NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12

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.

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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

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

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

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

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 :

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

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

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 _.

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

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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

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

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

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

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$\implies I = \frac{1}{4}\log(x^4 +1) + C$

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

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1554447107634823

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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

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

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

Answer:

Given to evaluate

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

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1554447156855423

1554447157709943

1554447158484136

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now the integral becomes

1554447160001344

Let tan x = f(x)

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

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

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

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

(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

(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.

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.

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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

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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.

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NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions For Class 12 Chapter 7 Miscellaneous Exercise

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

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

Question: Integrate the functions in Exercises 1 to 24.

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