4+1x31298B)x3+1x41298C)x4+1x3+1298D)x3+1x41298

Answer:

Given that the anti derivative of ddxf(x)=4x33x4

So, ddxf(x)=4x33x4

f(x)=4x33x4 dx

f(x)=4x33x4 dx

f(x)=4(x44)3(x33)+C

f(x)=x4+1x3+C

Now, to find the constant C;

we will put the condition given, f (2) = 0

f(2)=24+123+C=0

16+18+C=0

or C=1298

f(x)=x4+1x31298

Therefore the correct answer is A .

Class 12 Integrals NCERT solutions Exercise: 7.2

Question:1 Integrate the functions 2x1+x2

Answer:

Given to integrate 2x1+x2 function,

Let us assume 1+x2=t

we get, 2xdx=dt

2x1+x2dx=1tdt

=log|t|+C

=log|1+x2|+C now back substituting the value of t=1+x2

as (1+x2) is positive we can write

=log(1+x2)+C

Question:2 Integrate the functions (logx)2x

Answer:

Given to integrate (logx)2x function,

Let us assume log|x|=t

we get, 1xdx=dt

(log|x|)2x dx=t2dt

=t33+C

=(log|x|)33+C

Question:3 Integrate the functions 1x+xlogx

Answer:

Given to integrate 1x+xlogx function,

Let us assume 1+logx=t

we get, 1xdx=dt

1x(1+logx)dx=1tdt

=log|t|+C

=log|1+logx|+C

Question:4 Integrate the functions sinxsin(cosx)

Answer:

Given to integrate sinxsin(cosx) function,

Let us assume cosx=t

we get, sinxdx=dt

sinxsin(cosx)dx=sintdt

=(cost)+C

=cost+C

Back substituting the value of t we get,

=cos(cosx)+C

Question:5 Integrate the functions sin(ax+b)cos(ax+b)

Answer:

Given to integrate sin(ax+b)cos(ax+b) function,

sin(ax+b)cos(ax+b)=2sin(ax+b)cos(ax+b)2=sin2(ax+b)2

Let us assume 2(ax+b)=t

we get, 2adx=dt

sin2(ax+b)2dx=12sint2adt

=14a[cost]+C

Now, by back substituting the value of t,

=14a[cos2(ax+b)]+C

Question:6 Integrate the functions ax+b

Answer:

Given to integrate ax+b function,

Let us assume (ax+b)=t

we get, adx=dt

dx=1adt

(ax+b)12dx=1at12dt

Now, by back substituting the value of t,

=1a(t3232)+C

=2(ax+b)323a+C

Question:7 Integrate the functions xx+2

Answer:

Given function xx+2 ,

xx+2

Assume the (x+2)=t 19634

dx=dt

xx+2dx=(t2)tdt

=(t2)tdt

=(t322t12)dt

=t32dt2t12dt

=t52522(t3232)+C

=25t5243t32+C

Back substituting the value of t in the above equation.

or, 25(x+2)5243(x+2)32+C , where C is any constant value.

Question:8 Integrate the functions x1+2x2

Answer:

Given function x1+2x2 ,

x1+2x2 dx

Assume the 1+2x2=t

4xdx=dt

x1+2x2dx=t4dt

Or =14t12dt=14(t3232)+C

=16(1+2x2)32+C , where C is any constant value.

Question:9 Integrate the functions (4x+2)x2+x+1

Answer:

Given function (4x+2)x2+x+1 ,

(4x+2)x2+x+1dx

Assume the 1+x+x2=t

(2x+1)dx=dt

(4x+2)1+x+x2dx

=2tdt=2tdt

=2(t3232)+C

Now, back substituting the value of t in the above equation,

=43(1+x+x2)32+C , where C is any constant value.

Question:10 Integrate the functions 1xx

Answer:

Given function 1xx ,

1xxdx

Can be written in the form:

1xx=1x(x1)

Assume the (x1)=t

12xdx=dt

1x(x1)dx=2tdt

=2log|t|+C

=2log|x1|+C , where C is any constant value.

Question:11 Integrate the functions xx+4 , x > 0

Answer:

Given function xx+4 ,

xx+4dx

Assume the x+4=t so, x=t4

dx=dt

xx+4dx=t4tdt

t12dt4t12dt

=23t324(2t12)+C

=23(x+4)3216(x+4)12+C

, where C is any constant value.

Question:12 Integrate the functions (x31)1/3x5

Answer:

Given function (x31)1/3x5 ,

(x31)1/3x5dx

Assume the x31=t

3x2dx=dt

(x31)13x5dx=(x31)13x3.x2dx

=t13(t+1)dt3

=13(t43+t13)dt

=13[t7373+t4343]+C

=13[37t73+34t43]+C

=17(x31)73+14(x31)43+C , where C is any constant value.

Question:13 Integrate the functions x2(2+3x3)3

Answer:

Given function x2(2+3x3)3 ,

x2(2+3x3)3dx

Assume the 2+3x3=t

9x2dx=dt

x2(2+3x2)dx=19dtt3

=19(t22)+C

=118(1t2)+C

=118(2+3x3)2+C , where C is any constant value.

Question:14 Integrate the functions 1x(logx)m,x>0,m1

Answer:

Given function 1x(logx)m,x>0,m1 ,

Assume the logx=t

1xdx=dt

1x(logx)mdx=dttm

=(tm+11m)+C

=(logx)1m(1m)+C , where C is any constant value.

Question:15 Integrate the functions x94x2

Answer:

Given function x94x2 ,

Assume the 94x2=t

8xdx=dt

x94x2=181tdt

=18log|t|+C

Now back substituting the value of t ;

=18log|94x2|+C , where C is any constant value.

Question:16 Integrate the functions e2x+3

Answer:

Given function e2x+3 ,

Assume the 2x+3=t

2dx=dt

e2x+3dx=12etdt

=12et+C

Now back substituting the value of t ;

=12e2x+3+C , where C is any constant value.

Question:17 Integrate the functions xex2

Answer:

Given function xex2 ,

Assume the x2=t

2xdx=dt

xex2dx=121etdt

=12etdt

=12(et1)+C

=12ex2+C

=12ex2+C , where C is any constant value.

Question:18 Integrate the functions etan1x1+x2

Answer:

Given,

etan1x1+x2

Let's do the following substitution

tan1x=t11+x2dx=dt

etan1x1+x2dx=etdt=et+C

=etan1x+C

Question:19 Integrate the functions e2x1e2x+1

Answer:

Given function e2x1e2x+1 ,

Simplifying it by dividing both numerator and denominator by ex , we obtain

e2x1exe2x+1ex=exexex+ex

Assume the ex+ex=t

(exex)dx=dt

e2x1e2x+1dx=exexex+exdx

=dtt

=log|t|+C

Now, back substituting the value of t,

=log|ex+ex|+C , where C is any constant value.

Question:20 Integrate the functions e2xe2xe2x+e2x

Answer:

Given function e2xe2xe2x+e2x ,

Assume the e2x+e2x=t

(2e2x2e2x)dx=dt

e2xe2xe2x+e2xdx=dt2t

=121tdt

=12log|t|+C

Now, back substituting the value of t,

=12log|e2x+e2x|+C , where C is any constant value.

Question:21 Integrate the functions tan2(2x3)

Answer:

Given function tan2(2x3) ,

Assume the 2x3=t

2dx=dt

tan2(2x3)dx=tan2(t)2dt

=12(sec2t1)dt [tan2t+1=sec2t]

=12[tantt]+C

Now, back substituting the value of t,

=12[tan(2x3)2x+3]+C

or 12tan(2x3)x+C , where C is any constant value.

Question:22 Integrate the functions sec2(74x)

Answer:

Given function sec2(74x) ,

Assume the 74x=t

4dx=dt

sec2(74x)dx=14sec2tdt

=14(tant)+C

Now, back substituted the value of t.

=14tan(74x)+C , where C is any constant value.

Question:23 Integrate the functions sin1x1x2

Answer:

Given function sin1x1x2 ,

Assume the sin1x=t

11x2dx=dt

sin1x1x2dx=tdt

=t22+C

Now, back substituted the value of t.

=(sin1x)22+C , where C is any constant value.

Question:24 Integrate the functions 2cosx3sinx6cosx+4sinx

Answer:

Given function 2cosx3sinx6cosx+4sinx ,

or simplified as 2cosx3sinx2(3cosx+2sinx)

Assume the 3cosx+2sinx=t

(3sinx+2cosx)dx=dt

2cosx3sinx6cosx+4sinxdx=dt2t

=12dtt

=12log|t|+C

Now, back substituted the value of t.

=12log|3cosx+2sinx|+C , where C is any constant value.

Question:25 Integrate the functions 1cos2x(1tanx)2

Answer:

Given function 1cos2x(1tanx)2 ,

or simplified as 1cos2x(1tanx)2=sec2x(1tanx)2

Assume the (1tanx)=t

sec2xdx=dt

sec2x(1tanx)2dx=dtt2

=t2dt

=1t+C

Now, back substituted the value of t.

=11tanx+C

where C is any constant value.

Question:26 Integrate the functions cosxx

Answer:

Given function cosxx ,

Assume the x=t

12xdx=dt

cosxxdx=2costdt

=2sint+C

Now, back substituted the value of t.

=2sinx+C , where C is any constant value.

Question:27 Integrate the functions sin2xcos2x

Answer:

Given function sin2xcos2x ,

Assume the sin2x=t

2cos2xdx=dt

sin2xcos2xdx=12tdt

=12(t3232)+C

=13t32+C

Now, back substituted the value of t.

=13(sin2x)32+C , where C is any constant value.

Question:28 Integrate the functions cosx1+sinx

Answer:

Given function cosx1+sinx ,

Assume the 1+sinx=t

cosxdx=dt

cosx1+sinxdx=dtt

=t1212+C

=2t+C

Now, back substituted the value of t.

=21+sinx+C , where C is any constant value.

Question:29 Integrate the functions cotxlogsinx

Answer:

Given function cotxlogsinx ,

Assume the logsinx=t

1sinx.cosxdx=dt

cotxdx=dt

cotxlogsinxdx=tdt

=t22+C

Now, back substituted the value of t.

=12(logsinx)2+C , where C is any constant value.

Question:30 Integrate the functions sinx1+cosx

Answer:

Given function sinx1+cosx ,

Assume the 1+cosx=t

sinxdx=dt

sinx1+cosxdx=dtt

=log|t|+C

Now, back substituted the value of t.

=log|1+cosx|+C , where C is any constant value.

Question:31 Integrate the functions sinx(1+cosx)2

Answer:

Given function sinx(1+cosx)2 ,

Assume the 1+cosx=t

sinxdx=dt

sinx(1+cosx)2dx=dtt2

=t2dt

=1t+C

Now, back substituted the value of t.

=11+cosx+C , where C is any constant value.

Question:32 Integrate the functions 11+cotx

Answer:

Given function 11+cotx

Assume that I=11+cotxdx

Now solving the assumed integral;

I=11+cosxsinxdx

=sinxsinx+cosxdx

=122sinxsinx+cosxdx

=12(sinx+cosx)+(sinxcosx)(sinx+cosx)dx

=121dx+12sinxcosxsinx+cosxdx

=12(x)+12sinxcosxsinx+cosxdx

Now, to solve further we will assume sinx+cosx=t

Or, (cosxsinx)dx=dt

I=x2+12(dt)t

=x212log|t|+C

Now, back substituting the value of t,

=x212log|sinx+cosx|+C

Question:33 Integrate the functions 11tanx

Answer:

Given function 11tanx

Assume that I=11tanxdx

Now solving the assumed integral;

I=11sinxcosxdx

=cosxcosxsinxdx

=122cosxcosxsinxdx

=12(cosxsinx)+(cosx+sinx)(cosxsinx)dx

=121dx+12cosx+sinxcosxsinxdx

=12(x)+12cosx+sinxcosxsinxdx

Now, to solve further we will assume cosxsinx=t

Or, (sinxcosx)dx=dt

I=x2+12(dt)t

=x212log|t|+C

Now, back substituting the value of t,

=x212log|cosxsinx|+C

Question:34 Integrate the functions tanxsinxcosx

Answer:

Given function tanxsinxcosx

Assume that I=tanxsinxcosxdx

Now solving the assumed integral;

Multiplying numerator and denominator by cosx ;

I=tanx×cosxsinxcosx×cosxdx

=tanxtanxcos2xdx

=sec2xtanxdx

Now, to solve further we will assume tanx=t

Or, sec2xdx=dt

I=dtt

=2t+C

Now, back substituting the value of t,

=2tanx+C

Question:35 Integrate the functions (1+logx)2x

Answer:

Given function (1+logx)2x

Assume that 1+logx=t

1xdx=dt

=(1+logx)2xdx=t2dt

=t33+C

Now, back substituting the value of t,

=(1+logx)33+C

Question:36 Integrate the functions (x+1)(x+logx)2x

Answer:

Given function (x+1)(x+logx)2x

Simplifying to solve easier;

(x+1)(x+logx)2x=(x+1x)(x+logx)2

=(1+1x)(x+logx)2

Assume that x+logx=t

(1+1x)dx=dt

=(1+1x)(x+logx)2dx=t2dt

=t33+C

Now, back substituting the value of t,

=(x+logx)33+C

Question:37 Integrate the functions x3sin(tan1x4)1+x8

Answer:

Given function x3sin(tan1x4)1+x8

Assume that x4=t

4x3dx=dt

x3sin(tan1x4)1+x8dx=14sin(tan1t)1+t2dt ......................(1)

Now to solve further we take tan1t=u

11+t2dt=du

So, from the equation (1), we will get

x3sin(tan1x4)1+x8dx=14sinu du

=14(cosu)+C

Now back substitute the value of u,

=14cos(tan1t)+C

and then back substituting the value of t,

=14cos(tan1x4)+C

Question:38 Choose the correct answer 10x9+10xloge10dxx10+10xdxequals

(A)10xx10+C(B)10x+x10+C(C)(10xx10)1+C(D)log(10x+x10)+C

Answer:

Given integral 10x9+10xloge10dxx10+10xdx

Taking the denominator x10+10x=t

Now differentiating both sides we get

(10x9+10xloge10)dx=dt

10x9+10xloge10x10+10xdx=dtt

=logt+C

Back substituting the value of t,

=log(x10+10x)+C

Therefore the correct answer is D.

Question:39 Choose the correct answer dxsin2xcos2xequals

(A)tanx+cotx+C(B)tanxcotx+C(C)tanxcotx+C(D)tanxcot2x+C

Answer:

Given integral dxsin2xcos2x

dxsin2xcos2x=1sin2xcos2xdx

=sin2x+cos2xsin2xcos2xdx (sin2x+cos2x=1)

=sin2xsin2xcos2xdx+cos2xsin2xcos2xdx

=sec2xdx+cosec2xdx

=tanxcotx+C

Therefore, the correct answer is B.


Class 12 Integrals NCERT Solutions Exercise: 7.3

Question:1 Find the integrals of the functions sin2(2x+5)

Answer:

sin2(2x+5)

using the trigonometric identity

sin2x=1cos2x2

we can write the given question as

= 1cos2(2x+5)2 =1cos(4x+10)2
=1cos(4x+10)2dx=12dx12cos(4x+10)dx=x212[sin(4x+10)/4]=x2sin(4x+10)8+C

Question:2 Find the integrals of the functions sin3xcos4x

Answer:

Using identity sinAcosB=1/2[sin(A+B)+sin(AB)]

, therefore the given integral can be written as

sin3xcos4x=12sin(3x+4x)+sin(3x4x) dx

=12sin(7x)sin(x) dx=1/2[sin(7x)dxsinx dx]=12[(1/7)cos7x+cosx+C]=cosx2cos7x14+C

Question:3 Find the integrals of the functions cos2xcos4xcos6x

Answer:

Using identity
cosAcosB=12[cos(A+B)+cos(AB)]

cos2x.cos4x.cos6x=cos2x.12[(cos10x)+cos2x]dx

Again use the same identity mentioned in the first line

=12(cos2x.cos10x+cos2x.cos2x)dx=1212(cos12x+cos8x)dx+12(1+cos4x2)dx=sin12x48+sin8x32+sin4x16+x/4+C

Question:4 Find the integrals of the functions sin3(2x+1)

Answer:

sin3(2x+1)dx=sin2(2x+1).sin(2x+1)dx

The integral can be written as

=(1cos2(2x+1)).sin(2x+1)dx
Let
cos(2x+1)=tsin(2x+1)dx=dt/2

=12(1t2)dt=12[tt3/3]=t36t2

Now, replace the value of t, we get;

=cos3(2x+1)6cos(2x+1)2+C

Question:5 Find the integrals of the functions sin3xcos3x

Answer:

I=sin3x.cos3x dx

rewrite the integral as follows

=cos3x.sin2x.sinx dx=cos3x(1cos2x)sinx dx
Let cosx=tsinxdx=dt

=t3(1t2)dt=(t3t5)dt=[t44]+[t66]+C=cos6x6cos4x4+C ......(replace the value of t as cos x )

Question:6 Find the integrals of the functions sinxsin2xsin3x

Answer:

Using the formula
sinAsinB=12(cos(AB)cos(A+B))

we can write the integral as follows

sinx.sin2xsin3x dx=sinx12[cosxcos5x]dx
=12[sinx.cosxsinx.cos5x]dx=12sin2x2dx12sinx.cos5x dx=cos2x814[sin6xsin4x]=cos2x814[cos6x6+cos4x4]=cos2x8+cos6x24cos4x16+C

Question:7 Find the integrals of the functions sin4xsin8x

Answer:

Using identity

sinAsinB=12(cos(AB)cos(A+B))

we can write the following integral as

sin4xsin8x =
=12(cos4xcos12x)dx=12[cos4x dxcos12x dx]=sin4x8sin12x24+C

Question:8 Find the integrals of the functions 1cosx1+cosx

Answer:

We know the identities

1+cos2A=2cos2A1cos2A=2sin2A

Using the above relations we can write

1cosx1+cosx=sin2x/2cos2x/2=tan2x/2

=tan2x/2=(sec2x/21)dx
=(sec2x/2)dxdx=2[tanx/2]x+C

Question:9 Find the integrals of the functions cosx1+cosx

Answer:

The integral is rewritten using trigonometric identities

cosx1+cosx=cos2x/2sin2x/22cos2x/2=12[1tan2x/2]
=12[1tan2x/2]dx=121[sec2x21]=122sec2x2=xtanx2+c

Question:10 Find the integrals of the functions sin4x

Answer:

sin4x can be written as follows using trigonometric identities
=sin2x.sin2x=14(1cos2x)2=14(1+cos22x2cos2x)=14(1+12(1+cos4x)2cos2x)=3/8+cos4x8cos2x2

Therefore,
sin4x dx=38dx+18cos4x dx12cos2x dx
=3x8+sin4x32sin2x4+C

Question:11 Find the integrals of the functions cos42x

Answer:

cos42x=cos32xcos2x

now using the identity

cos3x=cos3x+3cosx4

cos32xcos2x=cos6x+3cos2x4cos2x=cos6xcos2x+3cos22x4

now using the below two identities

cosacosb=cos(a+b)+cos(ab)2and cos22x=1+cos4x2

the value

cos42x=cos32xcos2x=cos6xcos2x+3cos22x4=cos4x+cos8x8+341+cos4x2 .

the integral of the given function can be written as

cos42x=cos4x+cos8x8+341+cos4x2=38x+sin4x8+sin8x64+C

Question:12 Find the integrals of the functions sin2x1+cosx

Answer:

Using trigonometric identities we can write the given integral as follows.

sin2x1+cosx

=4sin2x2cos2x22cos2x2=2sin2x2=1cosx

sin22x1+cosx=(1cosx)dx
=1dxcosx dx=xsinx+C

Question:13 Find the integrals of the functions cos2xcos2αcosxcosα

Answer:

We know that,

cosAcosB=2sin(A+B2)sin(AB2)

Using this identity we can rewrite the given integral as

cos2xcos2αcosxcosα=2sin2x+2α2sin2x2α22sinx+α2sinxα2

=sin(x+α)sin(xα)sinx+α2sinxα2=[2sinx+α2cosx+α2][2sinxα2cosxα2]sinx+α2sinxα2=4cosx+α2cosxα2=2[cosx+cosα]

cos2xcos2αcosxcosα=2cosx dx+2cosα dx
=2[sinx+xcosα]+C

Question:14 Find the integrals of the functions cosxsinx1+sin2x

Answer:

cosxsinx1+2sinx=cosxsinx(sin2x+cos2x)+2sinx.cosx
=cosxsinx(sinx+cosx)2


sinx+cosx=t(cosxsinx)dx=dt

Now,
=dtt2=t2 dt=t1+C=1(sinx+cosx)+C

Question:15 Find the integrals of the functions tan32xsec2x

Answer:

tan32x.sec2x=tan22x.tan2x.sec2x
=(sec22x1).tan2x.sec2x=sec22x.tan2xtan2x.sec2x

Therefore integration of tan32xsec2x =
=sec22x.tan2x dxtan2x.sec2x dx=sec22x.tan2x dxsec2x/2+C .....................(i)
Let assume

sec2x=t
So, that 2sec2x.tan2x dx=dt
Now, the equation (i) becomes,

12t2 dtsec2x2+Ct36sec2x2+C=(sec2x)36sec2x2+C

Question:16 Find the integrals of the functions tan4x

Answer:

the given question can be rearranged using trigonometric identities

tan4x=(sec2x1).tan2x=sec2x.tan2xtan2x=sec2x.tan2xsec2x+1

Therefore, the integration of tan4x = =sec2x.tan2x dxsec2x dx+dx=(sec2x.tan2x dx)tanx+x+C ...................(i)
Considering only sec2x.tan2x dx
let tanx=tsec2x dx=dt

sec2xtan2x dx=t2 dt=t3/3=tan3x3

now the final solution is,

tan4x=tan3x3tanx+x+C

Question:17 Find the integrals of the functions sin3x+cos3xsin2xcos2x

Answer:

sin3x+cos3xsin2xcos2x

now splitting the terms we can write

=sin3xsin2x.cos2x+cos3xsin2x.cos2x=sinxcos2x+cosxsin2x=tanx.secx+cotxcosecx

Therefore, the integration of
sin3x+cos3xsin2xcos2x

=(tanxsecx+cotxcosecx)dx=secxcosec x+C

Question:18 Find the integrals of the functions cos2x+2sin2xcos2x

Answer:

The integral of the above equation is

=(cos2x+2sin2xcos2x)dx=(cos2x+(1cos2x)cos2x=1cos2x=sec2x dx=tanx+C

Thus after evaluation, the value of integral is tanx+ c

Question:19 Find the integrals of the functions 1sinxcos3x

Answer:

Let
We can write 1 = sin2x+cos2x
Then, the equation can be written as
I=sin2x+cos2xsinxcos3x

I=(tanx+1tanx)sec2xdx
put the value of tan x = t
So, that sec2xdx=dt

I=(t+1t)dt=t22+log|t|+C=log|tanx|+12tan2x+C

Question:20 Find the integrals of the functions cos2x(cosx+sinx)2

Answer:

we know that cos2x=cos2xsin2x
therefore,

cos2x(cosx+sinx)2
cos2x1+sin2xcos2x1+sin2x let 1+sin2x=t2cos2x dx=dt
Now the given integral can be written as

cos2x(cosx+sinx)2=121tdt
12log|t|+C12log|1+sin2x|+C=log|sin2x+cos2x+2sinxcosx|+C=12log|(sinx+cosx)2|+C=log|sinx+cosx|+C

Question:21 Find the integrals of the functions sin1(cosx)

Answer:

using the trigonometric identities we can evaluate the following integral as follows

\dpi100sin1(cosx)dx=sin1(sin(π2x))dx=(π2x)dx=πx2x22+C

Question:22 Find the integrals of the functions 1cos(xa)cos(xb)

Answer:

Using the trigonometric identities following integrals can be simplified as follows

1cos(xa)cos(xb)=1sin(ab)[sin(ab)cos(xa)cos(xb)]

=1sin(ab)[sin[(xb)(xa)]cos(xa)cos(xb)]

=1sin(ab)[sin(xb)cos(xa)cos(xb)sin(xa)cos(xa)cos(xb)]

=tan(xb)tan(xa)sin(ab)

=1sin(ab)tan(xb)tan(xa)dx
=1sin(ab)[log|cos(xb)|+log|cos(xa)|]=1sin(ab)(log|cos(xa)cos(xb)|)

Question:23 Choose the correct answer

sin2xcos2xdxsin2xcos2xdxisequalto stgdrffd

Answer:

The correct option is (A)

On reducing the above integral becomes sec2xcsc2x
sec2xcsc2x dx=tanx+cotx+C

Question:24 Choose the correct answer ex(1+x)cos2(exx)dxequals

(A)cot(exx)+C(B)tan(xex)+C(C)tan(ex)+C(D)cot(ex)+C

Answer:

The correct option is (B)

Let exx=t .
So, (ex.x+1.ex)dx=dt
(1+ x ) ex dx=dt

therefore,

ex(1+x)cos2(ex.x)dx=dtcos2t
=sec2tdt=tant+C=tan(ex.x)+C


NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.4

Question:1 Integrate the functions 3x2x6+1

Answer:

The given integral can be calculated as follows

Let x3=t
, therefore, 3x2dx=dt

3x2x6+1=dtt2+1

=tan1t+C=tan1(x3)+C

Question:2 Integrate the functions 11+4x2

Answer:

11+4x2
let suppose 2x = t
therefore 2dx = dt

11+4x2=12dt1+t2
=12[log|t+1+t2|]+C=12log|2x+4x2+1|+C .................using formula 1x2+a2dt=log|x+x2+a2|

Question:3 Integrate the functions 1(2x)2+1

Answer:

1(2x)2+1

let suppose 2-x =t
then, -dx =dt
1(2x)2+1dx=1t2+1dt

using the identity

1x2+1dt=log|x+x2+1|

=log|t+t2+1|+C=log|2x+(2x)2+1|+C=log|1(2x)+x24x+5|+C

Question:4 Integrate the functions 1925x2

Answer:

1925x2
Let assume 5x =t,
then 5dx = dt

1925x2=1519t2dt
=15132t2dt=15sin1(t3)+C=15sin1(5x3)+C

The above result is obtained using the identity

1a2x2dt=1asin1xa

Question:5 Integrate the functions 3x1+2x4

Answer:

3x1+2x4


Let 2x2=t
22xdx=dt

The integration can be done as follows

3x1+2x4=322dt1+t2
=322[tan1t]+C=322[tan1(2x2)]+C

Question:6 Integrate the functions x21x6

Answer:

x21x6

let x3=t
then 3x2dx=dt

using the special identities we can simplify the integral as follows

x21x6dx=13dt1t2
=13[12log|1+t1t|]+C=16log|1+x31x3|+C

Question:7 Integrate the functions x1x21

Answer:


We can write above eq as
x1x21 =xx21dx1x21dx ............................................(i)

for xx21dx let x21=t2xdx=dt

xx21dx=12dtt
=12t1/2dt=12[2t1/2]=t=x21
Now, by using eq (i)
=xx21dx1x21dx
=x211x21dx=x21log|x+x21|+C

Question:8 Integrate the functions x2x6+a6

Answer:

The integration can be down as follows

x2x6+a6
let x3=t3x2dx=dt

x2x6+a6=13dtt2+(a3)2
=13log|t+t2+a6|+C=13log|x3+x6+a6|+C ........................using dxx2+a2=log|x+x2+a2|

Question:9 Integrate the functions sec2xtan2x+4

Answer:

The integral can be evaluated as follows

sec2xtan2x+4
let tanx=tsec2xdx=dt

sec2xtan2x+4dx=dtt2+22
=log|t+t2+4|+C=log|tanx+tan2x+4|+C

Question:10 Integrate the functions 1x2+2x+2

Answer:

1x2+2x+2
the above equation can be also written as,
=1(1+x)2+12dx
let 1+x = t
then dx = dt
therefore,

=1t2+12dx=log|t+t2+1|+C=log|(1+x)+(1+x)2+1|+C=log|(1+x)+(x2+2x+2|+C

Question:11 Integrate the functions 19x2+6x+5

Answer:

19x2+6x+5
this denominator can be written as
9x2+6x+5=9[x2+23x+59]=9[(x+13)2+(23)2] Now,
191(x+13)2+(23)2dx=19[32tan1((x+1/3)2/3)]+C=16tan1(3x+12)]+C
......................................by using the form (1x2+a2=1atan1(xa))

Question:12 Integrate the functions 176xx2

Answer:

the denominator can be also written as,
76xx2=16(x2+6x+9)
=42(x+3)2

therefore

176xx2dx=142(x+3)2dx
Let x+3 = t
then dx =dt

142(x+3)2dx=142t2dt ......................................using formula 1a2x2=sin1(xa)
=sin1(t4)+C=sin1(x+34)+C

Question:13 Integrate the functions 1(x1)(x2)

Answer:

(x-1)(x-2) can be also written as
= x23x+2
= (x32)2(12)2

therefore

1(x1)(x2)dx=1(x32)2(12)2dx
let suppose
x3/2=tdx=dt
Now,

1(x32)2(12)2dx=1t2(12)2dt .............by using formula 1x2a2=log|x+x2+a2|
=log|t+t2(1/2)2|+C=log|(x32)+x23x+2|+C

Question:14 Integrate the functions 18+3xx2

Answer:

We can write denominator as
=8(x23x+9494)=414(x32)2

therefore
18+3xx2dx=1414(x32)2
let x3/2=tdx=dt


=1(412)2t2dt=sin1(t412)+C=sin1(2x341)+C

Question:15 Integrate the functions 1(xa)(xb)

Answer:

(x-a)(x-b) can be written as x2(a+b)x+ab
x2(a+b)x+ab+(a+b)24(a+b)24(x(a+b)22)2(ab)24

1(xa)(xb)dx=1(x(a+b)22)2(ab)24dx
let
x(a+b)2=tdx=dt
So,
=1t2(ab2)2dt=log|t+t2(ab2)2|+C=log|x(a+b2)+(xa)(xb)|+C

Question:16 Integrate the functions 4x+12x2+x3

Answer:

let
4x+1=Addx(2x2+x3)+B4x+1=A(4x+1)+B4x+1=4Ax+A+B

By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0

Let (2x2+x3)=t(4x+1)dx=dt

4x+12x2+x3dx=1tdt
=2t+C=22x2+x3+C

Question:17 Integrate the functions x+2x21

Answer:

let x+2=Addx(x21)+B=A(2x)+B
By comparing the coefficients and constant term on both sides, we get;

A=1/2 and B=2
then x+2=12(2x)+2

x+2x21dx=1/2(2x)+2x21dx
=12(2x)x21dx+2x21dx=12[2x21]+2log|x+x21|+C=x21+2log|x+x21|+C

Question:18 Integrate the functions 5x21+2x+3x2

Answer:

let
5x+2=Addx(1+2x+3x2)+B5x+2=A(2+6x)+B=2A+B+6Ax
By comparing the coefficients and constants we get the value of A and B

A = 5/6 and B = 11/3

NOW,
I=566x+23x2+2x+1dx113dx3x2+2x+1
I=I1113I2 ...........................(i)

put 3x2+2x+1=t(6x+2)dx=dt
Thus
I1=56dtt=56logt=56log(3x2+2x+1)+c1
I2=dx3x2+2x+1=13dx(x+1/3)2+(2/3)2
=12tan1(3x+12)+c2

I=I1+I2
I=56log(3x2+2x+1)11312tan1(3x+12)+C

Question:19 Integrate the functions 6x+7(x5)(x4)

Answer:

let
6x+7=Addx(x29x+20)+B=A(2x9)+B
By comparing the coefficients and constants on both sides, we get
A =3 and B =34

I=6x+7x29x+20dx=3(2x+9)x29x+20dx+34dxx29x+20 I=I1+I2 ....................................(i)

Considering I1

I1=2x9x29x+20dx let x29x+20=t(2x9)dx=dt

I1=dtt=2t=2x29x+20

Now consider I2

I2=dxx29x+20
here the denominator can be also written as
Dr = (x92)2(12)2

I2=dx(x92)2(12)2
=log|(x92)2+x29x+20|

Now put the values of I1 and I2 in eq (i)

I=3I1+34I2I=6x29x+20+34log|(x92)+x29x+20|+C

Question:20 Integrate the functions x+24xx2

Answer:

let
x+2=Addx(4xx2)+B=A(42x)+B
By equating the coefficients and constant term on both sides we get

A = -1/2 and B = 4

(x+2) = -1/2(4-2x)+4

x+24xx2dx=1242x4xx2+4dx4xx2 I=12I1+4I2 ....................(i)

Considering I1
42x4xx2dx
let 4xx2=t(42x)dx=dt
I1=dtt=2t=24xx2
now, I2

I2=dx4xx2=dx22(x2)2
=sin1(x22)

put the value of I1 and I2

I=4xx2+4sin1(x22)+C

Question:21 Integrate the functions x+2x2+2x+3

Answer:

x+2x2+2x+3
x+2x2+2x+3dx=122(x+2)x2+2x+3dx
=122x+2x2+2x+3dx+122x2+2x+3dx=122x+2x2+2x+3dx+1x2+2x+3dxI=12I1+I2 ...........(i)

take I1

2x+2x2+2x+3dx
let x2+2x+3=t(2x+2)dx=dt

I1=dtt=2t=2x2+2x+3
considering I2

=dxx2+2x+3=dx(x+1)2+(2)2
=log|(x+1)+x2+2x+3|
putting the values in equation (i)

I=x2+2x+3+log|(x+1)+x2+2x+3|+C

Question:22 Integrate the functions x+3x22x5

Answer:

Let (x+3)=Addx(x22x+5)+B=A(2x2)+B

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

x+3x22x+5dx=122x2x22x+5dx+41x22x+5dxI=I1+I2 ..............(i)

I1=2x2x22x5dx
put x22x5=t(2x2)dx=dt

=dtt=logt=log(x22x5)

I2=1x22x5dx=1(x1)2+(6)2dx=126log(x16x1+6)

I=I1+I2

=12log|x22x5|+26log(x16x1+6)+C

Question:23 Integrate the functions 5x+3x2+4x+10

Answer:

let
5x+3=Addx(x2+4x+10)+B=A(2x+4)+B
On comparing, we get

A =5/2 and B = -7

5x+3x2+4x+10dx=522x+4x2+4x+10dx7dxx2+4x+10dx I=5/2I17I2 ...........................................(i)

I12x+4x2+4x+10dx
put
x2+4x+10=t(2x+4)dx=dt

=dtt=2t=2x2+4x+10

I2=1x2+4x+10dx=1(x+2)2+(6)2dx=log|(x+2)+x2+4x+10|

I=5x2+4x+107log|(x+2)+x2+4x+10|+C

Question:24 Choose the correct answer

dxx2+2x+2equals

(A)xtan1(x+1)+C(B)tan1(x+1)+C(C)(x+1)tan1x+C(D)tan1x+C

Answer:

The correct option is (B)

dxx2+2x+2equals
the denominator can be written as (x+1)2+1
now, dx(x+1)2+1=tan1(x+1)+C

Question:25 Choose the correct answer dx9x4x2equals

A)19sin1(9x88)+CB)12sin1(8x99)+CC)13sin1(9x88)+CD)12sin1(9x88)+C

Answer:

The following integration can be done as

dx9x4x2equals
14(x294x)=14(x294x+81/6481/64)dx
=14[(x9/8)2(9/8)2]dx=121(x9/8)2+(9/8)2dx=12[sin1(x9/89/8)]+C=12sin1(8x99)+C

The correct option is (B)


NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.5

Question:1 Integrate the rational functions x(x+1)(x+2)

Answer:

Given function x(x+1)(x+2)

Partial function of this function:

x(x+1)(x+2)=A(x+1)+B(x+2)

x=A(x+2)+B(x+1)

Now, equating the coefficients of x and constant term, we obtain

A+B=1

2A+B=0

On solving, we get

A=1 and B=2

x(x+1)(x+2)=1(x+1)+2(x+2)

x(x+1)(x+2)dx=1(x+1)+2(x+2)dx

=log|x+1|+2log|x+2|+C

=log(x+2)2log|x+1|+C

=log(x+2)2(x+1)+C

Question:2 Integrate the rational functions 1x29

Answer:

Given function 1x29

The partial function of this function:

1(x+3)(x3)=A(x+3)+B(x3)

1=A(x3)+B(x+3)

Now, equating the coefficients of x and constant term, we obtain

A+B=1

3A+3B=1

On solving, we get

A=16 and B=16

1(x+3)(x3)=16(x+3)+16(x3)

1(x29)dx=(16(x+3)+16(x3))dx

=16log|x+3|+16log|x3|+C

=16log|x3x+3|+C

Question:3 Integrate the rational functions 3x1(x1)(x2)(x3)

Answer:

Given function 3x1(x1)(x2)(x3)

Partial function of this function:

3x1(x1)(x2)(x3)=A(x1)+B(x2)+C(x3)

3x1=A(x2)(x3)+B(x1)(x3)+C(x1)(x2) .(1)

Now, substituting x=1,2, and 3 respectively in equation (1), we get

A=1, B=5, and C=4

3x1(x1)(x2)(x3)=1(x1)5(x2)+4(x3)

That implies 3x1(x1)(x2)(x3)dx={1(x1)5(x2)+4(x3)}dx

=log|x1|5log|x2|+4log|x3|+C

Question:4 Integrate the rational functions x(x1)(x2)(x3)

Answer:

Given function x(x1)(x2)(x3)

Partial function of this function:

x(x1)(x2)(x3)=A(x1)+B(x2)+C(x3)

x=A(x2)(x3)+B(x1)(x3)+C(x1)(x2) .....(1)

Now, substituting x=1,2, and 3 respectively in equation (1), we get

A=12, B=2, and C=32

x(x1)(x2)(x3)=12(x1)2(x2)+32(x3)

That implies x(x1)(x2)(x3)dx={12(x1)2(x2)+32(x3)}dx

=12log|x1|2log|x2|+32log|x3|+C

Question:5 Integrate the rational functions 2xx2+3x+2

Answer:

Given function 2xx2+3x+2

Partial function of this function:

2xx2+3x+2=A(x+1)+B(x+2)

2x=A(x+2)+B(x+1) ...........(1)

Now, substituting x=1 and 2 respectively in equation (1), we get

A=2, B=4

2xx2+3x+2=2(x+1)+4(x+2)

That implies 2xx2+3x+2dx={2(x+1)+4(x+2)}dx

=4log|x+2|2log|x+1|+C

Question:6 Integrate the rational functions 1x2x(12x)

Answer:

Given function 1x2x(12x)

Integral is not a proper fraction so,

Therefore, on dividing (1x2) by x(12x) , we get

1x2x(12x)=12+12(2xx(12x))

Partial function of this function:

2xx(12x)=Ax+B(12x)

(2x)=A(12x)+Bx ...........(1)

Now, substituting x=0 and 12 respectively in equation (1), we get

A=2, B=3

2xx(12x)=2x+312x

No, substituting in equation (1) we get

1x2(12x)=12+12{23+3(12x)}

1x2x(12x)dx={12+12(2x+312x)}dx

=x2+log|x|+32(2)log|12x|+C

=x2+log|x|34log|12x|+C

Question:7 Integrate the rational functions x(x2+1)(x1)

Answer:

Given function x(x2+1)(x1)

Partial function of this function:

x(x2+1)(x1)=Ax+b(x2+1)+C(x1)

x=(Ax+B)(x1)+C(x2+1)

x=AxAx+BcB+Cx2+C

Now, equating the coefficients of x2,x and the constant term, we get

A+C=0

A+B=1 and B+C=0

On solving these equations, we get

A=12,B=12, and C=12

From equation (1), we get

x(x2+1)(x1)=(12x+12)x2+1+12(x1)

x(x2+1)(x1)

=12xx2+1dx+121x2+1dx+121x1dx

=142xx2+1dx+12tan1x+12log|x1|+C

Now, consider 2xx2+1dx ,

and we will assume (x2+1)=t2xdx=dt

So, 2xx2+1dx=dtt=log|t|=log|x2+1|

x(x2+1)(x1)=14log|x2+1|+12tan1x+12log|x1|+C or

12log|x1|14log|x2+1|+12tan1x+C

Question:8 Integrate the rational functions x(x+1)2(x+2)

Answer:

Given function x(x+1)2(x+2)

Partial function of this function:

x(x+1)2(x+2)=A(x1)+B(x1)2+C(x+2)

x=A(x1)(x+2)+B(x+2)+C(x1)2

Now, putting x=1 in the above equation, we get

B=13

By equating the coefficients of x2 and constant term, we get

A+C=0

2A+2B+C=0

then after solving, we get

A=29 and C=29

Therefore,

x(x1)2(x+2)=29(x1)+13(x1)229(x+2)

x(x1)2(x+2)dx=291(x1)dx+131(x1)2dx291(x+2)dx

=29log|x1|+13(1x1)29log|x+2|+C

29log|x1x+2|13(x1)+C

Question:9 Integrate the rational functions 3x+5x3x2x+1

Answer:

Given function 3x+5x3x2x+1

can be rewritten as 3x+5x3x2x+1=3x+5(x1)2(x+1)

Partial function of this function:

3x+5(x1)2(x+1)=A(x1)+B(x1)2+C(x+1)

3x+5=A(x1)(x+1)+B(x+1)+C(x1)2

3x+5=A(x21)+B(x+1)+C(x2+12x) ................(1)

Now, putting x=1 in the above equation, we get

B=4

By equating the coefficients of x2 and x , we get

A+C=0

B2C=3

then after solving, we get

A=12 and C=12

Therefore,

3x+5(x1)2(x+1)=12(x1)+4(x1)2+12(x+1)

3x+5(x1)2(x+1)dx=121(x1)dx+41(x1)2dx+121(x+1)dx

=12log|x1|+4(1x1)+12log|x+1|+C

=12log|x+1x1|4(x1)++C

Question:10 Integrate the rational functions 2x3(x21)(2x+3)

Answer:

Given function 2x3(x21)(2x+3)

can be rewritten as 2x3(x21)(2x+3)=2x3(x+1)(x1)(2x+3)

The partial function of this function:

2x3(x21)(2x3)=A(x+1)+B(x1)+C(2x+3)

(2x3)=A(x1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x1) (2x3)=A(2x2+x3)+B(2x2+5x+3)+C(x21) (2x3)=(2A+2B+C)x2+(A+5B)x+(3A+3BC)

Equating the coefficients of x2 and x , we get

B=110, A=52, and C=245

Therefore,

2x3(x21)(2x3)=52(x+1)110(x1)245(2x+3)

2x3(x21)(2x+3)dx=521(x+1)dx1101x1dx2451(2x+3)dx =52log|x+1|110log|x1|2410log|2x+3|

=52log|x+1|110log|x1|125log|2x+3|+C

=12log|x1|+4(1x1)+12log|x+1|+C

=12log|x+1x1|4(x1)++C

=29log|x1|+13(1x1)29log|x+2|+C

29log|x1x+2|13(x1)+C

Question:11 Integrate the rational functions 5x(x+1)(x24)

Answer:

Given function 5x(x+1)(x24)

can be rewritten as 5x(x+1)(x24)=5x(x+1)(x+2)(x2)

The partial function of this function:

5x(x+1)(x+2)(x2)=A(x+1)+B(x+2)+C(x2)

(5x)=A(x+2)(x2)+B(x+1)(x2)+C(x+1)(x+2)

Now, substituting the value of x=1,2, and 2 respectively in the equation above, we get

A=53, B=52, and C=56

Therefore,

5x(x+1)(x+2)(x2)=53(x+1)52(x+2)+56(x2)

5x(x+1)(x24)dx=531(x+1)dx521x+2dx+561(x2)dx =53log|x+1|52log|x+2|+56log|x2|+C

Question:12 Integrate the rational functions x3+x+1x21

Answer:

Given function x3+x+1x21

As the given integral is not a proper fraction.

So, we divide (x3+x+1) by x21 , we get

x3+x+1x21=x+2x+1x21

can be rewritten as 2x+1x21=A(x+1)+B(x1)

2x+1=A(x1)+B(x+1) ....................(1)

Now, substituting x=1 and x=1 in equation (1), we get

A=12 and B=32

Therefore,

x3+x+1(x21)=x+12(x+1)+32(x1)

x3+x+1(x21)dx=xdx+121(x+1)dx+321(x1)dx

=x22+12log|x+1|+32log|x1|+C

Question:13 Integrate the rational functions 2(1x)(1+x2)

Answer:

Given function 2(1x)(1+x2)

can be rewritten as 2(1x)(1+x2)=A(1x)+Bx+C1+x2

2=A(1+x2)+(Bx+C)(1x) ....................(1)

2=A+Ax2+BxBx2+CCx

Now, equating the coefficient of x2,x, and constant term, we get

AB=0 , BC=0 , and A+C=2

Solving these equations, we get

A=1,B=1, and C=1

Therefore,

2(1x)(1+x2)=1(1x)+x+11+x2

2(1x)(1+x2)dx=1(1x)dx+x1+x2dx+11+x2dx =1x1dx+122x1+x2dx+11+x2dx

=log|x1|+12log|1+x2|+tan1x+C

Question:14 Integrate the rational functions 3x1(x+2)2

Answer:

Given function 3x1(x+2)2

can be rewritten as 3x1(x+2)2=A(x+2)+B(x+2)2

3x1=A(x+2)+B

Now, equating the coefficient of x and constant term, we get

A=3 and 2A+B=1 ,

Solving these equations, we get

B=7

Therefore,

3x1(x+2)2=3(x+2)7(x+2)2

3x1(x+2)2dx=31(x+2)dx7x(x+2)2dx

3log|x+2|7(1(x+2))+C

3log|x+2|+7(x+2)+C

Question:15 Integrate the rational functions 1x41

Answer:

Given function 1x41

can be rewritten as 1x41=1(x21)(x2+1)=1(x+1)(x1)(1+x2)

The partial fraction of above equation,

1(x+1)(x1)(1+x2)=A(x+1)+B(x1)+Cx+D(x2+1)

1=A(x1)(x2+1)+B(x+1)(x2+1)+(Cx+D)(x21)

1=A(x3+xx21)+B(x3+x+x2+1)+Cx3+Dx2CxD 1=(A+B+C)x3+(A+B+D)x2+(A+BC)x+(A+BD)

Now, equating the coefficient of x3,x2,x and constant term, we get

A+B+C=0 and A+B+D=0

A+BC=0 and A+BD=1

Solving these equations, we get

A=14,B=14,C=0, and D=12

Therefore,

1x41=14(x+1)+14(x1)12(x2+1)

1x41dx=14log|x1|+14log|x1|12tan1x+C

=14log|x1x+1|12tan1x+C

Question:16 Integrate the rational functions 1x(xn+1)

[Hint: multiply numerator and denominator by xn1 and put xn=t ]

Answer:

Given function 1x(xn+1)

Applying Hint multiplying numerator and denominator by xn1 and putting xn=t

1x(xn+1)=xn1xn1x(xn+1)=xn1xn(xn+1)

Putting xn=t

xn1dx=dt

can be rewritten as 1x(xn+1)dx=xn1xn(xn+1)dx=1n1t(t+1)dt

Partial fraction of above equation,

1t(t+1)=At+B(t+1)

1=A(1+t)+Bt ................(1)

Now, substituting t=0,1 in equation (1), we get

A=1 and B=1

1t(t+1)=1t1(1+t)

1x(xn+1)dx=1n{1t1(t+1)}dx

=1n[log|t|log|t+1|]+C

=1n[log|xn|log|xn+1|]+C

=1nlog|xnxn+1|+C

Question:17 Integrate the rational functions cosx(1sinx)(2sinx)

[Hint : Put sinx=t ]

Answer:

Given function cosx(1sinx)(2sinx)

Applying the given hint: putting sinx=t

We get, cosxdx=dt

cosx(1sinx)(2sinx)dx=dt(1t)(2t)

Partial fraction of above equation,

1(1t)(2t)=A(1t)+B(2t)

1=A(2t)+B(1t) ................(1)

Now, substituting t=2 and 1 in equation (1), we get

A=1 and B=1

1(1t)(2t)=1(1t)1(2t)

cosx(1sinx)(2sinx)dx={11t1(2t)}dt

=log|1t|+log|2t|+C

=log|2t1t|+C

Back substituting the value of t in the above equation, we get

=log|2sinx1sinx|+C

Question:18 Integrate the rational functions (x2+1)(x2+2)(x2+3)(x2+4)

Answer:

Given function (x2+1)(x2+2)(x2+3)(x2+4)

We can rewrite it as: (x2+1)(x2+2)(x2+3)(x2+4)=1(4x2+10)(x2+3)(x2+4)

Partial fraction of above equation,

(4x2+10)(x2+3)(x2+4)=Ax+B(x2+3)+Cx+D(x2+4)

4x2+10=(Ax+B)(x2+4)+(Cx+D)(x2+3)

4x2+10=Ax3+4Ax+Bx2+4B+Cx3+3Cx+Dx2+3D

4x2+10=(A+C)x3+(B+D)x2+(4A+3C)x+(3D+4B)

Now, equating the coefficients of x3,x2,x and constant term, we get

A+C=0 , B+D=4 , 4A+3C=0 , 4B+3D=10

After solving these equations, we get

A=0,B=2,C=0,\and D=6

4x2+10(x2+3)(x2+4)=2(x2+3)+6(x2+4)

(x2+1)(x2+2)(x2+3)(x2+4)=1(2(x2+3)+6(x2+4))

(x2+1)(x2+2)(x2+3)(x2+4)dx={1+2(x2+3)6(x2+4)}dx

={1+2(x2+(3)2)6(x2+22)}dx

=x+2(13tan1x3)6(12tan1x2)+C

=x+23tan1x33tan1x2+C

Question:19 Integrate the rational functions 2x(x2+1)(x2+3)

Answer:

Given function 2x(x2+1)(x2+3)

Taking x2=t2xdx=dt

2x(x2+1)(x2+3)dx=dt(t+1)(t+3)

The partial fraction of above equation,

1(t+3)(t+3)=A(t+1)+B(t+3)

1=A(t+3)+B(t+1) ..............(1)

Now, substituting t=3 and t=1 in equation (1), we get

A=12 and B=12

1(t+3)(t+3)=12(t+1)12(t+3)

2x(x2+1)(x2+3)dx={12(t+1)12(t+3)}dt

=12log|t+1|12log|t+3|+C

=12log|t+1t+3|+C

=12log|x2+1x2+3|+C

Question:20 Integrate the rational functions 1x(x41)

Answer:

Given function 1x(x41)

So, we multiply numerator and denominator by x3 , to obtain

1x(x41)=x3x4(x41)

1x(x41)dx=x3x4(x41)dx

Now, putting x4=t

we get, 4x3dx=dt

Taking x2=t2xdx=dt

1x(x41)dx=14dtt(t1)

Partial fraction of above equation,

1t(t1)=At+B(t1)

1=A(t1)+Bt ..............(1)

Now, substituting t=0 and t=1 in equation (1), we get

A=1 and B=1

1t(t+1)=1t+1t1

1x(x4+1)dx=14{1t+1t1}dt

=14[log|t|+log|t1|]+C

=14log|t1t|+C

Back substituting the value of t,

=14log|x41x4|+C

Question:21 Integrate the rational functions 1(ex1) [Hint : Put ex=t ]

Answer:

Given function 1(ex1)

So, applying the hint: Putting ex=t

Then exdx=dt

1(ex1)dx=1t1×dtt=1t(t1)dt


Partial fraction of above equation,

1t(t1)=At+B(t1)

1=A(t1)+Bt ..............(1)

Now, substituting t=0 and t=1 in equation (1), we get

A=1 and B=1

1t(t+1)=1t+1t1

1t(t1)dt=log|t1t|+C

Now, back substituting the value of t,

=log|ex1ex|+C

Question:22 Choose the correct answer xdx(x1)(x2)equals

A)log|(x1)2x2|+CB)log|(x2)2x1|+CC)log|(x1x2)2|+CD)log|(x1)2(x2)|+C

Answer:

Given integral xdx(x1)(x2)

Partial fraction of above equation,

x(x1)(x2)=A(x1)+B(x2)

x=A(x+2)+B(x1) ..............(1)

Now, substituting x=1 and x=2 in equation (1), we get

A=1 and B=2

x(x1)(x2)=1(x1)+2(x2)

x(x1)(x2)dx={1(x1)+2(x2)}dx

=log|x1|+2log|x2|+C

=log|(x2)2x1|+C

Therefore, the correct answer is B.

Question:23 Choose the correct answer dxx(x2+1)equals

A)log|x|12log(x2+1)+CB)log|x|+12log(x2+1)+CC)log|x|+12log(x2+1)+CD)12log|x|+log(x2+1)+C

Answer:

Given integral dxx(x2+1)

Partial fraction of above equation,

1x(x2+1)=Ax+Bx+cx2+1

1=A(x2+1)+(Bx+C)x

Now, equating the coefficients of x2,x, and the constant term, we get

A+B=0 , C=0 , A=1

We have the values, A=1 and B=1, and C=0

1x(x2+1)=1x+xx2+1

1x(x2+1)dx={1x+xx2+1}dx

=log|x|12log|x2+1|+C

Therefore, the correct answer is A.


NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.6

Question:1 Integrate the functions xsinx

Answer:

Given function is
f(x)=xsinx
We will use integrate by parts method
xsinx=x.sinxdx(d(x)dx.sinxdx)dxxsinx=x.(cosx)(1.(cosx))dxxsinx=xcosx+sinx+C
Therefore, the answer is xcosx+sinx+C

Question:2 Integrate the functions xsin3x

Answer:

Given function is
f(x)=xsin3x
We will use integration by parts method
xsin3x=x.sin3xdx(d(x)dx.sin3xdx)dxxsin3x=x.(cos3x3)(1.(cos3x3))dxxsin3x=xcos3x3+sin3x9+C

Therefore, the answer is xcos3x3+sin3x9+C

Question:3 Integrate the functions x2ex

Answer:

Given function is
f(x)=x2ex
We will use integration by parts method
x2ex=x2.exdx(d(x2)dx.exdx)dxx2ex=x2.ex(2x.ex)dx
Again use integration by parts in (2x.ex)dx
(2x.ex)dx=2x.exdx(d(2x)dx.exdx)dx2x.exdx=2xex2.exdx2x.exdx=2xex2ex
Put this value in our equation
we will get,
x2.exdx=x2ex2xex+2ex+Cx2.exdx=ex(x22x+2)+C

Therefore, answer is ex(x22x+2)+C

Question:4 Integrate the functions xlogx

Answer:

Given function is
f(x)=x.logx
We will use integration by parts method
x.logxdx=logx.xdx(d(logx)dx.xdx)dxxlogxdx=logx.x22(1x.x22)dxxlogxdx=logx.x22x2dxxlogxdx=logx.x22x24+C

Therefore, the answer is x22logxx24+C

Question:5 Integrate the functions xlog2x

Answer:

Given function is
f(x)=x.log2x
We will use integration by parts method
x.log2xdx=log2x.xdx(d(log2x)dx.xdx)dxxlog2xdx=log2x.x22(22x.x22)dxxlog2xdx=log2x.x22x2dxxlog2xdx=log2x.x22x24+C

Therefore, the answer is log2x.x22x24+C

Question:6 Integrate the functions x2logx

Answer:

Given function is
f(x)=x2.logx
We will use integration by parts method
x2.logxdx=logx.x2dx(d(logx)dx.x2dx)dxx2logxdx=logx.x33(1x.x33)dxx2logxdx=logx.x33x23dxx2logxdx=logx.x33x39+C

Therefore, the answer is logx.x33x39+C

Question:7 Integrate the functions xsin1x

Answer:

Given function is
f(x)=x.sin1x
We will use integration by parts method
x.sin1xdx=sin1x.xdx(d(sin1x)dx.xdx)dxxsin1xdx=sin1x.x22(11x2.x22)dx
Now, we need to integrate (11x2.x22)dx
x221x2dx=12(1x21x211x2)dxx221x2dx=12(1x211x2)dxx221x2dx=12(1x2dx11x2dx)x221x2dx=12(x21x2+12sin1xsin1x)x221x2dx=x1x24sin1x4+C
Put this value in our equation

Therefore, the answer is xsin1xdx=sin1x4(2x21)x1x24

Question:8 Integrate the functions xtan1x

Answer:

Given function is
f(x)=x.tan1x
We will use integration by parts method
x.tan1xdx=tan1x.xdx(d(tan1x)dx.xdx)dxxtan1xdx=tan1x.x22(11+x2.x22)dxxtan1xdx=tan1x.x2212(x2+11+x211+x2)dxxtan1xdx=tan1x.x2212(111+x2)dxxtan1xdx=tan1x.x2212(xtan1x)+Cxtan1xdx=tan1x2(2x2+1)x2+C

Put this value in our equation
xsin1xdx=sin1x.x22x41x2sin1x4+Cxsin1xdx=sin1x4(2x21)x41x2

Therefore, the answer is tan1x2(2x2+1)x2+C

Question:9 Integrate the functions xcos1x

Answer:

Given function is
f(x)=x.cos1x
We will use integration by parts method
x.cos1xdx=cos1x.xdx(d(cos1x)dx.xdx)dxxcos1xdx=cos1x.x22(11x2.x22)dx
Now, we need to integrate (11x2.x22)dx
x221x2dx=12(1x21x211x2)dxx221x2dx=12(1x211x2)dxx221x2dx=12(1x2dx11x2dx)x221x2dx=12(x21x212cos1x+cos1x)x221x2dx=x1x24cos1x4+cos1x2+C
Put this value in our equation
xcos1xdx=cos1x.x22(x1x24cos1x4+cos1x2+C)xcos1xdx=cos1x4(2x21)x1x24

Therefore, the answer is cos1x4(2x21)x1x24

Question:10 Integrate the functions (sin1x)2

Answer:

Given function is
f(x)=(sin1x)2
we will use integration by parts method
(sin1x)2=(sin1x)2.1dx(d((sin1x)2)dx.1dx)dx(sin1x)2=(sin1x)2.x(sin1.2x1x2)dx(sin1x)2=(sin1x)2.x+[sin1x.2x1x2dx(d(sin1x)dx.2x1x2dx)].     =(sin1x)2.x+[sin1x.21x211x2.21x2dx].     =(sin1x)2.x+2sin1x1x22x+C
Therefore, answer is (sin1x)2.x+2sin1x1x22x+C

Question:11 Integrate the functions xcos11x2

Answer:

Consider xcos11x2dx=I

So, we have then: I=122x1x2.cos1xdx

After taking cos1x as a first function and (2x1x2) as second function and integrating by parts, we get

I=12[cos1x2x1x2dx{(ddxcos1x)2x1x2dx}dx] =12[cos1x.21x2+11x2.21x2dx]

=12[21x2cos1x2dx]

=12[21x2cos1x2x]+C

Or, -\left \sqrt{1-x^2}\cos^{-1}x +x\right +C

Question:12 Integrate the functions xsec2x

Answer:

Consider xsec2x

So, we have then: I=xsec2xdx

After taking x as a first function and sec2x as second function and integrating by parts, we get

I=xsec2xdx{(ddxx)sec2xdx}dx

=xtanx1.tanxdx

=xtanx+log|cosx|+C

Question:13 Integrate the functions tan1x

Answer:

Consider tan1x

So, we have then: I=1.tan1xdx

After taking tan1x as a first function and 1 as second function and integrating by parts, we get

I=tan1x1dx{(ddxtan1x)1.dx}dx

=tan1x.x11+x2.xdx

=xtan1x122x1+x2dx

=xtan1x12log|1+x2|+C

=xtan1x12log(1+x2)+C

Question:14 Integrate the functions x(logx)2

Answer:

Consider x(logx)2

So, we have then: I=x(logx)2dx

After taking (logx)2 as a first function and x as second function and integrating by parts, we get

I=(logx)2xdx{(ddx(logx)2)x.dx}dx

=(logx)2.x222logxx.x22dx

=(logx)2.x22xlogxdx

=(logx)2.x22(x2logx2x24)+C

Question:15 Integrate the functions (x2+1)logx

Answer:

Consider (x2+1)logx

So, we have then: I=(x2+1)logxdx=x2logxdx+logxdx

Let us take I=I1+I2 ....................(1)

Where, I1=x2logxdx and I2=logxdx

So, I1=x2logxdx

After taking logx as a first function and x2 as second function and integrating by parts, we get

I=logxx2dx{(ddxlogx)x2.dx}dx

=logx.x331x.x33dx

=logx.x33x39+C1 ....................(2)

I2=logxdx

After taking logx as a first function and 1 as second function and integrating by parts, we get

I2=logx1.dx{(ddxlogx)1.dx}dx

=logx.x1x.xdx

=xlogx1dx

=xlogxx+C2 ................(3)

Now, using the two equations (2) and (3) in (1) we get,

I=x33logxx39+C1+xlogxx+C2

=x33logxx39+xlogxx+(C1+C2)

=(x33+x)logxx39x+C

Question:16 Integrate the functions ex(sinx+cosx)

Answer:

Let suppose
I= ex(sinx+cosx)
f(x)=sinxf(x)=cosx
we know that,
I=ex[f(x)+f(x)]dx=ex[f(x)]+C
Thus, the solution of the given integral is given by

I=exsinx+C

Question:17 Integrate the functions xex(1+x)2

Answer:

xex(1+x)2
Let suppose
I=ex(x)(1+x)2dx
by rearranging the equation, we get
ex[11+x1(1+x)2]dx
let
f(x)=11+xf(x)=1(1+x)2
It is known that ex[f(x)+f(x)]=ex[f(x)]+C
therefore the solution of the given integral is

I=ex1+x+C

Question:18 Integrate the functions ex(1+sinx1+cosx)

Answer:

Let
I=ex(1+sinx1+cosx)
substitute 1=sin2x2+cos2x2 and sinx=2sinx2cosx2

ex(sin2x2+cos2x2+2sinx2cosx22cos2x2)=ex(12sec2x2+tanx2)
let
f(x)=tanx2f(x)=12sec2x2
It is known that ex[f(x)+f(x)]=ex[f(x)]+C
Therefore the solution of the given integral is

I=extanx2+C

Question:19 Integrate the functions ex(1x1x2)

Answer:

ex(1x1x2)
It is known that
ex[f(x)+f(x)]=ex[f(x)]+C

let
f(x)=1xf(x)=1x2
Therefore the required solution of the given above integral is
I=ex.1x+C

Question:20 Integrate the functions (x3)ex(x1)3

Answer:

(x3)ex(x1)3
It is known that ex[f(x)+f(x)]=ex[f(x)]+C

So, By adjusting the given equation, we get
(x3)ex(x1)3=ex(x12(x1)3)=ex(1(x1)22(x1)3)dx

to let
f(x)=1(x1)2f(x)=2(x1)3
Therefore the required solution of the given I=ex(x1)2+C integral is

Question:21 Integrate the functions e2xsinx

Answer:

Let
I=e2xsinx
By using integrating by parts, we get

=sinxe2xdx(ddxsinx.e2xdx) dx=sinx.e2x212e2x.cosx dx=sinx.e2x212[cosxe2xdx(ddxcosx.e2xdx) dx]=sinx.e2x212[cosx.ex2+12e2xsinxdx]=sinx.e2x214cosx.e2x14I54I=sinx.e2x214cosx.e2xI=e2x5[2sinxcosx]+C

Question:22 Integrate the functions sin1(2x1+x2)

Answer:

sin1(2x1+x2)

sin1(2x1+x2)dx
let
x=tanθdx=sec2θdθ

=sin1(2tanθ1+tanθ)sec2θdθ=sin1(sin2θ)sec2θdθ=2θsec2θdθ
Taking θ as a first function and sec2θ as a second function, by using by parts method

=2[θsec2θdθ(ddθθ.sec2θ dθ)dθ]=2[θtanθtanθ dθ]+C=2[θtanθ+log|cosθ|]+C=2xtan1x+2log(1+x2)1/2=2xtan1xlog(1+x2)+C

Question:23 Choose the correct answer

x2ex3dxequals

A)13ex3+CB)13ex2+CC)12ex3+CD)12ex2+C

Answer:

the integration can be done ass follows

let x3=t3x2dx=dt
Double exponent: use braces to clarify

Question:24 Choose the correct answer

exsec(1+tanx)dxequals

A)excosx+CB)exsecx+CC)exsinx+CD)extanx+C

Answer:

we know that,
I=ex[f(x)+f(x)]dx=ex[f(x)]+C
from above integral
let
f(x)=secxf(x)=secx.tanx
thus, the solution of the above integral is
I=exsecx+C

NCERT class 12 maths ch 7 question answer Exercise: 7.7

Question:1 Integrate the functions in Exercises 1 to 9.

4x2

Answer:

Given function 4x2 ,

So, let us consider the function to be;

I=4x2dx

=(2)2x2dx

Then it is known that, =a2x2dx=x2a2x2+a22sin1xa+C

Therefore, I=x24x2+42sin1x2+C

=x24x2+2sin1x2+C


Question:2 Integrate the functions in Exercises 1 to 9.

14x2

Answer:

Given function to integrate 14x2

Now we can rewrite as

=1(2x)2dx

As we know the integration of this form is [a2x2dx=x2a2x2+a22sin1xa]

=(2x2)12(2x)2+122sin12x12Coefficient of x in 2x+C

=12[x14x2+12sin12x]+C

=x214x2+14sin12x+C


Question:3 Integrate the functions in Exercises 1 to 9.

x2+4x+6

Answer:

Given function x2+4x+6 ,

So, let us consider the function to be;

I=x2+4x+6dx

=(x2+4x+4)+2dx=(x+2)2+(2)2dx

And we know that, x2+a2dx=x2x2+a2+a22log|x+x2+a2|+C

I=x+22x2+4x+6+22log|(x+2)+x2+4x+6|+C

=x+22x2+4x+6+log|(x+2)+x2+4x+6|+C


Question:4 Integrate the functions in Exercises 1 to 9.

x2+4x+1

Answer:

Given function x2+4x+1 ,

So, let us consider the function to be;

I=x2+4x+1dx

=(x2+4x+4)3dx=(x+2)2(3)2dx

And we know that, x2a2dx=x2x2a2a22log|x+x2a2|+C

I=x+22x2+4x+132log|(x+2)+x2+4x+1|+C


Question:5 Integrate the functions in Exercises 1 to 9.

14xx2

Answer:

Given function 14xx2 ,

So, let us consider the function to be;

I=14xx2dx

=1(x2+4x+44)dx=1+4(x+2)2dx

=(5)2(x+2)2dx

And we know that, a2x2dx=x2a2x2+a22sin1xa+C

I=x+2214xx2+52sin1(x+25)+C


Question:6 Integrate the functions in Exercises 1 to 9.

x2+4x5

Answer:

Given function x2+4x5 ,

So, let us consider the function to be;

I=x2+4x5dx

a =(x2+4x+4)9dx=(x+2)2(3)2dx

And we know that, x2a2dx=x2x2a2a22log|x+x2a2|+C

I=x+22x2+4x592log|(x+2)+x2+4x5|+C


Question:7 Integrate the functions in Exercises 1 to 9.

1+3xx2

Answer:

Given function 1+3xx2 ,

So, let us consider the function to be;

I=1+3xx2dx

=(1(x23x+9494)dx=(1+94)(x32)2dx =(132)2(x32)2dx

And we know that, a2x2dx=x2a2x2+a22sin1xa+C

I=x3221+3xx2+138sin1(x32132)+C

=2x341+3xx2+138sin1(2x313)+C


Question:8 Integrate the functions in Exercises 1 to 9.

x2+3x

Answer:

Given function x2+3x ,

So, let us consider the function to be;

I=x2+3xdx

=x2+3x+9494dx

=(x+32)2(32)2dx

And we know that, x2a2dx=x2x2a2a22log|x+x2a2|+C

I=x+322x2+3x942log|(x+32)+x2+3x|+C

=2x+34x2+3x98log|(x+32)+x2+3x|+C


Question:9 Integrate the functions in Exercises 1 to 9.

1+x29

Answer:

Given function 1+x29 ,

So, let us consider the function to be;

I=1+x29dx=139+x2dx

=1332+x2dx

And we know that, x2+a2dx=x2x2+a2+a22log|x+x2+a2|+C

I=13[x2x2+9+92log|x+x2+9|]+C

=x6x2+9+32log|x+x2+9|+C


Question:10 Choose the correct answer in Exercises 10 to 11.

1+x2dx is equal to

(A) x21+x2+12log|(x+1+x2)|+C

(B) 23(1+x2)32+C

(C) 23x(1+x2)32+C

(D) x221+x2+12x2log|x+1+x2|+C

Answer:

As we know that, x2+a2dx=x2x2+a2+a22log|x+x2+a2|+C

So, 1+x2dx=x2x2+1+12log|x+x2+1|+C

Therefore the correct answer is A.


Question:11 Choose the correct answer in Exercises 10 to 11.

x28x+7dx is equal to

(A) 12(x4)x28x+7+9log|x4+x28x+7|+C

(B) 12(x+4)x28x+7+9log|x+4+x28x+7|+C

(C) 12(x4)x28x+732log|x4+x28x+7|+C

(D) 12(x4)x28x+792log|x4+x28x+7|+C

Answer:

Given integral x28x+7dx

So, let us consider the function to be;

I=x28x+7dx=(x28x+16)(9)dx

=(x4)2(3)2dx

And we know that, x2a2dx=x2x2a2a22log|x+x2a2|+C

I=(x4)2x28x+792log|(x4)+x28x+7|+C

Therefore the correct answer is D.



NCERT class 12 maths ch 7 question answer - Exercise:7.8

Question:1 Evaluate the following definite integrals as a limit of sums.

abxdx

Answer:

We know that,
abf(x)dx=(ba)limx1n[f(a)+f(a+h)+...+f(a+(n1)h)]
abxdx=(ba)limx1n[a+(a+h)...(a+2h)..a+(n1)h]
=(ba)limx1n[(a+a...a+a)n+(h+2h+3h....(n1)h)]=(ba)limx1n[na+h(1+2+3..+n1)]=(ba)limx1n[na+h(n(n1)2)]=(ba)limx[a+n12h]=(ba)limx[a+(n1)(ba)2n]=(ba)limx[a+(11n)(ba)2]=(ba)[a+(ba)2]=(ba)(b+a)/2=(b2a2)2

This is how the integral is evaluated using limit of a sum

Question:2 Evaluate the following definite integrals as limit of sums.

05(x+1)dx

Answer:

We know that
let I=05(x+1)dx
abf(x)dx=(ba)limn1n[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]h=ban
Here a = 0, b = 5 and f(x)=(x+1)
therefore h=5n


05(x+1)dx=5limx1n[f(0)+f(5/n)+.....+f((n1)5/n)]

=5limx1n[1+(5/n+1)+....+(1+5(n1)n)]=5limx1n[(1+1..+1)n+5n(1+2+3+...+n1)]=5limx1n[n+5nn(n1)2]=5limx1n[n+5(n1)2]=5limx[1+5(11n)2]=5[1+52]=352

Question:3 Evaluate the following definite integrals as limit of sums.

23x2dx

Answer:

We know that

abf(x)dx=(ba)limn1n[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]h=ban
here a = 2 and b = 3 , so h = 1/n


23x2dx=(32)limx1n[f(2)+f(2+1n)+f(2+2n)+....+f(2+(n1)n)]

=(1)limx1n[22+(2+1n)2+......+(2+(n1)n)2]=limx1n[(22+22+...22)n+(1n)2+(2n)2+....(n1n)2+4(1n+2n+.....+n1n)=limx1n[4n+n(n1)(2n1)6n2+4n.n(n1)2]=limx1n[4n+(1(11n)(2n1)6)+4(n1)2]
=limx1n[4n+(1n(11n)(21n)6)+4(n1)2]=limx1n.n[4+(1(11n)(21n)6)+22n]=4+26+2=193

Question:4 Evaluate the following definite integrals as limit of sums.

14(x2x)dx

Answer:

Let
I=14(x2x)dx=14x2dx14xdxI=I1I2

14x2dx=(41)limx1n[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n1)h)]

=(41)limx1n[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n1)h)]=3limx1n[12+(1+3n)2+(1+2.3n)2+......+(1+(n1).3n)2]=3limx1n[(12+..12)n+(3n)2(12+22+32+....+(n1)2)+2.3n(1+2+3..+n1)]=3limx1n[n+9n2(n(n1)(2n1)6)+6n(n(n1)2)]

=3limx1n[n+9n2(n(n1)(2n1)6)+6n(n(n1)2)]=3limx[1+96(11n)(21n)+3(11n)]=3[1+96.2+3]=21

for the second part, we already know the general solution of abxdx=(b2a2)2
So, here a = 1 and b = 4
therefore 14xdx=(4212)2=152

So, I=21152=272

Question:5 Evaluate the following definite integrals as limit of sums.

. 11exdx

Answer:

let I=11exdx
We know that
abf(x)dx=(ba)limn1n[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]h=ban
Here a =-1, b = 1 and f(x)=ex
therefore h = 2/n
I=2.limx1n[f(1)+f(1+2n)+.....+f(1+(n1).2n)]
=2.limx1n[e1+e1+2n+e1+2.2n+...+e1+(n1).2n]=2.limx1n[e1(1+e2/n+e4/n+...+e(n1).2n)]=
By using sum of n terms of GP S=a(rn1)r1 ....where a = 1st term and r = ratio

=2limne1n[1.(e2n.n1)e2n1]=2limne1n(e21e2/n1)=e1(e21)lim2ne2/n12/n=e21e .........using [limx(ex1x)=1]

Question:6 Evaluate the following definite integrals as limit of sums.

04(x+e2x)dx

Answer:

It is known that,
04(x+e2x)dx=4limx1n[f(0)+f(h)+f(2h)+....+f(n1)h]
=4limx1n[(0+e0)+(h+e2h)+(2h+e4h)+......+((n1)h+e2(n1)h)]=4limx1n[h(1+2+3+.....+n1)+(e2nh1e2h1)]=4limx1n[4n(n(n1)2)+(e81e8/n1)]
=4limx[4.11n2+e818e8/n18n]=4(2)+4[(e818)]==8+e8/21/2=15+e82 ..........................( limx0ex1x=1 )

NCERT class 12 maths ch 7 question answer - Exercise:7.9

Question:1 Evaluate the definite integrals in Exercises 1 to 20.

11(x+1)dx

Answer:

Given integral: I=11(x+1)dx

Consider the integral (x+1)dx

(x+1)dx=x22+x

So, we have the function of x , f(x)=x22+x

Now, by Second fundamental theorem of calculus, we have

I=f(1)f(1)

=(12+1)(121)=12+112+1=2

Question:2 Evaluate the definite integrals in Exercises 1 to 20.

231xdx

Answer:

Given integral: I=231xdx

Consider the integral 231xdx

1xdx=log|x|

So, we have the function of x , f(x)=log|x|

Now, by Second fundamental theorem of calculus, we have

I=f(3)f(2)

=log|3|log|2|=log32

Question:3 Evaluate the definite integrals in Exercises 1 to 20.

12(4x35x2+6x+9)dx

Answer:

Given integral: I=12(4x35x2+6x+9)dx

Consider the integral I=(4x35x2+6x+9)dx

(4x35x2+6x+9)dx=4x445x33+6x22+9x

=x45x33+3x2+9x

So, we have the function of x , f(x)=x45x33+3x2+9x

Now, by Second fundamental theorem of calculus, we have

I=f(2)f(1)

={245(2)33+3(2)2+9(2)}{145(1)33+3(1)2+9(1)}

={16403+12+18}{153+3+9}

={46403}{1353}

={33353}={99353}

=643

Question:4 Evaluate the definite integrals in Exercises 1 to 20.

0π4sin2xdx

Answer:

Given integral: 0π4sin2xdx

Consider the integral sin2xdx

sin2xdx=cos2x2

So, we have the function of x , f(x)=cos2x2

Now, by Second fundamental theorem of calculus, we have

I=f(π4)f(0)

=cos2(π4)2+cos02

=120

=12

Question:5 Evaluate the definite integrals in Exercises 1 to 20.

0π2cos2xdx

Answer:

Given integral: 0π2cos2xdx

Consider the integral cos2xdx

cos2xdx=sin2x2

So, we have the function of x , f(x)=sin2x2

Now, by Second fundamental theorem of calculus, we have

I=f(π2)f(0)

=12{sin2(π2)sin0}

=12{00}=0

Question:6 Evaluate the definite integrals in Exercises 1 to 20.

45exdx

Answer:

Given integral: 45exdx

Consider the integral exdx

exdx=ex

So, we have the function of x , f(x)=ex

Now, by Second fundamental theorem of calculus, we have

I=f(5)f(4)

=e5e4

=e4(e1)

Question:7 Evaluate the definite integrals in Exercises 1 to 20.

0π4tanxdx

Answer:

Given integral: 0π4tanxdx

Consider the integral tanxdx

tanxdx=log|cosx|

So, we have the function of x , f(x)=log|cosx|

Now, by Second fundamental theorem of calculus, we have

I=f(π4)f(0)

=log|cosπ4|+log|cos0|

=log|cos12|+log|1|

=log|12|+0=log(2)12

=12log(2)

Question:8 Evaluate the definite integrals in Exercises 1 to 20.

π6π4cosecxdx

Answer:

Given integral: π6π4cosecxdx

Consider the integral cosecxdx

cosecxdx=log|cosecxcotx|

So, we have the function of x , f(x)=log|cosecxcotx|

Now, by Second fundamental theorem of calculus, we have

I=f(π4)f(π6)

=log|cosecπ4cotπ4|log|cosecπ6cotπ6|

=log|21|log|23|

=log(2123)

Question:9 Evaluate the definite integrals in Exercises 1 to 20.

01dx1x2

Answer:

Given integral: 01dx1x2

Consider the integral dx1x2

dx1x2=sin1x

So, we have the function of x , f(x)=sin1x

Now, by Second fundamental theorem of calculus, we have

I=f(1)f(0)

=sin1(1)sin1(0)

=π20

=π2

Question:10 Evaluate the definite integrals in Exercises 1 to 20.

01dx1+x2

Answer:

Given integral: 01dx1+x2

Consider the integral dx1+x2

dx1+x2=tan1x

So, we have the function of x , f(x)=tan1x

Now, by Second fundamental theorem of calculus, we have

I=f(1)f(0)

=tan1(1)tan1(0)

=π40

=π4

Question:11 Evaluate the definite integrals in Exercises 1 to 20.

23dxx21

Answer:

Given integral: 23dxx21

Consider the integral dxx21

dxx21=12log|x1x+1|

So, we have the function of x , f(x)=12log|x1x+1|

Now, by Second fundamental theorem of calculus, we have

I=f(3)f(2)

=12{log|313+1|log|212+1|}

=12{log|24|log|13|}

=12{log12log13}=12log32

Question:12 Evaluate the definite integrals in Exercises 1 to 20.

0π2cos2xdx

Answer:

Given integral: 0π2cos2xdx

Consider the integral cos2xdx

cos2xdx=1+cos2x2dx=x2+sin2x4

=12(x+sin2x2)

So, we have the function of x , f(x)=12(x+sin2x2)

Now, by Second fundamental theorem of calculus, we have

I=f(π2)f(0)

=12{(π2sinπ2)(0+sin02)}

=12{π2+000}

=π4

Question:13 Evaluate the definite integrals in Exercises 1 to 20.

23xdxx2+1

Answer:

Given integral: 23xdxx2+1

Consider the integral xdxx2+1

xdxx2+1=122xx2+1dx=12log(1+x2)

So, we have the function of x , f(x)=12log(1+x2)

Now, by Second fundamental theorem of calculus, we have

I=f(3)f(2)

=12{log(1+(3)2)log(1+(2)2)}

=12{log(10)log(5)}=12log(105)=12log2

Question:14 Evaluate the definite integrals in Exercises 1 to 20.

012x+35x2+1dx

Answer:

Given integral: 012x+35x2+1dx

Consider the integral 2x+35x2+1dx

Multiplying by 5 both in numerator and denominator:

2x+35x2+1dx=155(2x+3)5x2+1dx

=1510x+155x2+1dx

=1510x5x2+1dx+315x2+1dx

=1510x5x2+1+315(x2+15)dx

=15log(5x2+1)+35×115tan1x15

=15log(5x2+1)+35tan1(5x)

So, we have the function of x , f(x)=15log(5x2+1)+35tan1(5x)

Now, by Second fundamental theorem of calculus, we have

I=f(1)f(0)

={15log(1+5)+35tan1(5)}{15log(1)+35tan1(0)}

=15log6+35tan15

Question:15 Evaluate the definite integrals in Exercises 1 to 20.

01xex2dx

Answer:

Given integral: 01xex2dx

Consider the integral xex2dx

Putting x2=t which gives, 2xdx=dt

As, x0,t0 and as x1,t1 .

So, we have now:

I=1201etdt

=12etdt=12et

So, we have the function of x , f(x)=12et

Now, by Second fundamental theorem of calculus, we have

I=f(1)f(0)

=12e112e0=12(e1)

Question:16 Evaluate the definite integrals in Exercises 1 to 20.

125x2x2+4x+3

Answer:

Given integral: I=125x2x2+4x+3

So, we can rewrite the integral as;

I=125x2x2+4x+3=12(520x+15x2+4x+3)dx

=125dx1220x+15x2+4x+3dx

=[5x]121220x+15x2+4x+3dx

I=5I1 where I=1220x+15x2+4x+3dx . ................(1)

Now, consider I=1220x+15x2+4x+3dx

Take numerator 20x+15=Addx(x2+4x+3)+B

=2Ax+(4A+B)

We now equate the coefficients of x and constant term, we get

A=10\and B=25

I1=10122x+4x2+4x+3dx2512dxx2+4x+3

Now take denominator x2+4x+3=t

Then we have (2x+4)dx=dt

I1=10dtt25dx(x+2)212

=10logt25[12log(x+21x+2+1)]

=[10log(x2+4x+3)]1225[12log(x+1x+3)]12

=[10log1510log8]25[12log3512log24]

=[10log5+10log310log410log2]252[log3log5log2+log4] =(10+252)log5+(10252)log4+(10252)log3+(10+252)log2 =452log5452log452log3+52log2

=452log5452log32

Then substituting the value of I1 in equation (1), we get

I=5(452log5452log32)

=552(9log54log32)

Question:17 Evaluate the definite integrals in Exercises 1 to 20.

0π4(2sec2x+x3+2)dx

Answer:

Given integral: 0π4(2sec2x+x3+2)dx

Consider the integral (2sec2x+x3+2)dx

(2sec2x+x3+2)dx=2tanx+x44+2x

So, we have the function of x , f(x)=2tanx+x44+2x

Now, by Second fundamental theorem of calculus, we have

I=f(π4)f(0)

={(2tanπ4+14(π4)4+2π4)(2tan0+0+0)}

=2tanπ4+π445+π2

2+π2+π41024

Question:18 Evaluate the definite integrals in Exercises 1 to 20.

0π(sin2x2cos2x2)dx

Answer:

Given integral: 0π(sin2x2cos2x2)dx

Consider the integral (sin2x2cos2x2)dx

can be rewritten as: (cos2x2sin2x2)dx=0πcosxdx

=sinx

So, we have the function of x , f(x)=sinx

Now, by Second fundamental theorem of calculus, we have

I=f(π)f(0)

sinπsin0=00=0

Question:19 Evaluate the definite integrals in Exercises 1 to 20.

026x+3x2+4

Answer:

Given integral: 026x+3x2+4

Consider the integral 6x+3x2+4

can be rewritten as: 6x+3x2+4=32x+1x2+4dx

=32xx2+4dx+31x2+4dx

=3log(x2+4)+32tan1x2

So, we have the function of x , f(x)=3log(x2+4)+32tan1x2

Now, by Second fundamental theorem of calculus, we have

I=f(2)f(0)

={3log(22+4)+32tan1(22)}{3log(0+4)+32tan1(02)} =3log8+32tan113log432tan10

=3log8+32×π43log40

=3log84+3π8

or we have =3log2+3π8

Question:20 Evaluate the definite integrals in Exercises 1 to 20.

01(xex+sinπx4)dx

Answer:

Given integral: 01(xex+sinπx4)dx

Consider the integral (xex+sinπx4)dx

can be rewritten as: xexdx{(ddxx)exdx}dx+{cosπx4π4}

=xexexdx4ππcosx4

=xexex4ππcosx4

So, we have the function of x , f(x)=xexex4ππcosx4

Now, by Second fundamental theorem of calculus, we have

I=f(1)f(0)

=(1.etet4πcosπ4)(0.e0e04πcos0)

=ee4π(12)+1+4π

Question:21 Choose the correct answer in Exercises 20 and 21.

13dx1+x2

(A) π3

(B) 2π3

(C) π6

(D) π12

Answer:

Given definite integral 13dx1+x2

Consider dx1+x2=tan1x

we have then the function of x, as f(x)=tan1x

By applying the second fundamental theorem of calculus, we will get

13dx1+x2=f(3)f(1)

=tan13tan11

=π3π4

=π12

Therefore the correct answer is D.

Question:22 Choose the correct answer in Exercises 21 and 22.

023dx4+9x2 equals

(A) π6

(B) π12

(C) π24

(D) π4

Answer:

Given definite integral 023dx4+9x2

Consider dx4+9x2=dx22+(3x)2

Now, putting 3x=t

we get, 3dx=dt

Therefore we have, dx22+(3x)2=13dt22+t2

=13(12tan1t2)=16tan1(3x2)

we have the function of x , as f(x)=16tan1(3x2)

So, by applying the second fundamental theorem of calculus, we get

023dx4+9x2=f(23)f(0)

=16tan1(32.23)16tan10

=16tan110

=16×π4=π24

Therefore the correct answer is C.

NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.10

Question:1 Evaluate the integrals in Exercises 1 to 8 using substitution.

01xx2+1dx

Answer:

01xx2+1dx
let x2+1=txdx=dt/2
when x = 0 then t = 1 and when x =1 then t = 2
o1xx2+1dx=1212dtt
=12[log|t|]12=12log2

Question:2 Evaluate the integrals in Exercises 1 to 8 using substitution.

0π2sinϕcos5ϕdϕ

Answer:

0π2sinϕcos5ϕdϕ
let sinϕ=tcosϕdϕ=dt
when ϕ=0,t0 and ϕ=π/2,t1

using the above substitution we can evaluate the integral as

01t(1t2)dt=01t12(1+t42t2)dt=01t12dt+01t9/2dt201t5/2dt=[2t3/2/3+2t11/2/11+4t7/2/7]01=64231

Question:3 Evaluate the integrals in Exercises 1 to 8 using substitution.

01sin1(2x1+x2)dx

Answer:

01sin1(2x1+x2)dx
let
x=tanθdx=sec2θdθ
when x = 0 then θ=0 and when x = 1 then θ=π/4

=0π/4sin1(2tanθ1+tanθ)sec2θdθ=0π/4sin1(sin2θ)sec2θdθ=0π/42θsec2θdθ
Taking θ as a first function and sec2θ as a second function, by using by parts method

=2[θsec2θdθ(ddθθ.sec2θ dθ)dθ]0π/4=2[θtanθtanθ dθ]0π/4=2[θtanθ+log|cosθ|]0π/4=2[π/4+log(1/2)]=π/4log2

Question:4 Evaluate the integrals in Exercises 1 to 8 using substitution.

02xx+2 . (Put x+2=t2 )

Answer:

Let x+2=t2dx=2tdt
when x = 0 then t = 2 and when x=2 then t = 2

I=02xx+2dx

=222(t22)t2dt=222(t42t2)dt=2[t5/523t3]22=2[325163425+423]=162(2+1)15

Question:5 Evaluate the integrals in Exercises 1 to 8 using substitution.

0π2sinx1+cos2xdx

Answer:

0π2sinx1+cos2xdx=I
let cosx=tsinxdx=dt
when x=0 then t = 1 and when x= π/2 then t = 0

I=10dt1+t2=[tan1t]10=π/4

Question:6 Evaluate the integrals in Exercises 1 to 8 using substitution.

02dxx+4x2

Answer:

By adjusting, the denominator can also be written as (172)2(x12)2=x+4x2
Now,
02dx(172)2(x12)2
let x1/2=tdx=dt
when x= 0 then t =-1/2 and when x =2 then t = 3/2

1/23/2dt(172)2t2=12.172log172+t172t=117[log17/2+3/217/23/2log17/21/217/2+1/2]=117[log17+3173/.17+117+1]
=117[log(17+3+41717+3417)]=117[log(5+17517)]
On rationalisation, we get

=117log21+5174

Question:7 Evaluate the integrals in Exercises 1 to 8 using substitution.

11dxx2+2x+5

Answer:

11dxx2+2x+5
the Dr can be written as x2+2x+5=(x+1)2+22
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

02dtt2+22=12[tan1t2]02=12(π/4)=π8

Question:8 Evaluate the integrals in Exercises 1 to 8 using substitution.

12(1x12x2)e2xdx

Answer:

12(1x12x2)e2xdx
let 2x=t2dx=dt
when x = 1 then t = 2 and when x = 2 then t= 4

=1224(2t2t2)etdt
let
1t=f(t)f(t)=1t2
24(1t1t2)etdt=24et[f(t)+f(t)]dt
=[etf(t)]24=[et.1t]24=e44e22=e2(e22)4

Question:9 Choose the correct answer in Exercises 9 and 10.

The value of the integral 131(xx3)13x4dx is

(A) 6

(B) 0

(C) 3

(D) 4

Answer:

The value of integral is (A) = 6

131(xx3)13x4dx
131(1x21)13x3dx
let
1x21=tdxx3=dt/2
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

=1280t1/3dt=12.34[t4/3]80=38[24]=6

Question:10 Choose the correct answer in Exercises 9 and 10.

If f(x)=0xtsintdt , then f(x) is

(A) cosx+xsinx

(B) xsinx

(C) xcosx

(D) sinx+xcosx

Answer:

The correct answer is (B) = xsinx

f(x)=0xtsintdt
by using by parts method,
=tsintdt(ddttsintdt)dt=[t(cost)+sint]0x

f(x)=xcosx+sinx
So, f(x)=cosx+xsinx+cosx=xsinx

NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0π2cos2xdx

Answer:

We have I = 0π2cos2xdx ............................................................. (i)

By using

 0a f(x)dx =  0a f(ax)dx

We get :-

I = 0π2cos2xdx = 0π2cos2 (π2x)dx

or

I = 0π2sin2xdx ................................................................ (ii)

Adding both (i) and (ii), we get :-

0π2cos2xdx + 0π2sin2xdx = 2I

or 0π2 (cos2x + sin2x)dx = 2I

or 0π21.dx = 2I

or 2I = [x]0Π2 = Π2

or I = Π4

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. 0π2sinxsinx+cosxdx

Answer:

We have I = 0π2sinxsinx+cosxdx .......................................................................... (i)

By using ,

 0a f(x)dx =  0a f(ax)dx

We get,

I = 0π2sinxsinx+cosxdx = 0π2sin(π2x)sin(π2x)+cos(π2x)dx


or I = 0π2cosxcosx+sinxdx .......................................................(ii)

Adding (i) and (ii), we get,

2I = 0π2sinx + cosxsinx+cosxdx

or 2I = 0π21.dx


or 2I = [x]0Π2 = Π2

Thus I = Π4

Question:€‹3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0π2sin32xdxsin32x+cos32x

Answer:

We have I = 0π2sin32xdxsin32x+cos32x ..................................................................(i)

By using :

 0a f(x)dx =  0a f(ax)dx

We get,

I = 0π2sin32(π2x)dxsin32(π2x)+cos32(π2x)


or I = 0π2cos32xdxsin32x+cos32x . ............................................................(ii)

Adding (i) and (ii), we get :

2I = 0π2 (sin32x+cos32x)dxsin32x+cos32x

or 2I =0π21.dx

or 2I =[x]0π2 = π2

Thus I = π4

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. 0π2cos5xdxsin5x+cos5x

Answer:

We have I = 0π2cos5xdxsin5x+cos5x ..................................................................(i)

By using :

 0a f(x)dx =  0a f(ax)dx

We get,

I = 0π2cos5(π2x)dxsin5(π2x)+cos5(π2x)

or I = 0π2sin5xdxsin5x+cos5x . ............................................................(ii)

Adding (i) and (ii), we get :

2I = 0π2 (sin5x + cos5x)dxsin5x+cos5x

or 2I =0π21.dx

or 2I =[x]0π2 = π2

Thus I = π4

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

55|x+2|dx

Answer:

We have, I = 55|x+2|dx

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

I = 52(x+2)dx + 25(x+2)dx

or I = [x22 + 2x]52 + [x22 + 2x]25

This gives I = 29

Question:6€‹ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

28|x5|dx

Answer:

We have, I = 28|x5|dx

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

I = 25(x5)dx + 58(x5)dx

or I = [x22  5x]25 + [x22  5x]58

This gives I = 9

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

01x(1x)ndx

Answer:

We have I = 01x(1x)ndx

U sing the property : -

 0a f(x)dx =  0a f(ax)dx

We get : -

I = 01x(1x)ndx = 01(1x)(1(1x))ndx

or I = 01(1x)xn dx

or I = 01(xn  xn+1) dx

or = [xn+1n+1  xn+2n+2]01

or = [1n+1  1n+2]

or I = 1(n+1)(n+2)

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0π4log(1+tanx)dx

Answer:

We have I = 0π4log(1+tanx)dx

By using the identity

 0a f(x)dx =  0a f(ax)dx

We get,

I = 0π4log(1+tanx)dx = 0π4log(1+tan(π4x))dx

or I = 0π4log(1+1tanx1+tanx)dx

or I = 0π4log(21+tanx)dx

or I = 0π4log2dx  0π4log(1+tanx)dx

or I = 0π4log2dx  I

or 2I = [xlog2]0Π4

or I = Π8log2

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

02x2xdx

Answer:

We have I = 02x2xdx

By using the identity

 0a f(x)dx =  0a f(ax)dx

We get :

I = 02x2xdx = 02(2x)2(2x)dx

or I = 02(2x)xdx

or I = 02(2x  x32dx

or = [43x32  25x52]02

or = 43(2)32  25(2)52

or I = 16215

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0π2(2logsinxlogsin2x)dx

Answer:

We have I = 0π2(2logsinxlogsin2x)dx

or I = 0π2(2logsinxlog(2sinxcosx))dx

or I = 0π2(logsinxlogcosx  log2)dx ..............................................................(i)

By using the identity :

 0a f(x)dx =  0a f(ax)dx

We get :

I = 0π2(logsin(π2x)logcos(π2x)  log2)dx

or I = 0π2(logcosxlogsinx  log2)dx ....................................................................(ii)

Adding (i) and (ii) we get :-

2I = 0π2(log2 log2)dx

or I = log2[Π2]

or I = Π2log12

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

π2π2sin2xdx

Answer:

We have I = π2π2sin2xdx

We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.

Also,

I = aaf(x)dx = 20af(x)dx

So,

I = 20π2sin2xdx = 20π2(1cos2x)2dx

or = [x  sin2x2]0Π2

or I = Π2

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0πxdx1+sinx

Answer:

We have I = 0πxdx1+sinx ..........................................................................(i)

By using the identity :-

 0a f(x)dx =  0a f(ax)dx

We get,

I = 0πxdx1+sinx = 0π(Πx)dx1+sin(Πx)

or I = 0π(Πx)dx1+sinx ............................................................................(ii)


Adding both (i) and (ii) we get,

2I = 0πΠ1+sinxdx

or 2I = Π0π1sinx(1+sinx)(1sinx)dx = Π0π1sinxcos2xdx

or 2I = Π0π(sec2  tanxsecx)xdx

or I = Π

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

π2π2sin7xdx

Answer:

We have I = π2π2sin7xdx

We know that sin7x is an odd function.

So the following property holds here:-

aaf(x)dx = 0

Hence

I = π2π2sin7xdx = 0

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

02πcos5xdx

Answer:

We have I = 02πcos5xdx

I t is known that :-

02af(x)dx = 20af(x)dx If f (2a - x) = f(x)

= 0 If f (2a - x) = - f(x)

Now, using the above property

cos5(Πx) = cos5x

Therefore, I = 0

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx

Answer:

We have I = 0π2sinxcosx1+sinxcosxdx ................................................................(i)

By using the property :-

 0a f(x)dx =  0a f(ax)dx

We get ,

I = 0π2sin(π2x)cos(π2x)1+sin(π2x)cos(π2x)dx

or I = 0π2cosxsinx1+sinxcosxdx ......................................................................(ii)

Adding both (i) and (ii), we get


2I = 0π201+sinxcosxdx

Thus I = 0

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0πlog(1+cosx)dx

Answer:

We have I = 0πlog(1+tanx)dx .....................................................................................(i)

By using the property:-

 0a f(x)dx =  0a f(ax)dx

We get,

or

I = 0πlog(1+cos(Πx))dx

I = 0πlog(1cosx)dx ....................................................................(ii)


Adding both (i) and (ii) we get,

2I = 0πlog(1+cosx)dx + 0πlog(1cosx)dx

or 2I = 0πlog(1cos2x)dx = 0πlogsin2xdx

or 2I = 20πlogsinxdx

or I = 0πlogsinxdx ........................................................................(iii)

or I = 20π2logsinxdx ........................................................................(iv)

or I = 20π2logcosxdx .....................................................................(v)

Adding (iv) and (v) we get,

I = πlog2

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

0axx+axdx

Answer:

We have I = 0axx+axdx ................................................................................(i)

By using, we get


 0a f(x)dx =  0a f(ax)dx

We get,

I = 0axx+axdx = 0a(ax)(ax)+xdx .................................................................(ii)


Adding (i) and (ii) we get :

2I = 0ax + axx+axdx

or 2I = [x]0a=a

or I = a2

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

04|x1|dx

Answer:

We have, I = 04|x1|dx

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

I = 01(x1)dx + 14(x1)dx

or I = [x  x22 ]01 + [x22  x]14

This gives I = 5

Question:19 Show that 0af(x)g(x)dx=20af(x)dx if f and g are defined as f(x)=f(ax) and g(x)+g(ax)=4

Answer:

Let I = 0af(x)g(x)dx ........................................................(i)

This can also be written as :

I = 0af(ax)g(ax)dx

or I = 0af(x)g(ax)dx ................................................................(ii)

Adding (i) and (ii), we get,

2I = 0af(x)g(ax)dx+ 0af(x)g(x)dx

2I = 0af(x)4dx

or I = 20af(x)dx

Question:20 Choose the correct answer in Exercises 20 and 21.

The value of is π2π2(x3+xcosx+tan5x+1)dx is

(A) 0

(B) 2

(C) π

(D) 1

Answer:

We have

I = π2π2(x3+xcosx+tan5x+1)dx

This can be written as :

I = π2π2x3dx+ π2π2xcosx+ π2π2tan5x+ π2π21dx

Also if a function is even function then aaf(x) dx = 20af(x) dx

And if the function is an odd function then : aaf(x) dx = 0

Using the above property I become:-

I = 0+0+0+20Π21.dx

or I = 2[x]0Π2

or I = Π

Question:€‹21 Choose the correct answer in Exercises 20 and 21.

The value of 0π2log(4+3sinx4+3cosx)dx is

Answer:

We have

I = 0π2log(4+3sinx4+3cosx)dx .................................................................................(i)


By using :

 0a f(x)dx =  0a f(ax)dx

We get,

I = 0π2log(4+3sinx4+3cosx)dx = 0π2log(4+3sin(π2x)4+3cos(π2x))dx

or I = 0π2log(4+3cosx4+3sinx)dx .............................................................................(ii)

Adding (i) and (ii), we get:

2I = 0π2log(4+3sinx4+3cosx)dx + 0π2log(4+3cosx4+3sinx)dx

or 2I = 0π2log1.dx

Thus I = 0


NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise

Question:1 Integrate the functions in Exercises 1 to 24.

1xx3

Answer:

Firstly we will simplify the given equation :-

1xx3 = 1(x)(1x)(1+x)

Let

1(x)(1x)(1+x)= Ax + B1x + C1+x

By solving the equation and equating the coefficients of x 2 , x and the constant term, we get

A = 1, B = 12, C = 12

Thus the integral can be written as :

1(x)(1x)(1+x)dx= 1xdx + 1211xdx + 1211+xdx

= logx  12log(1x) + 12log(1+x)

or = 12logx21x2 + C


Question:2 Integrate the functions in Exercises 1 to 24.

1x+a+x+b

Answer:

At first we will simplify the given expression,

1x+a+x+b = 1x+a+x+b×x+ax+bx+ax+b

or = x+ax+bab

Now taking its integral we get,

1x+a+x+b = 1ab(x+ax+b)dx

or = 1ab[(x+a)3232  (x+b)3232]

or = 23(ab)[(x+a)32  (x+b)32] + C


Question:3€‹ Integrate the functions in Exercises 1 to 24.

1xaxx2 [Hint: Put x=at ]

Answer:

Let

x=at dx  dx = at2dh

Using the above substitution we can write the integral is

1xaxx2 = 1ata.at  (at)2at2dt

or

= 1a1(t1)dt

or

= 1a (2t1) + C

or = 1a (2ax  1) + C

or = 2a axx + C

Question:4 Integrate the functions in Exercises 1 to 24.

1x2(x4+1)34

Answer:

For the simplifying the expression, we will multiply and dividing it by x -3 .

We then have,

x3x2x3(x4+1)34 = 1x5[x4 + 1x4]34

Now, let

1x4 = t  1x5dx = dt4

Thus,

1x2(x4+1)34 = 1x5(1+ 1x434 )dx

or = 14(1+t)34dt

= 14(1+1x4)1414 + C

= [1+1x4]14 + C

Question:5 Integrate the functions in Exercises 1 to 24.

1x12+x13 [Hint: 1x12+x13=1x13(1+ x16) , put x=t6 ]

Answer:

Put x=t6  dx=6t5dt

We get,

1x12+x13dx = 6t5t3+t2dt

or = 6t31+tdt

or = 6{(t2t+1)11+t}dt

or = 6[(t33)(t22)+tlog(1+t)]

Now put x=t6 in the above result :

= 2x3x13+6x166log(1x16) + C

Question:6 Integrate the functions in Exercises 1 to 24.

5x(x+1)(x2+9)

Answer:

Let us assume that :

5x(x+1)(x2+9) = A(x+1) + Bx+cx2+9

Solving the equation and comparing coefficients of x 2 , x and the constant term.

We get,

A = 12 ; B = 12 ; C = 92

Thus the equation becomes :

5x(x+1)(x2+9) = 12(x+1) + x2+92x2+9

or

5x(x+1)(x2+9) = [12(x+1) + x+92(x2+9)]dx

or = 12log|x+1|+12xx2+9dx+921x2+9dx

or = 12log|x+1|+142xx2+9dx+921x2+9dx

or = 12log|x+1|+14log(x2+9)+32tan1x3 + C

Question:7 Integrate the functions in Exercises 1 to 24.

sinxsin(xa)

Answer:

We have,

I = sinxsin(xa)

Assume :- (xa) = t dx=dt

Putting this in above integral :

sinxsin(xa)dx = sin(t+a)sintdt

or = sintcosa + costsinasintdt

or = (cosa + cottsina)dt

or = tcosa + sinalog|sint| + C

or = sinalog|sin(xa)|+xcosa + C

Question:€‹9 Integrate the functions in Exercises 1 to 24.

cosx4sin2x

Answer:

We have the given integral

I = cosx4sin2x

Assume sinx=t cosxdx=dt

So, this substitution gives,

cosx4sin2x = dt(2)2(t)2

= sin1t2 + C

or = sin1(sinx2) + C

Question:10 Integrate the functions in Exercises 1 to 24.

sin8xcos8x12sinxcos2x

Answer:

We have

I = sin8xcos8x12sinxcos2x

Simplifying the given expression, we get :

sin8xcos8x12sinxcos2x = (sin4x+cos4x)(sin4xcos4x)12sinxcos2x

or = (sin4x+cos4x)(sin2xcos2x)(sin2x+cos2x)12sinxcos2x

or = (sin4x+cos4x)(cos2xsin2x)12sinxcos2x

or = cos2xsin2x = cos2x

Thus,

I = sin8xcos8x12sinxcos2x = cos2x dx

and = sin2x2 + C

Question:11 Integrate the functions in Exercises 1 to 24.

1cos(x+a)cos(x+b)

Answer:

For simplifying the given equation, we need to multiply and divide the expression by sin(ab) .

Thus we obtain :

1cos(x+a)cos(x+b) = 1sin(ab)×sin(ab)cos(x+a)cos(x+b)

or =1sin(ab)×sin[(x+a)(x+b)]cos(x+a)cos(x+b)

or =1sin(ab)×(sin(x+a)cos(x+a)sin(x+b)cos(x+b))

or =1sin(ab)×(tan(x+a)  tan(x+b))

Thus integral becomes :

1cos(x+a)cos(x+b) = 1sin(ab)×(tan(x+a)  tan(x+b))dx

or = 1sin(ab)×[log|cos(x+a)|+log|cos(x+b)|] + C

or = 1sin(ab)×log[cos(x+b)cos(x+a)] + C

Question:12€‹ Integrate the functions in Exercises 1 to 24.

x31x8

Answer:

Given that to integrate

x31x8

Let x4=t4x3dx=dt

x31x8dx=1411t2dt

=14sin1t+C=14sin1x4+C

the required solution is 14sin1(x4)+C

Question:13 Integrate the functions in Exercises 1 to 24.

ex(1+ex)(2+ex)

Answer:

we have to integrate the following function

ex(1+ex)(2+ex)

Let 1+ex=texdx=dt

using this we can write the integral as

ex(1+ex)(2+ex)dx=1t(1+t)dt=(1+t)tt(1+t)dt

=(1t1t+1)dt

=1tdt1t+1dt

=logtlog(1+t)+C=log(1+ex)log(2+ex)+C=log(ex+1ex+2)+C

Question:14 Integrate the functions in Exercises 1 to 24.

1(x2+1)(x2+4)

Answer:

Given,

1(x2+1)(x2+4)

Let I=1(x2+1)(x2+4)

Now, Using partial differentiation,

1(x2+1)(x2+4)=Ax+B(x2+1)+Cx+D(x2+4)

1(x2+1)(x2+4)=(Ax+B)(x2+4)+(Cx+D)(x2+1)(x2+1)(x2+4)
1=(Ax+B)(x2+4)+(Cx+D)(x2+1)1=Ax3+4Ax+Bx2+4B+Cx3+Cx+Dx2+D(A+C)x3+(B+D)x2+(4A+C)x+(4B+D)=1

Equating the coefficients of x,x2,x3 and constant value,

A + C = 0 C = -A

B + D = 0 B = -D

4A + C =0 4A = -C 4A = A A = 0 = C

4B + D = 1 4B – B = 1 B = 1/3 = -D

Putting these values in equation, we have

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155444707593450

1554447076674147

1554447078146237

I=13tan1x16tan1x2+C

Question:15 Integrate the functions in Exercises 1 to 24.

cos3xelogsinx

Answer:

Given,

cos3xelogsinx

I=cos3xelogsinx (let)

Let cosx=tsinxdx=dtsinxdx=dt

using the above substitution the integral is written as

cos3xelogsinxdx=cos3x.sinxdx

155444708344321

1554447084197794

1554447084982414

155444708573581

I=cos4x4+C

Question:16Integrate the functions in Exercises 1 to 24.

e3logx(x4+1)1

Answer:

Given the function to be integrated as

e3logx(x4+1)1
=elogx3(x4+1)1=x3x4+1

Let I=e3logx(x4+1)1

Let x4=t4x3dx=dt

I=e3logx(x4+1)1=x3x4+1

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1554447092538842

I=14log(x4+1)+C

Question:€‹17 Integrate the functions in Exercises 1 to 24.

f(ax+b)[f(ax+b)]n

Answer:

Given,

f(ax+b)[f(ax+b)]n

Let I=f(ax+b)[f(ax+b)]n

Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt

Now we can write the ntegral as

f(ax+b)[f(ax+b)]n=1atndt

=1a.tn+1n+1+C=1a.(f(ax+b))n+1n+1+C

I=(f(ax+b))n+1a(n+1)+C


Question:18 Integrate the functions in Exercises 1 to 24.

1sin3xsin(x+α)

Answer:

Given,

1sin3xsin(x+α)

Let I=1sin3xsin(x+α)

We know the identity that

sin (A+B) = sin A cos B + cos A sin B

1sin3xsin(x+α)=1sin3x(sinxcosα+cosxsinα)

=1sin3x.sinx(cosα+cotxsinα)=1sin4x(cosα+cotxsinα)

cosec2x(cosα+cotxsinα)

1554447105407729

1554447106158144


1554447106897484


1554447107634823


1554447108399148


155444710916564


1554447109907513


1554447110653730


1554447111419992


1554447112159711


1554447112971688


Question:19 Integrate the functions in Exercises 1 to 24.

. sin1xcos1xsin1x+cos1x,x[0,1]

Answer:

We have

I = sin1xcos1xsin1x+cos1x dx

or = sin1x(Π2sin1x)Π2 dx

or = 2Π( 2sin1xΠ2) dx

or = (4Πsin1x1) dx

or = 4Πsin1x1 dx  1 dx + C

or = 4Πsin1x dx  x+ C

Thus I = 4ΠI  x+ C


Now we will solve I'.

I = sin1x dx

Put x = t 2 .

Differentiating the equation wrt x, we get

dx = 2t dt

Thus sin1x dx = sin1t 2t dt

or = 2t sin1t  dt

Using integration by parts, we get :

= 2[sin1tt dt  ((ddtsin1t)t dt)] dt

or = t2sin1t  t21t2 dt + C

We know that

t21t2 dt = t21t2  12 sin1t

Thus it becomes :

I = t2sin1t + t21t2  12 sin1t

So I come to be :-

I = 4ΠI  x+ C

I = sin1x[2(2x1)Π] + 2xx2Π  x + C

Question:€‹20 Integrate the functions in Exercises 1 to 24.

1x1+x

Answer:

Given,

1x1+x = I (let)

Let x=cos2θdx=2sinθcosθdθ

Andx=cosθθ=cos1x

using the above substitution we can write the integral as

I=1cos2θ1+cos2θ(2sinθcosθ)dθ=1cosθ1+cosθ(2sinθcosθ)dθ

=tan2θ2(2sinθcosθ)dθ=tan2θ2(2.2sinθ2cosθ2cosθ)dθ=4sin2θ2cosθdθ

=4sin2θ2(2cos2θ21)dθ

1554447146338657

1554447147081166

1554447147826729

1554447148562798

1554447149316945

1554447150058843

1554447150798562

1554447151538561

1554447152324930

1554447153875620

Question:21 Integrate the functions in Exercises 1 to 24.

2+sin2x1+cos2xex

Answer:

Given to evaluate

2+sin2x1+cos2xex

2+sin2x1+cos2xex

1554447156109418

1554447156855423

1554447157709943

1554447158484136

1554447159245239

now the integral becomes

1554447160001344

Let tan x = f(x)

f(x)=sec2xdx

1554447160771541

1554447161556589

1554447162309720

Question:22 Integrate the functions in Exercises 1 to 24.

x2+x+1(x+1)2(x+2)

Answer:

Given,

x2+x+1(x+1)2(x+2)

using partial fraction we can simplify the integral as

Let x2+x+1(x+1)2(x+2)=Ax+1+B(x+1)2+Cx+2

x2+x+1(x+1)2(x+2)=A(x+1)(x+2)+B(x+2)+C(x+1)2(x+1)2(x+2)x2+x+1(x+1)2(x+2)=A(x2+3x+2)+B(x+2)+C(x2+2x+1)(x+1)2(x+2)

x2+x+1=A(x2+3x+2)+B(x+2)+C(x2+2x+1)=(A+C)x2+(3A+B+2C)x+(2A+2B+C)

Equating the coefficients of x, x 2 and constant value, we get:

A + C = 1

3A + B + 2C = 1

2A+2B+C =1

Solving these:

A= -2, B=1 and C=3

x2+x+1(x+1)2(x+2)=2x+1+1(x+1)2+3x+2

x2+x+1(x+1)2(x+2)=2x+1dx+1(x+1)2dx+3x+2dx=2log(x+1)1(x+1)+3log(x+2)+C

Question:23 Integrate the functions in Exercises 1 to 24.

tan11x1+x

Answer:

We have

I = tan11x1+x

Let us assume that : x = cos2Θ

Differentiating wrt x,

dx = 2sin2Θ dΘ

Substituting this in the original equation, we get

tan11x1+x = tan11cos2Θ1+cos2Θ×2sin2Θ dΘ

or = 2tan1(sinΘcosΘ)×sin2Θ dΘ

or = 2Θsin2Θ dΘ

Using integration by parts , we get

= 2(Θsin2Θ dΘ dΘdΘsin2Θ dΘ )

or = 2(Θ(cos2Θ2)1.cos2Θ2 dΘ )

or = 2(Θcos2Θ2+sin2Θ4)

Putting all the assumed values back in the expression,

= 2(12(12cos1x)+1x24)

or = 12(xcos1x  1x2) + C

Question:24 Integrate the functions in Exercises 1 to 24.

x2+1[log(x2+1)2logx]x4

Answer:

x2+1[log(x2+1)2logx]x4

Here let's first reduce the log function.

=x2+1x4[log(x2+1)logx2]dx

=x2(1+1x2)x4[log(x2+1)x2]dx

=(1+1x2)x3[log(1+1x2)]dx

Now, let

t=1+1x2

dt=2x3dx

So our function in terms if new variable t is :

I=12[logt]t12dt

now let's solve this By using integration by parts

I=12[(logt)t32321tt3232dt]

I=13t32logt+13t12dt

I=13t32logt+13t3232

I=29t3213t32logt+c

I=13t32[23logt]+c

I=13(1+1x2)32[23log(1+1x2)]+c

Question:25 Evaluate the definite integrals in Exercises 25 to 33.

π2πex(1sinx1cosx)dx

Answer:

π2πex(1sinx1cosx)dx

Since, we have ex multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

ex(f(x)+f(x))dx=exf(x)

So,

π2πex(1sinx1cosx)dx

=π2πex(12sinx2cosx22sin2x2)dx

=π2πex(12sin2x22sinx2cosx22sin2x2)dx

=π2πex(12cosec2x2cotx2)dx

=π2πex(cotx2+12cosec2x2)dx

Here let's use the property

ex(f(x)+f(x))dx=exf(x)

so,

=π2πex(cotx2+12cosec2x2)dx

=[excotx2]π2π

=[eπcotπ2][eπ2cotπ4]

=eπ2

Question:26 Evaluate the definite integrals in Exercises 25 to 33.

0π4sinxcosxcos4x+sin4x

Answer:

0π4sinxcosxcos4x+sin4x

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let' divide both numerator and denominator by cos4x

=0π4sinxcosxcosxcosxsos2x1+sin4xcos4x

=0π4tanxsec2x1+tan4x

Now lets change the variable

t=tan2xdt=2tanxsec2xdx

the limits will also change since the variable is changing

whenx=0,t=tan20=0

whenx=π4,t=tan2π4=1

So, the integration becomes:

I=1201dt1+t2

I=12[tan1t]01

I=12[tan11]12[tan10]

I=12[π4]0

I=π8

Question:27 Evaluate the definite integrals in Exercises 25 to 33.

0π2cos2xdxcos2x+4sin2x

Answer:

Lets first simplify the function.

0π2cos2xdxcos2x+4sin2x=0π2cos2xdxcos2x+4(1cos2x)=0π2cos2xdx43cos2x

130π243cos2x443cos2xdx=130π243cos2x43cos2xdx130π2443cos2xdx

=130π21dx130π2443cos2xdx=13[x]0π2130π2443cos2xdx

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

=13[π20]130π24sec2x4sec2x3dx

AS we can write square of sec in term of tan,


=13[π20]130π24sec2x4(1+tan2x)3dx

=13[π20]130π24sec2x1+4tan2xdx

Now let's calculate the integral of the second function, (we already have calculated the first function)

=130π24sec2x1+4tan2xdx

let

t=2tanx,dt=2sec2xdx

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

when x=π/2,t=2tanπ/2=

our function in terms of t is

=23011+t2dt

=[tan1t]0=[tan1tan10]

=π2

Hence our total solution of the function is

=π6+23π2=π6

Question:28 Evaluate the definite integrals in Exercises 25 to 33.

π6π3sinx+cosxsin2x

Answer:

π6π3sinx+cosxsin2x

Here first let convert sin2x as the angle of x ( sinx, and cosx)

=π6π3sinx+cosx2sinxcosx

Now let's remove the square root form function by making a perfect square inside the square root

=π6π3sinx+cosx(1+12sinxcosx)

=π6π3sinx+cosx(1(sin2x+cos2x2sinxcosx)

=π6π3sinx+cosx(1(sinxcosx)2

Now let

, t=sinxcosxdt=(cosx+sinx)dx

since we are changing the variable, limit of integration will change

whenx=π/6,t=sinπ/6cosπ/6=(13)/2whenx=π/3,t=sinπ/3cos/pi/3=(31)/2

our function in terms of t :

=1323121(1t2)dt

=[sin1t]132312=2sin1(312)

Question:29 Evaluate the definite integrals in Exercises 25 to 33.

01dx1+xx

Answer:

01dx1+xx

First, let's get rid of the square roots from the denominator,

=01dx1+xx1+x+x1+x+x

=011+x+x1+xxdx

=01(1+x+x)dx

$\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx$

=01(1+x)12dx+01x12dx

=[23(1+x)32]01+[23(x)32]01

=[23(1+1)32][23]+[23(1)32][0]

=423

Question:30 Evaluate the definite integrals in Exercises 25 to 33.

0π4sinx+cosx9+16sin2xdx

Answer:

0π4sinx+cosx9+16sin2xdx

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when x=π/4 t=cosπ/4sinπ/4=0

Now,

(cosxsinx)2=t2

cos2x+sin2x2cosxsinx=t2

1sin2x=t2

sin2x=1t2

Hence our function in terms of t becomes,

10dt9+16(1t2)=10dt9+1616t2=10dt2516t2=10dt52(4t)2)

=14[12(5)log5+4t54t]10

=140[log(1)log(19)]

=log940

Question:31 Evaluate the definite integrals in Exercises 25 to 33.

0π2sin2xtan1(sinx)dx

Answer:

Let I =

0π2sin2xtan1(sinx)dx

=0π22sinxcosxtan1(sinx)dx

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

t=sinxdt=cosxdx

Now the important step here is to change the limit of the integration as we are changing the variable.so,

whenx=0,t=sin0=0whenx=π2,t=sinπ2=1

So our function becomes,

I=201(tan1t)tdt

Now, let's integrate this by using integration by parts method,

I=2[tan1tt2211+t2t22dt]01

I=2[tan1tt2212t21+t2dt]01

I=2[tan1tt2212(1+t2)11+t2dt]01

I=2[tan1tt2212(111+t2)dt]01

I=2[tan1tt2212(ttan1t)]01

I=2[tan1tt2212(t)+12tan1t)]01

I=2[12(tan1t(t2+1)t)]01

I=[(tan1t(t2+1)t)]01

I=[(tan1(1)(12+1)1)][(tan1(0)(02+1)0)] I=2tan111=2×π41

I=π21

Question:32 Evaluate the definite integrals in Exercises 25 to 33.

0πxtanxsecx+tanxdx

Answer:

Let I = 0πxtanxsecx+tanxdx -(i)

Replacing x with ( π -x),

I=π0(πx)tan(πx)sec(πx)+tan(πx)(dx)=π0(πx)()tanxsecxtanxdx

I=0π(πx)tanxsecx+tanxdx - (ii)

Adding (i) and (ii)

I+I=0π(xtanxsecx+tanx+(πx)tanxsecx+tanx)dx

2I=0ππtanxsecx+tanxdx

2I=0ππsinxcosx1cosx+sinxcosxdx2I=π0πsinx1+sinxdx2I=π0π(1+sinx)11+sinxdx2I=π0π[111+sinx]dx

2I=π0π[111+sinx]dx2I=π0π1dxπ0π11+sinx.(1sinx)(1sinx)dx2I=π0π1dxπ0π[sec2xsecxtanx]dx2I=π[x]0ππ[secxtanx]0π

2I=π[π0]π[tanπsecπtanπ+sec0]2I=π[π2]I=π2[π2]

Question:33 Evaluate the definite integrals in Exercises 25 to 33.

14[|x1|+|x2|+|x3|]dx

Answer:

Given integral 14[|x1|+|x2|+|x3|]dx

So, we split it in according to intervals they are positive or negative.

=14|x1|dx+14|x2|dx+14|x3|dx

=I1+I2+I3

Now,

I1=14|x1|dx=14(x1)dx

as (x1) is positive in the given x -range [1,4]

=[x22x]14=[4224][1221]

=[84][12]=4+12=92

Therefore, I1=92

I2=14|x2|dx=12(2x)dx+24(x2)dx

as (x2)0 is in the given x -range [2,4] and 0 in the range [1,2]

=[2xx22]12+[x222x]24

={[2(2)222][2(1)122]}+{[4222(4)][2222(2)]}

=[422+12]+[882+4]

=12+2=52

Therefore, I2=52

I3=14|x3|dx=13(3x)dx+34(x3)dx

as (x3)0 is in the given x -range [3,4] and 0 in the range [1,3]

=[3xx22]13+[x223x]34

={[3(3)322][3(1)122]}+{[4223(4)][3223(3)]}

=[9923+12]+[81292+9]

=[64]+12=52

Therefore, I3=52

So, We have the sum =I1+I2+I3

I=92+52+52=192

Question:34 Prove the following (Exercises 34 to 39)

. 13dxx2(x+1)=23+log23

Answer:

L.H.S = 13dxx2(x+1)

We can write the numerator as [(x+1) -x]

13dxx2(x+1)=13(x+1)xx2(x+1)dx

=13[1x21x(x+1)]dx=131x2dx13(x+1)xx(x+1)dx

=131x2dx13[1x1(x+1)]dx=131x2dx131xdx+131(x+1)dx=[1x]13[logx]13+[log(x+1)]13

=[13+1][log3log1]+[log4log2]=23+log(43.2)

=log(23)+23 = RHS

Hence proved.

Question:35 Prove the following (Exercises 34 to 39)

01xexdx=1

Answer:

Let I=xexdx

Integrating I by parts

I=xexdx((d(x)dx)exdx)dxI=xexexdxI=xexex+c

Applying Limits from 0 to 1

01xexdx=[xexex+c]01I=[ee+c][01+c]I=1

Hence proved I = 1

Question:36 Prove the following (Exercises 34 to 39)

11x17cos4xdx=0

Answer:

Let x17cos4x=g(x)

g(x)=(x)17cos4(x)=x17cos4x=g(x)

The Integrand g(x) therefore is an odd function and therefore

11g(x)dx=0

Question:37 Prove the following (Exercises 34 to 39)

0π2sin3xdx=23

Answer:

Let I=0π2sin3xdxI=0π2sinx(1sin2x)dxI=0π2sinxdx0π2cos2xsinxdxI=I1I2

I1=[cosx]0π2I1=0(1)=1

For I 2 let cosx=t, -sinxdx=dt

The limits change to 0 and 1

I2=10t2dtI2=[t33]10I2=0(13)I2=13

I 1 -I 2 =2/3

Hence proved.

Question:€‹38 Prove the following (Exercises 34 to 39)

0π42tan3xdx=1log2

Answer:

The integral is written as

Let I=2tan3xdxI=2tan2xtanxdxI=2(sec2x1)tanxdxI=2tanxsec2xdx2tanxdxI=2tdt2log(cosx)+c       (t=tanx)I=t22log(cosx)+cI=tan2x2log(cosx)+c

[I]0π4=[tan2x2log(cosx)]0π4

[I]0π4=(12log2)(02log1)

[I]0π4=1log2

Hence Proved

Question:39 Prove the following (Exercises 34 to 39)\

01sin1xdx=π21

Answer:

Let I=sinxdx

Integrating by parts we get

I=sinx1dx(d(sinx)dx1dx)I=xsinx+c11x2xdxI=I1I2

For I 2 take 1-x 2 = t 2 , -xdx=tdt

I2=11x2xdxI2=1ttdtI2=t+cI2=1x2+c

[I]01=[I1I2]01

=[xsinx(1x2)]01=[xsinx+1x2]01=[1π2+0][0+1]=π21

Hence Proved

Question:40 Evaluate 01e23xdx as a limit of a sum.

Answer:

As we know

abf(x)dx=(ba)limn1n[f(a)+f(a+h)+f(a+2h)........+f(a+(n1)h)]

where b-a=hn

In the given problem b=1, a=0 and f(x)=e23x
01e23xdx=(10)limn1n(e2+e23h+e23(2h).....+e23(n1)h)=e2limn1n(1+e3h+e6h....+e3(n1)h)=e2limn1n(1(e3h)n1e3h)=e2limn1n(1e3n×n1e3n)

=e2limn1n(1e31e3n)=e2(1e3)3limn3ne3n1=e2(1e3)3

=e2e13

Question:41 Choose the correct answers in Exercises 41 to 44.

. dxex+ex is equal to

(A) tan1(ex)+c

(B) tan1(ex)+c

(C) log(exex)+C

(D) log(ex+ex)+C

Answer:

dxex+ex

the above integral can be re arranged as

=exe2x+1dx

let e x =t, e x dx=dt

dxex+ex

=1t2+1dt=tan1t+c=tan1(ex)+c

(A) is correct

Question:42​​​​​​​ Choose the correct answers in Exercises 41 to 44.

. cos2x(sinx+cosx)2dx is equal to

(A) 1sinx+cosx+C

(B) log|sinx+cosx|+C

(C) log|sinxcosx|+C

(D) 1(sinx+cosx)2+C

Answer:

cos2x(sinx+cosx)2=cos2xsin2x(sinx+cosx)2=(sinx+cosx)(cosxsinx)(sinx+cosx)2=(cosxsinx)(sinx+cosx) cos2x=cos 2 x-sin 2 x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

cos2x(sinx+cosx)2dx=dtt=log|t|+c=log|cosx+sinx|+c

B is correct

Question:43​​​​​​​ Choose the correct answers in Exercises 41 to 44.

If f(a+bx)=f(x) , then abxf(x)dx is equal to

(A) a+b2abf(bx)dx

(B) a+b2abf(b+x)dx

(C) ba2abf(x)dx

(D) a+b2abf(x)dx

Answer:

Let abxf(x)dx=I

As we know abf(x)dx=abf(a+bx)dx

Using the above property we can write the integral as

I=ab(a+bx)f(a+bx)dxI=ab(a+bx)f(x)dxI=(a+b)abf(x)dxabxf(x)dxI=(a+b)abf(x)dxI2I=(a+b)abf(x)dxI=a+b2abf(x)dx

Answer (D) is correct

Question:€‹44 Choose the correct answers in Exercises 41 to 44.

The value of 01tan1(2x11+xx2)dx is

(A) 1

(B) 0

(C) -1

(D) π4

Answer:

Let I=01tan1(2x11+xx2)dx

tan1(2x11+xx2)=tan1(x(1x)1+x(1x))=tan1xtan1(1x) as tan1(ab1+ab)=tan1atan1b

Now the integral can be written as

I=01(tan1xtan1(1x))dxI=01(tan1(1x)tan1(1(1x)))dxI=01(tan1(1x)tan1x)dxI=I2I=0I=0

(B) is correct.

If you are looking for integrals class 12 NCERT solutions of exercises then they are listed below.

About NCERT solutions for class 12 maths chapter 7 integrals

The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration.

Integrals has 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own. If you are not able to do, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals. In this chapter, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercise. Here, the NCERT solutions for class 12 maths chapter 7 integrals are solved and explained in detail to develop a grip on the topic. Here, you will learn two types of integrals: Definite integral and Indefinite integral and also learn their properties and formulas.


Definite Integral

Indefinite Integral

Definition
A definite Integral has upper and lower limits if 'a' and 'b' are the limits or boundaries. The definite integral of f(x) is a number, not function .

An integral without upper limit and lower limit. It is also an antiderivative. The indefinite integral of f(x) is a function not number.

Expression
abf(x)dx

F(x)=f(x)dx

NCERT Solutions for Class 12 Maths Chapter 7 Integrals - Topics

7.1 Introduction

7.2 Integration as an Inverse Process of Differentiation

7.2.1 Geometrical interpretation of indefinite integral

7.2.2 Some properties of indefinite integral

7.2.3 Comparison between differentiation and integration

7.3 Methods of Integration

7.3.1 Integration by substitution

7.3.2 Integration using trigonometric identities

7.4 Integrals of Some Particular Functions

7.5 Integration by Partial Fractions

7.6 Integration by Parts

7.7 Definite Integral

7.7.1 Definite integral as the limit of a sum

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

7.8.2 First fundamental theorem of integral calculus

7.8.3 Second fundamental theorem of integral calculus

7.9 Evaluation of Definite Integrals by Substitution

7.10 Some Properties of Definite Integral

NCERT solutions for class 12 maths - Chapter wise

Key Features of NCERT Solutions for Class 12 Maths Chapter 7 Integrals

The NCERT chapter 7 class 12 maths offers several key features to aid students in their understanding and mastery of this topic. Some of these features include

  1. Detailed explanation: The chapter 7 class 12th maths solutions provide a comprehensive and in-depth explanation of the concepts of integrals, making it easier for students to grasp the subject.

  2. Step-by-Step Solutions: The integration class 12 ncert solutions break down complex problems into simpler steps, making it easier for students to follow and understand the process.

  3. Clear Diagrams: The integrals class 12 solutions make use of clear and informative diagrams to help students visualize the concepts, making it easier for them to comprehend the material.

  4. Plenty of Practice Problems: The ch 7 maths class 12 include a large number of practice problems to help students strengthen their knowledge and skills.

  5. Accurate answers: The class 12 ch 7 maths ncert solutions are checked and verified by experts to ensure accuracy, helping students avoid mistakes and improve their grades.

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Also read,

More about NCERT Solutions for Class 12 Maths Chapter 7 Integrals

The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration.

  • Integrals have 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own.

  • If you are not able to, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals.

  • In this NCERT Class 12 Maths solutions chapter 7, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercises. Here, the Class 12 Maths Chapter 7 NCERT solutions are solved and explained in detail to develop a grip on the topic.

NCERT solutions for class 12 - subject wise

How to use NCERT solutions for class 12 maths chapter 7 Integrals

  • You are expected to remember all formulas of differentiation, and then can start with basic integration

  • Try to relate differentiation formulas with integrations formulas because it will help you to remember all the integration formulas

  • When you are done with basic integration, you should learn different methods of integration for different types of functions. If you find difficulties in learning the method you should learn one method at a time to solve a particular type of problem

  • After learning the different methods of integration, you should be able to determine which method will be used to solve a particular type of problem

  • When you become good with indefinite integrals, move to the definite integrals and learn some properties to solve definite integrals. NCERT Class 12 Maths solutions chapter 7 integrals will help you for the same

  • This chapter requires a lot of practice. First, solve all the NCERT textbook questions, then, you can take the help of NCERT solutions for class 12 maths chapter 7 integrals.

  • If you have solved all NCERT questions, you can solve CBSE previous year papers also to get familiar with the type of questions which are asked in previous years.

  • NCERT solutions class 12 maths chapter 12 pdf download will be made available soon. Till then you can save the webpage and practice these solutions offline.

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Students consider Integration which is integrals and applications of integration are the most difficult chapter in CBSE class 12 maths but with the regular practice of NCERT problems you will be able to have a strong grip on this chapter also. it's true that there is no substitute for hard work but the right strategy and quality study material are also essential to get command of this chapter, therefore, NCERT exercises are recommended for practice.  

2. How are the NCERT solutions helpful in the board exam?

NCERT solutions are created by the expert team of careers360 who know how best to write answers in the board exam in order to get good marks. Integrating their techniques can benefit in obtaining meritorious marks. Sometimes students do not understand where s/he are making a mistake, NCERT solutions can help them to understand that. the practice of a lot of questions and their solution make you confident and help you in getting an in-depth understanding of concepts. Therefore NCERT solutions are very helpful for students.

3. What is the weightage of the chapter Integrals for CBSE board exam?

Integrals have 13 % weightage in CBSE class 12th board final examination.  having 13% weightage, Integral become very students for CBSE aspirant but it demand lot of practice. Interested students can refer to integrals class 12 solutions.

4. Is the NCERT Solutions for class 12 maths integration challenging to comprehend?

The NCERT Solutions for maths ch 7 maths class 12 is not complicated to understand. It is a fascinating topic in Class 12 that is also relevant at higher education levels. A solid understanding of integral formulas will enable students to solve integration problems effectively. A thorough knowledge of derivatives is crucial for comprehending the concepts of integral calculus with ease. For ease, students can study integrals class 12 ncert solutions pdf both online and offline.

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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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I hope this information helps you.







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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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