NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

Edited By Team Careers360 | Updated on Nov 3, 2021 - 2:35 p.m. IST

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. Apart from class 12 maths chapter 7 miscellaneous exercise solutions, you can also refer below exercises for wide understanding of the topic.

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

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

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

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

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

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

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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

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

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

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

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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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now the integral becomes

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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.

Also see-

NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

Question: Can this exercise be done with self study only ?

Answer:

Yes, but for that basic concepts must be clear.

Question: What is the weightage from miscellaneous exercise in the examination ?

Answer:

Around 5 marks of questions are asked in the examination.

Question: Is it mandatory to solve every question of Miscellaneous exercise ?

Answer:

No, but questions on different concepts must be solved.

Question: Can shortcuts be used while solving the questions.?

Answer:

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

Question: How much time will it take topcomlete miscellaneous exercise for the first time ?

Answer:

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

Question: What is the level of questions in Miscellaneous exercise ?

Answer:

Advanced level problems are dealt with in this exercise.

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NCERT Solutions

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

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.

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