Integration is considered an important pillar of calculus, and it has different applications in real life. You will use integration to find the area under the curves, volumes of irregular forms, and total cumulative quantities such as money or work completed. It is imperative in the field of physics to comprehend forces such as friction and gravity, calculate the centre of mass, and analyse motion. As we have studied all the integration concepts in this exercise, you will find many new and advanced questions that will give more clarity on the topic. These NCERT Solutions for Class 12 are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation.
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Integrals connect functions to their rates of change, making them an important concept in calculus. The main benefit of NCERT Solutions for Class 12 Maths is that they explain answers clearly, making learning simpler and more effective. Many teachers recommend NCERT Solutions because they closely match the exam pattern. For syllabus, notes, and PDF, refer to this NCERT article.
Question 1: Integrate the functions in Exercises 1 to 23.
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 x2, 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 23.
$\frac{1}{\sqrt{x+a} + \sqrt{x+b}}$
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 23.
$\frac{1}{x\sqrt{ax-x^2}}$ [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}}\ $
$=\ \int \frac{1}{\frac{a}{t}\sqrt{a.\frac{a}{t}\ -\ (\frac{a}{t})^2}} \frac{-a}{t^2}dt$
$=\ \frac{-1}{a}\int \frac{1}{\sqrt{(t-1)}}dt$
$=\ \frac{-1}{a}\ (2\sqrt{t-1})\ +\ C$
$=\ \frac{-1}{a}\ (2\sqrt{\frac{a}{x}\ -\ 1})\ +\ C$
$=\ \frac{-2}{a}\ \sqrt{\frac{a-x}{x}}\ +\ C$
Question 4: Integrate the functions in Exercises 1 to 23.
$\frac{1}{x^2(x^4 + 1)^\frac{3}{4}}$
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,
$\int \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}\ =\ \int \frac{1}{x^5}\left ( 1+\ \frac{1}{x^4}^{\frac{-3}{4}}\ \right )dx$
$=\ \frac{-1}{4} \int (1+t)^{\frac{-3}{4}}dt$
$=\ \frac{-1}{4} \frac{(1+\frac{1}{x^4})^{\frac{1}{4}}}{\frac{1}{4}}\ +\ C$
$=\ - \left [ 1+\frac{1}{x^4} \right ]^{\frac{1}{4}}\ +\ C$
Question 5: Integrate the functions in Exercises 1 to 23.
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 23.
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 x2, 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.
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 9: Integrate the functions in Exercises 1 to 23.
$\frac{\cos x}{\sqrt{4 - \sin^2 x}}$
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 23.
$\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}$
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 23.
$\frac{1}{\cos(x+a)\cos(x+b)}$
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 23.
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 23.
$\frac{e^x}{(1 + e^x)(2 + e^x)}$
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 23.
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
$\implies I = \frac{1}{3}tan^{-1}x - \frac{1}{6}tan^{-1}\frac{x}{2} + C$
Question 15: Integrate the functions in Exercises 1 to 23.
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
$\therefore \int cos^3xe^{\log sinx}dx = \int cos^3x.sinx dx$
$\begin{aligned} & =\int \mathrm{t}^3 \cdot(-\mathrm{dt}) \\ & =-\int \mathrm{t}^3 \cdot \mathrm{dt}\end{aligned}$
$I = -\frac{cos^4x}{4} + C$
Question 16: Integrate the functions in Exercises 1 to 23.
Answer:
Given the function to be integrated as $e^{3 \log x}\left(x^4+1\right)^{-1}$
$=e^{\log x^3}\left(x^4+1\right)^{-1}=\frac{x^3}{x^4+1}$
$I=\int e^{3 \log x}\left(x^4+1\right)^{-1}$
Let $x^4=t \Longrightarrow 4 x^3 d x=d t$
$\begin{aligned} & I=\int e^{3 \log x}\left(x^4+1\right)^{-1}=\int \frac{x^3}{x^4+1} \\ & =\frac{1}{4} \cdot \int \frac{1}{\mathrm{t}+1} \cdot \mathrm{dt} \\ & =\frac{1}{4} \log (\mathrm{t}+1)+\mathrm{C} \\ & \Longrightarrow I=\frac{1}{4} \log \left(x^4+1\right)+C\end{aligned}$
Question 17: Integrate the functions in Exercises 1 to 23.
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 integral 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 23.
$\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$
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
$\begin{aligned} & \sin (A+B)=\sin A \cos B+\cos A \sin B \\ & \therefore \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}} \\ & =\frac{1}{\sqrt{\sin ^3 x \cdot \sin x(\cos \alpha+\cot x \sin \alpha)}}=\frac{1}{\sqrt{\sin ^4 x(\cos \alpha+\cot x \sin \alpha)}} \\ & \frac{\operatorname{cosec}^2 x}{\sqrt{(\cos \alpha+\cot x \sin \alpha)}}\end{aligned}$
$\begin{aligned} & \Rightarrow \int \frac{1}{\sqrt{\sin ^3 \mathrm{x} \sin (\mathrm{x}+\alpha)}} \mathrm{dx}=\int \frac{\operatorname{cosec}^2 \mathrm{x}}{\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}} \mathrm{dx} \\ & =\int \frac{1}{\sqrt{\mathrm{t}}} \cdot-\frac{\mathrm{dt}}{\sin \alpha} \\ & =-\frac{1}{\sin \alpha} \int \frac{1}{\sqrt{\mathrm{t}}} \cdot \mathrm{dt} \\ & =-\frac{1}{\sin \alpha} \int \mathrm{t}^{-\frac{1}{2}} \cdot \mathrm{dt} \\ & =-\frac{1}{\sin \alpha}\left[\frac{\mathrm{t}^{\frac{1}{2}}}{\frac{1}{2}}\right]+\mathrm{C} \\ & =-\frac{2}{\sin \alpha}[\sqrt{\mathrm{t}}]+\mathrm{C} \\ & =-\frac{2}{\sin \alpha}[\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}]+\mathrm{C}\end{aligned}$
$=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{(\cos \alpha \sin x+\cos x \sin \alpha)}{\sin x}}\right]+C$
Question 19: Integrate the functions in Exercises 1 to 23.
$\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}$
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$
$\\ = -4\int \sin^2\frac{\theta}{2}(2cos^2\frac{\theta}{2} -1) d\theta$
$=-2 \cdot \int \sin ^2 \theta d \theta+4 \int \sin ^2\left(\frac{\theta}{2}\right) d \theta$
$\begin{aligned} & =\theta+\frac{2 \cdot \sqrt{1-\cos ^2 \theta} \cdot \cos \theta}{2}-2 \sqrt{1-\cos ^2 \theta}+C \\ & =\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C \\ & \Rightarrow I=\cos ^{-1} \sqrt{x}+\sqrt{x-x^2}-2 \sqrt{1-x}+C\end{aligned}$
Question 20: Integrate the functions in Exercises 1 to 23.
$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$
Answer:
Given to evaluate
$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$
$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$
$\begin{aligned} & =\left(\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right) e^x \\ & =\left(\sec ^2 x+\tan x\right) e^x\end{aligned}$
Now the integral becomes
$\Rightarrow \int \frac{2+\sin 2 x}{1+\cos 2 x} e^x d x=\int\left(\sec ^2 x+\tan x\right) e^x d x$
Let tan x = f(x)
$\implies f'(x) = sec^2x dx$
$\Rightarrow \mathrm{I}=\mathrm{e}^{\mathrm{x}} \tan \mathrm{x}+\mathrm{C}$
Question 21: Integrate the functions in Exercises 1 to 23.
$\frac{x^2 + x + 1}{(x+1)^2 (x+2)}$
Answer:
Given,
$\frac{x^2 + x + 1}{(x+1)^2 (x+2)}$
Using partial fractions, 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, x2, 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 22: Integrate the functions in Exercises 1 to 23.
$\tan^{-1}\sqrt{\frac{1-x}{1+x}}$
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 23: Integrate the functions in Exercises 1 to 23.
$\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}$
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 24: 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 25: Evaluate the definite integrals in Exercises 24 to 31.
$\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}$
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 26: Evaluate the definite integrals in Exercises 24 to 31.
$\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}$
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 $x=\frac{\pi}2,t=2\tan{\frac{\pi}2}=\infty$
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 27: Evaluate the definite integrals in Exercises 24 to 31.
$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$
Answer:
$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$
Here first, let's 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, the 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 28: Evaluate the definite integrals in Exercises 24 to 31.
$\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}$
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 29: Evaluate the definite integrals in Exercises 24 to 31.
$\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx$
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 30: Evaluate the definite integrals in Exercises 24 to 31.
$\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx$
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 31: Evaluate the definite integrals in Exercises 24 to 31
$\int_1^4[|x-1| + |x-2| + |x-3|]dx$
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 32: Prove the following (Exercises 32 to 37)
$\int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}$
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 33: Prove the following (Exercises 32 to 37)
Answer:
$Let\ I=\int xe^{x}dx$
Integrating 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 34: Prove the following (Exercises 32 to 37)
$\int_{-1}^1x^{17}\cos^4 x dx=0$
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 35: Prove the following (Exercises 32 to 37)
$\int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}$
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 I2 let cosx=t, -sinxdx=dt
The limits change to 0 and 1
$\\I_{2}=-\int_{1}^{0}t^{2}dt\\ I_{2}=-[\frac{t^{3}}{3}]{_{1}}^{0}\\ I_{2}=0-(-\frac{1}{3})\\ I_{2}=\frac{1}{3}$
I1 -I 2 = 2/3
Hence proved.
Question 36: Prove the following (Exercises 32 to 37)
$\int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2$
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 37: Prove the following (Exercises 32 to 37)
$\int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1$
Answer:
$Let \ I=\int sin^{-1}xdx$
Integrating by parts we get
$I= sin^{-1}x\int 1\cdot dx-\int (\frac{\mathrm{d} (sin^{-1}x)}{\mathrm{d} x}\int 1\cdot dx)$
$I=xsin^{-1}x+c-\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx$
$I=I_{1}-I_{2}$
For I2 take 1-x2 = t2 , -xdx=tdt
$\\I_{2}=\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx$
$I_{2}=-\int\frac{1}{t}tdt \\ I_{2}=-t+c\\ I_{2}=-\sqrt{1-x^{2}}+c$
$[I]_{0}^{1}=[I_{1}-I_{2}]_{0}^{1}\\$
$\\=[xsin^{-1}x-(-\sqrt{1-x^{2}})]_{0}^{1}$
$=[xsin^{-1}x+\sqrt{1-x^{2}}]_{0}^{1}$
$=[1\cdot \frac{\pi }{2}+0]-[0+1]$
$=\frac{\pi }{2}-1$
Hence Proved.
Question 38: $\int\frac{dx}{e^x + e^{-x}}$ is equal to
Answer:
$\int\frac{dx}{e^x + e^{-x}}$
The above integral can be rearranged 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 39: $\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx$ 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 40: 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.
Exercise-wise NCERT Solutions of Integrals Class 12 Maths Chapter 7 are provided in the link below.
Some properties of indefinite integrals are as follows:
1. $\int[f(x)+g(x)] d x=\int f(x) d x+\int g(x) d x$
2. For any real number $k, \int k f(x) d x=k \int f(x) d x$
Let the area function be defined by $\mathrm{A}(x)=\int_a^x f(x) d x$ for all $x \geq a$, where the function $f$ is assumed to be continuous on $[a, b]$. Then $\mathrm{A}^{\prime}(x)=f(x)$ for all $x \in[a, b]$.
Let $f$ be a continuous function of $x$ defined on the closed interval $[a, b]$ and let F be another function such that $\frac{d}{d x} \mathrm{~F}(x)=f(x)$ for all $x$ in the domain of $f$, then $\int_a^b f(x) d x=[\mathrm{F}(x)+\mathrm{C}]_a^b=\mathrm{F}(b)-\mathrm{F}(a)$.
This is called the definite integral of $f$ over the range $[a, b]$, where $a$ and $b$ are called the limits of integration, $a$ being the lower limit and $b$ the upper limit.
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