Careers360 Logo
NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

Edited By Team Careers360 | Updated on Dec 04, 2023 01:21 PM IST

NCERT Solutions For Class 12 Chapter 7 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 7 class 12 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for class 12 maths chapter 7 miscellaneous exercise provides questions based on integration of square root functions, trigonometric functions etc. Class 12 maths chapter 7 miscellaneous exercise is last but not the least as you can find some of the questions in previous years from this exercise only. class 12 maths chapter 7 miscellaneous exercise can be found difficult for some students but after giving sufficient time, you will be able to deal with the questions. NCERT solutions for class 12 maths chapter 7 with all the other exercises can be found in NCERT Class 12th book.

Miscellaneous exercise class 12 chapter 7 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise

Question:1 Integrate the functions in Exercises 1 to 24.

\frac{1}{x - x^3}

Answer:

Firstly we will simplify the given equation :-

\frac{1}{x - x^3}\ =\ \frac{1}{(x)(1-x)(1+x)}

Let

\frac{1}{(x)(1-x)(1+x)} =\ \frac{A}{x}\ +\ \frac{B}{1-x}\ +\ \frac{C}{1+x}

By solving the equation and equating the coefficients of x 2 , x and the constant term, we get

A\ =\ 1,\ B\ =\ \frac{1}{2},\ C\ =\ \frac{-1}{2}

Thus the integral can be written as :

\int \frac{1}{(x)(1-x)(1+x)}dx =\ \int \frac{1}{x}dx\ +\ \frac{1}{2}\int \frac{1}{1-x}dx\ +\ \frac{-1}{2}\int \frac{1}{1+x}dx

=\ \log x\ -\ \frac{1}{2}\log(1-x)\ +\ \frac{-1}{2}\log (1+x)

or =\ \frac{1}{2} \log \frac{x^2}{1-x^2}\ +\ C


Question:2 Integrate the functions in Exercises 1 to 24.

\frac{1}{\sqrt{x+a} + \sqrt{x+b}}

Answer:

At first we will simplify the given expression,

\frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\times\frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}}

or =\ \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b}

Now taking its integral we get,

\int \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{a-b}\int (\sqrt{x+a} -\sqrt{x+b})dx

or =\ \frac{1}{a-b}\left [ \frac{(x+a)^{\frac{3}{2}}} {\frac{3}{2}}\ -\ \frac{(x+b)^{\frac{3}{2}}} {\frac{3}{2}} \right ]

or =\ \frac{2}{3(a-b)}\left [ (x+a)^{\frac{3}{2}}\ -\ (x+b)^{\frac{3}{2}} \right ]\ +\ C


Question:3 Integrate the functions in Exercises 1 to 24.

\frac{1}{x\sqrt{ax-x^2}} [Hint: Put x = \frac{a}{t} ]

Answer:

Let

x = \frac{a}{t}\ dx\ \Rightarrow \ dx\ =\ \frac{-a}{t^2}dh

Using the above substitution we can write the integral is

\int \frac{1}{x\sqrt{ax-x^2}}\ =\ \int \frac{1}{\frac{a}{t}\sqrt{a.\frac{a}{t}\ -\ (\frac{a}{t})^2}} \frac{-a}{t^2}dt

or

=\ \frac{-1}{a}\int \frac{1}{\sqrt{(t-1)}}dt

or

=\ \frac{-1}{a}\ (2\sqrt{t-1})\ +\ C

or =\ \frac{-1}{a}\ (2\sqrt{\frac{a}{x}\ -\ 1})\ +\ C

or =\ \frac{-2}{a}\ \sqrt{\frac{a-x}{x}}\ +\ C

Question:4 Integrate the functions in Exercises 1 to 24.

. \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}

Answer:

For the simplifying the expression, we will multiply and dividing it by x -3 .

We then have,

\frac{x^{-3}}{x^2 x^{-3}(x^4 + 1)^\frac{3}{4}}\ =\ \frac{1}{x^5}\left [ \frac{x^4\ +\ 1}{x^4} \right ]^{\frac{-3}{4}}

Now, let

\frac{1}{x^4}\ =\ t\ \Rightarrow \ \frac{1}{x^5}dx\ =\ \frac{-dt}{4}

Thus,

\int \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}\ =\ \int \frac{1}{x^5}\left ( 1+\ \frac{1}{x^4}^{\frac{-3}{4}}\ \right )dx

or =\ \frac{-1}{4} \int (1+t)^{\frac{-3}{4}}dt

=\ \frac{-1}{4} \frac{(1+\frac{1}{x^4})^{\frac{1}{4}}}{\frac{1}{4}}\ +\ C

=\ - \left [ 1+\frac{1}{x^4} \right ]^{\frac{1}{4}}\ +\ C

Question:5 Integrate the functions in Exercises 1 to 24.

\frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}} [Hint: \frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}} = \frac{1}{x^\frac{1}{3}(1 + \ x^\frac{1}{6})} , put x = t^6 ]

Answer:

Put x = t^6\ \Rightarrow \ dx = 6t^5dt

We get,

\int \frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}}dx\ =\ \int \frac{6t^5}{t^3+t^2}dt

or =\ 6\int \frac{t^3}{1+t}dt

or =\ 6\int \left \{ (t^2-t+1)-\frac{1}{1+t} \right \}dt

or =\ 6 \left [ \left ( \frac{t^3}{3} \right ) -\left ( \frac{t^2}{2} \right )+t - \log(1+t) \right ]

Now put x = t^6 in the above result :

=\ 2\sqrt{x} -3x^{\frac{1}{3}}+ 6x^{\frac{1}{6}} - 6 \log \left ( 1-x^\frac{1}{6} \right )\ +\ C


Question:6 Integrate the functions in Exercises 1 to 24.

\frac{5x}{(x+1)(x^2 + 9)}

Answer:

Let us assume that :

\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{A}{(x+1)}\ +\ \frac{Bx + c}{x^2 + 9}

Solving the equation and comparing coefficients of x 2 , x and the constant term.

We get,

A\ =\ \frac{-1}{2}\ ;\ B\ =\ \frac{1}{2}\ ;\ C\ =\ \frac{9}{2}

Thus the equation becomes :

\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{-1}{2(x+1)}\ +\ \frac{\frac{x}{2}+\frac{9}{2}}{x^2 + 9}

or

\int \frac{5x}{(x+1)(x^2 + 9)}\ =\ \int \left [ \frac{-1}{2(x+1)}\ +\ \frac{x+9}{2(x^2 + 9}) \right ]dx

or =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{2} \int \frac{x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx

or =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \int \frac{2x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx

or =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \log {(x^2 +9)} +\frac{3}{2} \tan^{-1}\frac{x}{3}\ +\ C


Question:7 Integrate the functions in Exercises 1 to 24.

\frac{\sin x}{\sin (x-a)}

Answer:

We have,

I\ =\ \frac{\sin x}{\sin (x-a)}

Assume :- (x-a)\ =\ t \Rightarrow \ dx=dt

Putting this in above integral :

\int \frac{\sin x}{\sin (x-a)}dx\ =\ \int \frac{\sin (t+a)}{\sin t}dt

or =\ \int \frac{\sin t \cos a\ +\ \cos t \sin a }{\sin t}dt

or =\ \int (\cos a\ +\ \cot t \sin a)dt

or =\ t\cos a\ +\ \sin a \log |\sin t|\ +\ C

or =\ \sin a \log \left | \sin(x-a) \right | + x\cos a\ +\ C


Question: Integrate the functions in Exercises 1 to 24.

\frac{\cos x}{\sqrt{4 - \sin^2 x}}

Answer:

We have the given integral

I\ =\ \frac{\cos x}{\sqrt{4 - \sin^2 x}}

Assume \sin x = t\ \Rightarrow \cos x dx = dt

So, this substitution gives,

\int \frac{\cos x}{\sqrt{4 - \sin^2 x}}\ =\ \int \frac{dt}{\sqrt{(2)^2 - (t)^2}}

=\ \sin^{-1}\frac{t}{2}\ +\ C

or =\ \sin^{-1}\left ( \frac{\sin x}{2} \right )\ +\ C


Question:10 Integrate the functions in Exercises 1 to 24.

\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}

Answer:

We have

I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}

Simplifying the given expression, we get :

\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ \frac{(\sin^4x + \cos^4x)(\sin^4x - \cos^4x) }{1- 2\sin^ x\cos^2 x}

or =\ \frac{(\sin^4x + \cos^4x)(\sin^2x - \cos^2x)(\sin^2x + \cos^2x) }{1- 2\sin^ x\cos^2 x}

or =\ -\frac{(\sin^4x + \cos^4x)(\cos^2x - \sin^2x) }{1- 2\sin^ x\cos^2 x}

or =\ -\cos^2x - \sin^2x\ =\ -\cos 2x

Thus,

I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ -\int \cos 2x\ dx

and =\ - \frac{\sin 2x}{2}\ +\ C

Question:11 Integrate the functions in Exercises 1 to 24.

\frac{1}{\cos(x+a)\cos(x+b)}


Answer:

For simplifying the given equation, we need to multiply and divide the expression by \sin (a-b) .

Thus we obtain :

\frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin(a-b)}\times\frac{\sin (a-b)}{\cos(x+a)\cos(x+b)}

or = \frac{1}{ \sin (a-b)}\times \frac{\sin{\left [ (x+a) - (x+b) \right ]}}{\cos (x+a) \cos (x+b)}

or = \frac{1}{ \sin (a-b)}\times \left ( \frac{\sin (x+a) }{\cos (x+a) } - \frac{\sin(x+b)}{\cos (x+b)} \right )

or = \frac{1}{ \sin (a-b)}\times \left ( \tan(x+a)\ -\ \tan(x+b) \right )

Thus integral becomes :

\int \frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin (a-b)} \times \int \left ( \tan(x+a)\ -\ \tan(x+b) \right )dx

or =\ \frac{1}{\sin (a-b)} \times \left [ -\log \left | \cos (x+a) \right | + \log \left | \cos(x+b) \right | \right ]\ +\ C

or =\ \frac{1}{\sin (a-b)} \times \log \left [ \frac{\cos(x+b) }{cos(x+a)} \right ]\ +\ C


Question:12 Integrate the functions in Exercises 1 to 24.

\frac{x^3}{\sqrt{1-x^8}}


Answer:

Given that to integrate

\frac{x^3}{\sqrt{1-x^8}}

Let x^4 = t \implies 4x^3dx = dt

\therefore \int \frac{x^3}{\sqrt{1-x^8}}dx = \frac{1}{4}\int\frac{1}{\sqrt {1-t^2}}dt

= \frac{1}{4}sin^{-1}t + C= \frac{1}{4}sin^{-1}{x^4} + C

the required solution is \frac{1}{4}sin^{-1}{(x^4)} + C


Question:13 Integrate the functions in Exercises 1 to 24.

\frac{e^x}{(1 + e^x)(2 + e^x)}

Answer:

we have to integrate the following function

\frac{e^x}{(1 + e^x)(2 + e^x)}

Let 1+e^x = t \implies e^xdx = dt

using this we can write the integral as

\therefore \int\frac{e^x}{(1 + e^x)(2 + e^x)}dx = \int\frac{1}{t(1+t)}dt = \int\frac{(1+t)-t}{t(1+t)}dt

\\ = \int\left ( \frac{1}{t}-\frac{1}{t+1} \right )dt

\\ = \int\frac{1}{t}dt - \int\frac{1}{t+1}dt

\\ = \log t - \log (1+t) + C \\ = \log (1+e^x) - \log (2+e^x) + C \\ = \log\left ( \frac{e^x + 1}{e^x + 2} \right ) + C

Question:14 Integrate the functions in Exercises 1 to 24.

\frac{1}{(x^2 + 1)(x^2 +4)}


Answer:

Given,

\frac{1}{(x^2 + 1)(x^2 +4)}

Let I = \int\frac{1}{(x^2 + 1)(x^2 +4)}

Now, Using partial differentiation,

\frac{1}{(x^2 + 1)(x^2 +4)} = \frac{Ax + B}{(x^2 + 1)} + \frac{Cx +D}{(x^2 +4)}

\implies \frac{1}{(x^2 + 1)(x^2 +4)} = \frac{(Ax + B)(x^2 +4) + (Cx +D)(x^2 + 1)}{(x^2 + 1)(x^2 +4)}
\\ \implies1 = (Ax + B)(x^2 + 4)+(Cx + D)(x^2 + 1) \\ \implies 1 = Ax^3 +4Ax+ Bx^2 + 4B+ Cx^3 + Cx + Dx^2 + D \\ \implies (A+C)x^3 +(B+D)x^2 +(4A+C)x + (4B+D) = 1

Equating the coefficients of x, x^2, x^3 and constant value,

A + C = 0 \implies C = -A

B + D = 0 \implies B = -D

4A + C =0 \implies 4A = -C \implies 4A = A \implies A = 0 = C

4B + D = 1 \implies 4B – B = 1 \implies B = 1/3 = -D

Putting these values in equation, we have

155444707519314

155444707593450

1554447076674147

1554447078146237

\implies I = \frac{1}{3}tan^{-1}x - \frac{1}{6}tan^{-1}\frac{x}{2} + C


Question:15 Integrate the functions in Exercises 1 to 24.

\cos^3 x \;e^{\log\sin x}

Answer:

Given,

\cos^3 x \;e^{\log\sin x}

I = \int \cos^3 x \;e^{\log\sin x} (let)

Let cos x = t \implies -sin x dx = dt \implies sin x dx = -dt

using the above substitution the integral is written as

\therefore \int cos^3xe^{\log sinx}dx = \int cos^3x.sinx dx

155444708344321

1554447084197794

1554447084982414

155444708573581

I = -\frac{cos^4x}{4} + C


Question:16 Integrate the functions in Exercises 1 to 24.

e^{3\log x} (x^4 + 1)^{-1}

Answer:

Given the function to be integrated as

e^{3\log x} (x^4 + 1)^{-1}
= e^{\log x^3}(x^4 + 1)^{-1} = \frac{x^3}{x^4 + 1}

Let I = \int e^{3\log x} (x^4 + 1)^{-1}

Let x^4 = t \implies 4x^3 dx = dt

I = \int e^{3\log x} (x^4 + 1)^{-1} = \int \frac{x^3}{x^4 + 1}

1554447091784564

1554447092538842

\implies I = \frac{1}{4}\log(x^4 +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 = \int 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

\int f'(ax +b)[f(ax +b)]^n = \frac{1}{a}\int t^ndt

\\ = \frac{1}{a}.\frac{t^{n+1}}{n+1} + C \\ = \frac{1}{a}.\frac{(f(ax+b))^{n+1}}{n+1} + C

\implies I = \frac{(f(ax+b))^{n+1}}{a(n+1)} + C


Question:18 Integrate the functions in Exercises 1 to 24.

. \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}

Answer:

Given,

\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}

Let I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}

We know the identity that

sin (A+B) = sin A cos B + cos A sin B


\therefore \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}}

= \frac{1}{\sqrt{\sin^3 x . \sin x(\cos \alpha + \cot x \sin \alpha)}} = \frac{1}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}}

\frac{cosec^2 x}{\sqrt{(\cos \alpha + \cot x \sin \alpha)}}


1554447105407729


1554447106158144


1554447106897484


1554447107634823


1554447108399148


155444710916564


1554447109907513


1554447110653730


1554447111419992


1554447112159711


1554447112971688


Question:19 Integrate the functions in Exercises 1 to 24.

. \frac{\sin^{-1}\sqrt x - \cos^{-1}\sqrt x}{\sin^{-1}\sqrt x + \cos^{-1}\sqrt x}, \;\;\; x\in [0,1]

Answer:

We have

I\ =\ \int \frac{\sin^{-1}\sqrt x - \cos^{-1}\sqrt x}{\sin^{-1}\sqrt x + \cos^{-1}\sqrt x}\ dx

or =\ \int \frac{\sin^{-1}\sqrt x - \left ( \frac{\Pi }{2} - \sin^{-1}\sqrt x \right )}{\frac{\Pi }{2}}\ dx

or =\ \frac{2}{\Pi } \int \left ( \ 2sin^{-1}\sqrt x - \frac{\Pi }{2} \right )\ dx

or =\ \int \left (\frac{4}{\Pi } \sin^{-1}\sqrt x - 1 \right )\ dx

or =\ \frac{4}{\Pi }\int \sin^{-1}\sqrt x - 1 \ dx\ -\ \int 1 \ dx\ +\ C

or =\ \frac{4}{\Pi }\int \sin^{-1}\sqrt x \ dx\ -\ x +\ C

Thus I\ =\ \frac{4}{\Pi }I'\ -\ x +\ C


Now we will solve I'.

I'\ =\ \int \sin^{-1}\sqrt x \ dx

Put x = t 2 .

Differentiating the equation wrt x, we get

dx\ =\ 2t\ dt

Thus \int \sin^{-1}\sqrt x \ dx\ =\ \int \sin^{-1} t\ 2t \ dt

or =\ 2 \int t\ \sin^{-1} t\ \ dt

Using integration by parts, we get :

=\ 2 \left [ \sin^{-1}t \int t\ dt\ -\ \int \left ( \left ( \frac{d}{dt} \sin^{-1} t \right ) \int t\ dt \right ) \right ]\ dt

or =\ t^2 \sin^{-1}t\ -\ \int \frac{t^2}{\sqrt{1-t^2}}\ dt\ +\ C'

We know that

\int \frac{- t^2}{\sqrt{1-t^2}}\ dt\ =\ \frac{t}{2}\sqrt{1-t^2}\ -\ \frac{1}{2}\ \sin^{-1}t

Thus it becomes :

I'\ =\ t^2\sin^{-1} t\ +\ \frac{t}{2}\sqrt{1-t^2}\ -\ \frac{1}{2}\ \sin^{-1}t

So I come to be :-

I\ =\ \frac{4}{\Pi }I'\ -\ x +\ C

I\ =\ \sin^{-1}\sqrt{x} \left [ \frac{2(2x-1)}{\Pi } \right ]\ +\ \frac{2\sqrt{x-x^2}}{\Pi }\ -\ x\ +\ C

Question:20 Integrate the functions in Exercises 1 to 24.

\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}

Answer:

Given,

\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}} = I (let)

Let x= cos^2\theta \implies dx = -2sin\theta cos\theta d\theta

And \sqrt x= cos\theta \implies \theta = \cos^{-1}\sqrt x

using the above substitution we can write the integral as

\\ I = \int \sqrt{\frac{1-\sqrt {cos^2\theta}}{1 +\sqrt {cos^2\theta}}}(-2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{\frac{1-cos\theta}{1 +cos\theta}}(2\sin\theta\cos\theta)d\theta

\\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2. 2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\cos\theta)d\theta \\ = -4\int \sin^2\frac{\theta}{2}\cos\theta d\theta

\\ = -4\int \sin^2\frac{\theta}{2}(2cos^2\frac{\theta}{2} -1) d\theta

1554447146338657

1554447147081166

1554447147826729

1554447148562798

1554447149316945

1554447150058843

1554447150798562

1554447151538561

1554447152324930

1554447153875620


Question:21 Integrate the functions in Exercises 1 to 24.

\frac{2 + \sin 2x}{1 + \cos 2x}e^x

Answer:

Given to evaluate

\frac{2 + \sin 2x}{1 + \cos 2x}e^x

\frac{2 + \sin 2x}{1 + \cos 2x}e^x

1554447156109418

1554447156855423

1554447157709943

1554447158484136

1554447159245239

now the integral becomes

1554447160001344

Let tan x = f(x)

\implies f'(x) = sec^2x dx

1554447160771541

1554447161556589

1554447162309720


Question:22 Integrate the functions in Exercises 1 to 24.

\frac{x^2 + x + 1}{(x+1)^2 (x+2)}

Answer:

Given,

\frac{x^2 + x + 1}{(x+1)^2 (x+2)}

using partial fraction we can simplify the integral as

Let \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}

\\ \implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A(x+1)(x+2) + B(x+2) + C(x+1)^2}{(x+1)^2 (x+2)} \\ \implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A(x^2 + 3x+2) + B(x+2) + C(x^2 + 2x+1)}{(x+1)^2 (x+2)}

\\ \implies x^2 + x + 1 = A(x^2 + 3x+2) + B(x+2) + C(x^2 + 2x+1) \\ = (A+C)x^2 + (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

\implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}

\\ \implies \int \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \int\frac{-2}{x+1}dx+\int\frac{1}{(x+1)^2}dx+\int\frac{3}{x+2}dx \\ = -2\log(x+1) - \frac{1}{(x+1)} + 3\log (x+2) + C


Question:23 Integrate the functions in Exercises 1 to 24.

\tan^{-1}\sqrt{\frac{1-x}{1+x}}

Answer:

We have

I\ =\ \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}

Let us assume that : x\ =\ \cos 2\Theta

Differentiating wrt x,

dx\ =\ -2 \sin 2\Theta\ d\Theta

Substituting this in the original equation, we get

\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\ =\ \int \tan^{-1}\sqrt{\frac{1-cos2\Theta }{1+cos2\Theta }}\times -2\sin 2\Theta \ d\Theta

or =\ -2\int \tan^{-1} (\frac{sin\Theta }{cos\Theta })\times \sin 2\Theta \ d\Theta

or =\ -2\int \Theta \sin 2\Theta \ d\Theta

Using integration by parts , we get

=\ -2\left ( \Theta \int \sin 2\Theta \ d\Theta\ - \int \frac{d\Theta }{d\Theta } \int \sin 2\Theta \ d\Theta\ \right )

or =\ -2\left ( \Theta \left ( \frac{-\cos 2\Theta }{2} \right ) - \int 1.\frac{-\cos 2\Theta }{2} \ d\Theta\ \right )

or =\ -2\left ( \frac{-\Theta \cos 2\Theta }{2}+ \frac{\sin 2\Theta }{4} \right )

Putting all the assumed values back in the expression,

=\ -2\left ( -\frac{1}{2}\left ( \frac{1}{2} \cos^{-1} x \right )+ \frac{\sqrt{1-x^2} }{4} \right )

or =\ \frac{1}{2}\left ( x \cos^{-1} x\ -\ \sqrt{1-x^2} \right )\ +\ C

Question:24 Integrate the functions in Exercises 1 to 24.

\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}


Answer:

\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}

Here let's first reduce the log function.

=\frac{\sqrt{x^2+1}}{x^4}\left [ \log (x^2+1)-\log x^2 \right ]dx

=\frac{\sqrt{x^2\left ( 1+\frac{1}{x^2} \right )}}{x^4}\left [ \log\frac{ (x^2+1)}{x^2} \right ]dx

=\int\frac{\sqrt{\left ( 1+\frac{1}{x^2} \right )}}{x^3}\left [ \log\left ( 1+\frac{1}{x^2} \right ) \right ]dx

Now, let

t=1+\frac{1}{x^2}

dt=\frac{-2}{x^3}dx

So our function in terms if new variable t is :

I=\frac{-1}{2}\int \left [\log t \right ]\cdot t^{\frac{1}{2}}dt

now let's solve this By using integration by parts

I=\frac{-1}{2}\int \left [(\log t)\frac{t^\frac{3}{2}}{\frac{3}{2}} -\int \frac{1}{t}\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt\right ]

I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\int t^{\frac{1}{2}}dt

I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}

I=\frac{2}{9}t^{\frac{3}{2}}-\frac{1}{3}t^{\frac{3}{2}}logt+c

I=\frac{1}{3}t^{\frac{3}{2}}\left [ \frac{2}{3}-\log t \right ]+c

I=\frac{1}{3}\left ( 1+\frac{1}{x^2} \right )^{\frac{3}{2}}\left [ \frac{2}{3}-\log \left ( 1+\frac{1}{x^2} \right ) \right ]+c


Question:25 Evaluate the definite integrals in Exercises 25 to 33.

\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

Answer:

\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

Since, we have e^x multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

\int e^x(f(x)+f'(x))dx=e^xf(x)

So,

\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}} \right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2sin^2\frac{x}{2}} -\frac{2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}}\right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2}cosec^2\frac{x}{2}-cot\frac{x}{2}\right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx

Here let's use the property

\int e^x(f(x)+f'(x))dx=e^xf(x)

so,

=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx

=\left [ -e^xcot\frac{x}{2} \right ]_\frac{\pi}{2}^\pi

=\left [ -e^\pi cot\frac{\pi}{2} \right ]-\left [ -e^{\frac{\pi}{2}} cot\frac{\pi}{4} \right ]

=e^{\frac{\pi}{2}}

Question:26 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

Answer:

\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

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 cos^4x

=\int_0^\frac{\pi}{4}\frac{\frac{\sin x\cos x}{cosxcosxsos^2x} }{1+\frac{\sin^4 x}{\cos^4x}}

=\int_0^\frac{\pi}{4}\frac{tanxsec^2x}{1+tan^4x}

Now lets change the variable

\\t=tan^2x \\dt=2tanxsec^2xdx

the limits will also change since the variable is changing

when\:x=0,t=tan^20=0

when\:x=\frac{\pi}{4},t=tan^2\frac{\pi}{4}=1

So, the integration becomes:

I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^2}

I=\frac{1}{2}\left [ tan^{-1}t \right ]_0^1

I=\frac{1}{2}\left [ tan^{-1}1 \right ]-\frac{1}{2}\left [ tan^{-1}0\right ]

I=\frac{1}{2}\left [ \frac{\pi}{4} \right ]-0

I=\frac{\pi}{8}

Question:27 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}

Answer:

Lets first simplify the function.

\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4(1-\cos^2 x)}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{4-3\cos^2 x}

\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x-4\:\: }{4-3\cos^2 x }dx=\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x\:\: }{4-3\cos^2 x }dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx

\\=\frac{-1}{3}\int_0^\frac{\pi}{2}1dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx \\ \\ \\=\frac{-1}{3} \left [ x \right ]_0^{\frac{\pi}{2}}-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4\:\: }{4-3\cos^2 x }dx


As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4sec^2x-3 }dx

AS we can write square of sec in term of tan,


=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4(1+tan^2x)-3 }dx

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx

Now let's calculate the integral of the second function, (we already have calculated the first function)

=-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx

let

\\t=2tanx, \\dt=2sec^2xdx

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=\pi/2,t=2tan{\pi/2}=\infty

our function in terms of t is


=-\frac{2}{3}\int_0^\infty\frac{1 }{1+t^2 }dt

=\left [ tan^{-1} t\right ]_0^\infty=\left [ tan^{-1} \infty-tan^{-1} 0\right ]

=\frac{\pi}{2}

Hence our total solution of the function is

\\=-\frac{\pi}{6}+\frac{2}{3}*\frac{\pi}{2}\\=\frac{\pi}{6}

Question:28 Evaluate the definite integrals in Exercises 25 to 33.

\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}

Answer:

\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}

Here first let convert sin2x as the angle of x ( sinx, and cosx)

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{2sinxcosx}}

Now let's remove the square root form function by making a perfect square inside the square root

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{-(-1+1-2sinxcosx)}}

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sin^2x+cos^2x-2sinxcosx)}}

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sinx-cosx)^2}}

Now let

, \\t=sinx-cosx \\dt=(cosx+sinx)dx

since we are changing the variable, limit of integration will change

\\when\: x=\pi/6, t=sin\pi/6-cos\pi/6=(1-\sqrt{3})/2 \\ when x= \pi/3,t=sin\pi/3-cos/pi/3=(\sqrt{3}-1)/2

our function in terms of t :

\\=\int_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \frac{1}{\sqrt{(1-t^2)}}dt

\\=\left [ sin^{-1}t \right ]_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \\ \\ \\=2sin^{-1}\left (\frac{\sqrt{3}-1}{2} \right )

Question:29 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}

Answer:

\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}

First, let's get rid of the square roots from the denominator,

\\=\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}*\frac{\sqrt{1+x} +\sqrt x}{\sqrt{1+x} +\sqrt x}

\\=\int_0^1\frac{\sqrt{1+x}+\sqrt{x}}{{1+x} -x}dx

\\=\int_0^1({\sqrt{1+x}+\sqrt{x}})dx

\\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx

\\=\int_0^1(1+x)^\frac{1}{2}dx+\int_0^1x^\frac{1}{2}dx

\\=\left [ \frac{2}{3}(1+x)^{\frac{3}{2}} \right ]_0^1+\left [ \frac{2}{3}(x)^{\frac{3}{2}} \right ]_0^1

\\=\left [ \frac{2}{3}(1+1)^{\frac{3}{2}} \right ]-\left [ \frac{2}{3} \right ]+\left [ \frac{2}{3}(1)^{\frac{3}{2}} \right ]-\left [ 0 \right ]

\\=\frac{4\sqrt{2}}{3}

Question:30 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx

Answer:

\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx

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=\pi/4 t=\cos\pi/4-\sin\pi/4=0

Now,

(\cos x-\sin x)^2=t^2

\cos ^2x+\sin^2 x-2\cos x \sin x =t^2

1-\sin 2x =t^2

\sin 2x =1-t^2

Hence our function in terms of t becomes,

\int_{-1}^{0}\frac{dt}{9+16(1-t^2)}=\int_{-1}^{0}\frac{dt}{9+16-16t^2}=\int_{-1}^{0}\frac{dt}{25-16t^2}=\int_{-1}^{0}\frac{dt}{5^2-(4t)^2)}

= \frac{1}{4}\left [\frac{1}{2(5)}\log \frac{5+4t}{5-4t} \right ]_{-1}^0

= \frac{1}{40}\left[ \log (1)-\log (\frac{1}{9})\right ]

=\frac{\log 9}{40}

Question:31 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx

Answer:

Let I =

\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx

=\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(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=sinx \\dt=cosxdx

Now the important step here is to change the limit of the integration as we are changing the variable.so,

\\when\:x=0,t=sin0=0 \\when\:x=\frac{\pi}{2},t=sin\frac{\pi}{2}=1

So our function becomes,

I=2\int_{0}^{1}(tan^{-1}t)tdt

Now, let's integrate this by using integration by parts method,

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\int\frac{1}{1+t^2}\cdot\frac{t^2}{2}dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{t^2}{1+t^2}\cdot dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{(1+t^2)-1}{1+t^2}\cdot dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\left ( 1-\frac{1}{1+t^2} \right )\cdot dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t-tan^{-1}t) \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t)+\frac{1}{2}tan^{-1}t) \right ]_0^1

I=2\left [ \frac{1}{2} \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1

I=\left [ \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1

I=\left [ \left (tan^{-1}(1)\cdot(1^2+1)-1 \right )\right ]-\left [ \left (tan^{-1}(0)\cdot(0^2+1)-0 \right )\right ] I=2tan^{-1}1-1=2\times \frac{\pi}{4}-1

I=\frac{\pi}{2}-1

Question:32 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx

Answer:

Let I = \int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx -(i)

Replacing x with ( \pi -x),

\\ I = \int_\pi^0\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x) + \tan (\pi -x)} (-dx) \\ = -\int_\pi^0\frac{(\pi -x)(-)\tan x}{-\sec x - \tan x} dx

\\ \implies I = \int^\pi_0\frac{(\pi -x)\tan x}{\sec x + \tan x} dx - (ii)

Adding (i) and (ii)

I + I = \int^\pi_0\left(\frac{x\tan x}{\sec x + \tan x} + \frac{(\pi -x)\tan x}{\sec x + \tan x} \right) dx

\implies 2I = \int^\pi_0\frac{\pi\tan x}{\sec x + \tan x} dx

\\ \implies 2I = \int^\pi_0\frac{\pi \frac{sin x}{cos x} }{\frac{1}{cos x} + \frac{sin x}{cos x}} dx \\ \implies 2I =\pi \int^\pi_0\frac{ sin x }{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\frac{ (1 +sin x ) -1}{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx

\\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx \\ \implies 2I =\pi \int^\pi_01 dx - \pi \int^\pi_0\frac{1}{1+sin x}.\frac{(1-sin x)}{(1 - sin x)}dx \\ \implies 2I =\pi\int^\pi_01 dx - \pi \int^\pi_0[\sec^2 x - \sec x \tan x]dx \\ \implies 2I =\pi[x]^\pi_0 - \pi[\sec x - \tan x]^\pi_0

\\ \implies 2I =\pi[\pi - 0] - \pi[tan \pi - sec \pi- tan \pi + sec 0] \\ \implies 2I =\pi[\pi -2] \\ \implies I =\frac{\pi}{2}[\pi -2]

Question:33 Evaluate the definite integrals in Exercises 25 to 33.

\int_1^4[|x-1| + |x-2| + |x-3|]dx

Answer:

Given integral \int_1^4[|x-1| + |x-2| + |x-3|]dx

So, we split it in according to intervals they are positive or negative.

= \int_{1}^4 |x-1| dx + \int_{1}^4 |x-2| dx + \int^4_{1} |x-3| dx

= I_{1}+I_{2}+I_{3}

Now,

I_{1} = \int^4_{1}|x-1| dx = \int^4_{1} (x-1)dx

\because as (x-1) is positive in the given x -range [1,4]

=\left [ \frac{x^2}{2}-x\right ]^4_{1} = \left [ \frac{4^2}{2}-4 \right ] - \left [ \frac{1^2}{2}-1 \right ]

= \left [ 8-4 \right ] - [-\frac{1}{2}] = 4+\frac{1}{2} = \frac{9}{2}

Therefore, I_{1} = \frac{9}{2}

I_{2} = \int^4_{1}|x-2| dx = \int^2_{1} (2-x)dx +\int^4_{2} (x-2)dx

\because as (x-2)\geq 0 is in the given x -range [2,4] and \leq 0 in the range [1,2]

=\left [ 2x - \frac{x^2}{2}\right ] ^2_{1} + \left [ \frac{x^2}{2} -2x\right ] ^4_{2}

= \left \{ \left [ 2(2)-\frac{2^2}{2} \right ] - \left [ 2(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-2(4) \right ] - \left [ \frac{2^2}{2}-2(2) \right ] \right \}

= [4-2-2+\frac{1}{2}] +[8-8-2+4]

= \frac{1}{2}+2 =\frac{5}{2}

Therefore, I_{2} = \frac{5}{2}

I_{3} = \int^4_{1}|x-3| dx = \int^3_{1} (3-x)dx +\int^4_{3} (x-3)dx

\because as (x-3)\geq 0 is in the given x -range [3,4] and \leq 0 in the range [1,3]

=\left [ 3x - \frac{x^2}{2}\right ] ^3_{1} + \left [ \frac{x^2}{2} -3x\right ] ^4_{3}

= \left \{ \left [ 3(3)-\frac{3^2}{2} \right ] - \left [ 3(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-3(4) \right ] - \left [ \frac{3^2}{2}-3(3) \right ] \right \}

= [9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]

= [6-4]+\frac{1}{2} =\frac{5}{2}

Therefore, I_{3} = \frac{5}{2}

So, We have the sum = I_{1}+I_{2}+I_{3}

I = \frac{9}{2}+\frac{5}{2}+\frac{5}{2} = \frac{19}{2}

Question:34 Prove the following (Exercises 34 to 39)

. \int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}

Answer:

L.H.S = \int_1^3\frac{dx}{x^2(x+1)}

We can write the numerator as [(x+1) -x]

\therefore \int_1^3\frac{dx}{x^2(x+1)} = \int_1^3\frac{(x+1)-x}{x^2(x+1)}dx

\\ = \int_1^3\left [ \frac{1}{x^2} - \frac{1}{x(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{(x+1)-x}{x(x+1)}dx

\\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\left [ \frac{1}{x} - \frac{1}{(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{1}{x}dx + \int_1^3\frac{1}{(x+1)}dx \\ = \left [ -\frac{1}{x} \right ]^3_1 - \left [ \log x \right ]^3_1 +\left [ \log(x+1) \right ]^3_1

\\ = \left [ -\frac{1}{3} +1 \right ] - \left [ \log 3 - \log 1 \right ] +\left [\log 4 - \log 2 \right ] \\ = \frac{2}{3} + \log \left ( \frac{4}{3.2}\right ) \\

= \log \left(\frac{2}{3} \right ) +\frac{2}{3} = RHS

Hence proved.

Question:35 Prove the following (Exercises 34 to 39)

\int_0^1 xe^xdx =1

Answer:

Let\ I=\int xe^{x}dx

Integrating I by parts

\\I=x\int e^{x}dx-\int ( (\frac{\mathrm{d} (x)}{\mathrm{d} x})\int e^{x}dx)dx\\ I=xe^{x}-\int e^{x}dx\\ I=xe^{x}-e^{x}+c

Applying Limits from 0 to 1

\\\int_{0}^{1}xe^{x}dx=[xe^{x}-e^{x}+c]_{0}^{1}\\ I=[e-e+c]-[0-1+c]\\ I=1

Hence proved I = 1

Question:36 Prove the following (Exercises 34 to 39)

\int_{-1}^1x^{17}\cos^4 x dx=0

Answer:

Let \ x^{17}cos^{4}x=g(x)

g(-x)= (-x)^{17}cos^{4}(-x)=-x^{17}cos^{4}x=-g(x)

The Integrand g(x) therefore is an odd function and therefore

\int_{-1}^{1}g(x)dx=0


Question:37 Prove the following (Exercises 34 to 39)

\int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}

Answer:

\\Let\ I= \int_{0}^{\frac{\pi }{2}}sin^{3}xdx\\ I=\int_{0}^{\frac{\pi }{2}}sinx(1-sin^{2}x)dx\\ I=\int_{0}^{\frac{\pi }{2}}sinxdx-\int_{0}^{\frac{\pi }{2}}cos^{2}xsinxdx\\ I=I_{1}-I_{2}

\\I_{1}=[-cosx]_{0}^{\frac{\pi }{2}}\\ I_{1}=-0-(-1)=1

For I 2 let cosx=t, -sinxdx=dt

The limits change to 0 and 1

\\I_{2}=-\int_{1}^{0}t^{2}dt\\ I_{2}=-[\frac{t^{3}}{3}]{_{1}}^{0}\\ I_{2}=0-(-\frac{1}{3})\\ I_{2}=\frac{1}{3}

I 1 -I 2 =2/3

Hence proved.

Question:38 Prove the following (Exercises 34 to 39)

\int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2

Answer:

The integral is written as

\\Let\ I=\int 2tan^{3}xdx\\ I=\int 2tan^{2}x\cdot tanxdx\\ I=\int 2(sec^{2}x-1)tanxdx\\ I=2\int tanxsec^{2}xdx-2\int tanxdx\\ I=2\int tdt-2log(cosx)+c\ \ \ \ \ \ \ (t=tanx) \\I=t^{2}-2log(cosx)+c\\ I=tan^{2}x-2log(cosx)+c

[I]_{0}^{\frac{\pi }{4}}=[tan^{2}x-2log(cosx)]_{0}^{\frac{\pi }{4}}\\

[I]_{0}^{\frac{\pi }{4}}=(1-2log\sqrt{2})-(0-2log1)

[I]_{0}^{\frac{\pi }{4}}=1-log2

Hence Proved

Question:39 Prove the following (Exercises 34 to 39)\

\int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1

Answer:

Let \ I=\int sin^{-}xdx

Integrating by parts we get

\\ I= sin^{-}x\int 1\cdot dx-\int (\frac{\mathrm{d} (sin^{-}x)}{\mathrm{d} x}\int 1\cdot dx)\\ I=xsin^{-}x+c-\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I=I_{1}-I_{2}

For I 2 take 1-x 2 = t 2 , -xdx=tdt

\\I_{2}=\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I_{2}=-\int\frac{1}{t}tdt \\ I_{2}=-t+c\\ I_{2}=-\sqrt{1-x^{2}}+c

[I]_{0}^{1}=[I_{1}-I_{2}]_{0}^{1}\\

\\=[xsin^{-}x-(-\sqrt{1-x^{2}})]_{0}^{1}\\ =[xsin^{-}x+\sqrt{1-x^{2}}]_{0}^{1}\\ =[1\cdot \frac{\pi }{2}+0]-[0+1]\\ =\frac{\pi }{2}-1

Hence Proved

Question:40 Evaluate \int_0^1e^{2-3x}dx as a limit of a sum.

Answer:

As we know

\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)........+f(a+(n-1)h)]

where b-a=hn

In the given problem b=1, a=0 and f(x)=e^{2-3x}
\\\int_{0}^{1}e^{2-3x}dx=(1-0)\lim_{n\rightarrow \infty }\frac{1}{n}(e^{2}+e^{2-3h}+e^{2-3(2h)}.....+e^{2-3(n-1)h})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(1+e^{-3h}+e^{-6h}....+e^{-3(n-1)h})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-(e^{-3h})^{n}}{1-e^{-3h}})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-e^{-\frac{3}{n}\times n}}{1-e^{-\frac{3}{n}}})\\

\\=e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-e^{-3}}{1-e^{-\frac{3}{n}}})\\ =\frac{e^{2}(1-e^{-3})}{3}\lim_{n\rightarrow \infty }\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\\ =\frac{e^{2}(1-e^{-3})}{3}

=\frac{e^{2}-e^{-1}}{3}

Question:41 Choose the correct answers in Exercises 41 to 44.

. \int\frac{dx}{e^x + e^{-x}} is equal to

(A) \tan^{-1}(e^x) + c

(B) \tan^{-1}(e^{-x}) + c

(C) \log (e^x - e^{-x}) + C

(D) \log (e^x + e^{-x}) + C

Answer:

\int\frac{dx}{e^x + e^{-x}}

the above integral can be re arranged as

\\=\int \frac{e^{x}}{e^{2x}+1}dx\\

let e x =t, e x dx=dt

\int\frac{dx}{e^x + e^{-x}}

\\=\int \frac{1}{t^{2}+1}dt\\ =tan^{-1}t+c\\ =tan^{-1}(e^{x})+c

(A) is correct

Question:42 Choose the correct answers in Exercises 41 to 44.

. \int\frac{\cos 2x}{(\sin x + \cos x)^2}dx is equal to

(A) \frac{-1}{\sin x + \cos x} + C

(B) \log |{\sin x + \cos x} |+ C

(C) \log |{\sin x- \cos x} |+ C

(D) \frac{1}{(\sin x + \cos x)^2} + C

Answer:

\\\frac{\cos 2x}{(\sin x + \cos x)^2}\\ =\frac{cos^{2}x-sin^{2}x}{(\sin x + \cos x)^2}\\ =\frac{(\sin x + \cos x)(\cos x-\sin x)}{(\sin x + \cos x)^2} \\=\frac{(\cos x-\sin x)}{(\sin x + \cos x)} cos2x=cos 2 x-sin 2 x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

\\\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx\\ =\int \frac{dt}{t}\\ =log|t|+c \\=log|cosx+sinx|+c

B is correct

Question:43 Choose the correct answers in Exercises 41 to 44.

If f(a+b-x) = f(x) , then \int_a^bxf(x)dx is equal to

(A) \frac{a+b}{2}\int^b_af(b-x)dx

(B) \frac{a+b}{2}\int^b_af(b+x)dx

(C) \frac{b-a}{2}\int^b_af(x)dx

(D) \frac{a+b}{2}\int^b_af(x)dx

Answer:

Let\ \int_a^bxf(x)dx=I

As we know \int_a^bf(x)dx=\int_a^bf(a+b-x)dx

Using the above property we can write the integral as

\\I=\int_{a}^{b}(a+b-x)f(a+b-x)dx\\ I=\int_{a}^{b}(a+b-x)f(x)dx\\ I=(a+b)\int_{a}^{b}f(x)dx-\int_{a}^{b}xf(x)dx\\ I=(a+b)\int_{a}^{b}f(x)dx-I\\ 2I=(a+b)\int_{a}^{b}f(x)dx\\ I=\frac{a+b}{2}\int_{a}^{b}f(x)dx

Answer (D) is correct

Question: 44 Choose the correct answers in Exercises 41 to 44.

The value of \int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx is

(A) 1

(B) 0

(C) -1

(D) \frac{\pi}{4}

Answer:

\\Let\ I=\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx\\

\\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )\\ =tan^{-1}\left ( \frac{x-(1-x)}{1+x(1-x)} \right )\\ =tan^{-1}x-tan^{-1}(1-x) as tan^{-1}\left ( \frac{a-b}{1+ab} \right )=tan^{-1}a-tan^{-1}b

Now the integral can be written as

\\I=\int_{0}^{1} \left ( tan^{-1}x-tan^{-1}(1-x) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}(1-(1-x)) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}x\right )dx\\ I=-I\\ 2I=0\\ I=0

(B) is correct.


More about NCERT solutions for class 12 maths chapter 7 miscellaneous exercise

The NCERT class 12 maths chapter integrals is most important in the whole maths syllabus and it has applications in Physics and chemistry also. NCERT solutions for class 12 maths chapter 7 miscellaneous exercise deals with quite tricky and interesting questions which can be realised while solving the questions. NCERT solutions for class 12 maths chapter 7 miscellaneous exercise will take some longer time to complete but one should remain focussed while solving it as it is quite important. class 12 maths chapter 7 miscellaneous solutions is a good source to practice well.

Benefits of ncert solutions for class 12 maths chapter 7 miscellaneous exercises

  • The class 12th maths chapter 7 exercise has great importance and it should be done in step by step manner.
  • Class 12 maths chapter 7 miscellaneous exercises Will take more effort than other exercises, Hence should be done patiently.
  • These class 12 maths chapter 7 miscellaneous exercises along with previous exercises are good to go for the exam.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Key Features Of NCERT Solutions For Class 12 Chapter 7 Miscellaneous Exercise

  • Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 7, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 chapter 7 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 7 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this class 12 maths ch 7 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for class 12 chapter 7 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 7 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!


Frequently Asked Question (FAQs)

1. What is the level of questions in Miscellaneous exercise ?

Advanced level problems are dealt with in this exercise.  

2. How much time will it take topcomlete miscellaneous exercise for the first time ?

It can take 3 to 4 hours for the first time.

3. Can shortcuts be used while solving the questions.?

In boards, step by step method is used but shortcuts can be used in JEE and NEET.

4. Is it mandatory to solve every question of Miscellaneous exercise ?

No, but questions on different concepts must be solved.

5. What is the weightage from miscellaneous exercise in the examination ?

Around 5 marks of questions are asked in the examination.

6. Can this exercise be done with self study only ?

Yes, but for that basic concepts must be clear.

Articles

Upcoming School Exams

View All School Exams
Get answers from students and experts

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

Back to top