Exercise 19.10
Question 1$\int x \cdot e^{2 x} d x$
Sol :
$I=\int x \cdot e^{2 x} d x$
Let u=x and $v=e^{2x}$
then $\frac{d u}{d x}=1$
$\int v d x=\int e^{2 x} d x$
$=\frac{e^{2 x}}{2} d x$
Now , $\int u v d x=v \int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$I=x \cdot \frac{e^{2 x}}{2}-\int 1 \times \frac{e^{2 x}}{2} d x$
$=\frac{x}{2} \cdot e^{2 x}-\frac{1}{2} \int e^{2 x} d x$
$I=\frac{x}{2} \cdot e^{2x}-\frac{1}{2} \cdot \frac{e^{2 x}}{2}+c$
$=\frac{x}{2} \cdot e^{2 x}-\frac{1}{4} \cdot e^{2 x}+c$
$I=\frac{e^{2 x}}{4}(2 x-1)+c$
Question 2
$\int x \cos x d x$
Sol :
$I=\int x \cos x d x$
Let u=x and v=cosx
then $\frac{d u}{d x}=1$ , $\int v d x=\int \cos x d x=\sin x$
Now , $I=\int u v d x$
$=v \int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$I=x \sin x-\int 1 \times \sin x d x$
I=xsinx-(-cosx)+c
I=xsinx+cosx+c
Question 3
$\int x \cdot \sin 2 x d x$
Sol :
$I=\int x \sin 2 x d x$
Let u=x and v=sin2x
then $\frac{d u}{d x}=1$ , $\int v d x=\frac{-\cos 2 x}{2}$
Now , $I=\int u v d x$
$=U \int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$I=x\left(\frac{-\cos 2 x}{2}\right)-\int 1 \times \left(\frac{-\cos 2 x}{2}\right) d x$
$I=-\frac{x}{2} \cos 2 x+\frac{1}{2} \int \cos 2 x d x$
$I=\frac{-x}{2} \cos 2 x+\frac{1}{2} \frac{\sin 2 x}{2}+c$
$I=\frac{-x \cos 2 x}{2}+\frac{\sin 2 x}{4}+c$
Question 4
$\int x \cdot \sin x d x$
Sol :
$I=\int x \sin x d x$
Let u=x and v=sinx
then $\frac{d u}{d x}=1$ , $\int v d x=-\cos x$
Now , $I=\int uvd x$
$=u\int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$I=x(-\cos x)-\int 1 \times(-\cos x) d x$
I=-xcosx+sinx+c
Question 5
$\int x \cos 2 x d x$
Sol :
$I=\int x \cdot \cos 2 x d x$
Let u=x and v=cos2x
then $\frac{d u}{d x}=1$ ,$\int v d x=\frac{\sin 2 x}{2}$
Now , $\int x \cos 2 x d x$
$=x \int \cos 2 x d x-\int\left(\frac{d x}{d x} \times \int \cos 2 x\right) d x$
$I=\frac{x \sin 2 x}{2}-\int \frac{\sin 2 x}{2} d x$
$I=\frac{x \sin 2 x}{2}-\frac{1}{2}\left(\frac{-\cos 2 x}{2}\right)+c$
$=\frac{x \sin 2 x}{2}+\frac{1}{4} \cos 2 x+c$
Question 6
$\int x \cdot \cos (2 x+3) d x$
Sol :
$I=\int x \cdot \cos (2 x+3) d x$
Let u=x and v=cos(2x+3)
then $\frac{d u}{d x}=1$ , $\int v d x=\int \cos (2 x+3) d x$
$=\frac{\sin (2 x+3)}{2}$
Now , $I=\int u v d x$
$=u \int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$\therefore I=\int x \cos (2 x+3) d x$
$=x \int \cos (2 x+3) d x-\int\left(\frac{d x}{d x} \times \int \cos (2 x+3)\right) d x$
$I=\frac{x \cdot \sin (2 x+3)}{2}-\int 1 \times \frac{\sin (2 x+3)}{2} d x$
$I=\frac{x \sin (2 x+3)}{2}-\frac{1}{2}\left(\frac{-\cos (2 x+3)}{2}\right)+c$
$I=\frac{x \sin (2 x+3)}{2}+\frac{1}{4} \cos (2 x+3)+c$
Question 7
$\int(x+1) \cos x d x$
Sol :
$I=\int(x+1) \cos x d x$
Let u=(x+1) and v=cosx
then $\frac{d u}{d x}=1$ , $\int v d x=\sin x$
Now , $I=\int u v d x$
$=v \int v d x-\int\left(\frac{d y}{d x} \times \int v d x\right) d x$
$I=(x+1) \sin x-\int 1 \times \sin x d x$
I=(x+1)sinx-(-cosx)+c
I=(x+1)sinx+cosx+c
Question 8
$\int x \sin x \cdot \cos x d x$
Sol :
$I=\int x \sin x \cdot \cos x d x=\frac{1}{2} \int x \cdot 2 \sin x \cdot \cos x d x$
$I=\frac{1}{2} \int x \sin 2 x d x$
Let u=x and v=sin2x
then $\frac{d u}{d x}=1$ , $\int v d x=\frac{-\cos 2 x}{2}$
Now , $I=\int uv d x$
$=U \int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$I=\frac{1}{2}\left[x\left(\frac{-\cos 2 x}{2}\right)-\int 1 \times \left(-\frac{\cos 2 x}{2}\right) d x\right]$
$I=\frac{-x \cos 2 x}{2}+\frac{1}{2} \frac{\sin 2 x}{2}+c$
$I=\frac{1}{2}\left[\frac{-x \cos 2 x}{2}+\frac{1}{2} \frac{\sin 2 x}{2}+c\right]$
$I=\frac{1}{2}\left[\frac{-x \cos 2 x}{2}+\frac{\sin 2 x}{4}\right]+c$
$I=\frac{\sin 2 x}{8}-\frac{x \cos 2 x}{4}+c$
Question 9
$\int x \sin 2 x \cdot \cos 2 x d x$
Sol :
$I=\int x \sin 2 x \cdot \cos 2 x d x$
$=\frac{1}{2} \int x \cdot 2 \sin 2 x \cdot \cos 2 xdx$
$I=\frac{1}{2} \int x \sin 4 x d x$
Let u=x and v=sin4x
then $\frac{d u}{d x}=1$ , $\int v d x=\frac{-\cos 4 x}{4}$
Now , $I=\int u v d x$
$=v \int v d x-\int\left(\frac{d u}{d x} \times \int v d x\right) d x$
$I=\frac{1}{2}\left[x\left(\frac{-\cos 4 x}{4}\right)-\int 1 \times\left(-\frac{\cos 4 x}{4}\right) d x\right]$
$I=\frac{1}{2}\left[-\frac{x \cos 4 x}{4}+\frac{1}{4} \frac{\sin 4 x}{4}+c\right]+c$
$I=\frac{1}{2}\left[\frac{-x \cos 4 x}{4}+\frac{1}{16} \sin 4 x\right]+c$
$I=\frac{\sin 4 x}{32}-\frac{x \cos 4 x}{8}+c$
$I=\frac{1}{32}[\sin 4 x-4 x \cos 4 x]+c$
Question 10
$\int x \cdot \cos x \cdot \cos 2 x d x$
Sol :
$I=\int x \cdot \cos x \cdot \cos 2 x d x$
$=\frac{1}{2} \int x \cdot 2 \cos 2 x \cdot \cos x d x$
∵2cosA.cosB=cos(A-B)+cos(A+B)
$I=\frac{1}{2} \int x(\cos x+\cos 3 x) d x$
$I=\frac{1}{2} \int x \cos x d x+\frac{1}{2} \int x \cdot \cos 3 x d x$
$I=\frac{1}{2}\left[x \int \cos x d x-\int\left(\frac{d(x)}{d x}\times \int \cos x d x\right) d x\right]+\frac{1}{2}\left[x \cdot \int \cos 3 x d x\right. -\int\left(\frac{d(x)}{d x} \times \int \cos 3 x d x\right) d x$
$I=\frac{1}{2}\left[x \sin x-\int \sin x d x\right]+\frac{1}{2}\left[x \frac{\sin 2 x}{3}-\int \frac{\sin 3 x}{3} d x\right]+c$
$I=\frac{1}{2}[x \sin x+\cos x]+\frac{1}{2}\left[\frac{x \sin 3 x}{3}+\frac{1}{3} \times \frac{\cos 3 x}{3}\right]+c$
$I=\frac{1}{2}\left[x \sin x+\cos x+\frac{x \sin 3 x}{3}+\frac{\cos 3 x}{9}\right]+c$
Question 11
$\int x \sin x \cdot \sin 2 x d x$
Sol :
$I=\int x \sin x \cdot \sin 2 x d x$
$=\frac{1}{2} \int x \cdot 2 \sin 2 x \cdot \sin x d x$
$I=\frac{1}{2} \int x(\cos x-\cos 3 x) d x$
$=\frac{1}{2} \int x \cos x d x-\frac{1}{2} \int x \cos 3 x d x$
$I=\frac{1}{2}\left[x \sin x-\int 1 \times \sin x d x\right]-\frac{1}{2}\left[x \frac{\sin 3 x}{3}-\int 1 \times \frac{\sin 3 x}{3} d x\right.$
$I=\frac{1}{2}[x \sin x+\cos x]-\frac{1}{2}\left[\frac{x \sin 3 x}{3}+\frac{\cos 3 x}{9}\right]+c$
$I=\frac{1}{2}\left[x \sin x+\cos x-\frac{x \sin 3 x}{3}-\frac{\cos 3 x}{9}\right]+c$
Question 12
$\int x \sin 2 x \cdot \cos x d x$
Sol :
$I=\int x \sin 2 x \cdot \cos x d x$
$I=\frac{1}{2} \int x \cdot 2 \sin 2 x \cdot \cos x d x$
[2sinA.cosB=sin(A+B)+sin(A-B)]
$I=\frac{1}{2} \int x(\sin 3 x+\sin x) d x$
$I=\frac{1}{2} \int x \sin 3 x d x+\frac{1}{2} \int x \sin x d x$
$I=\frac{1}{2}\left[x\left(\frac{-\cos 3 x}{3}\right)-\int 1 \times\left(\frac{-\cos 3 x}{3}\right) d x\right]+\frac{1}{2}\left[x(-\cos x)-\int 1 \times(-\cos x)\right]dx$
$I=\frac{1}{2}\left[\frac{-x \cos 3 x}{3}+\frac{\sin 3 x}{9}\right]+\frac{1}{2}[-x \cos x+\sin x]+c$
$I=\frac{\sin 3 x}{18}-\frac{x \cos 3 x}{6}+\frac{\sin x}{2}-\frac{x \cos x}{2}+c$
Question 13
$\int x \sin 5 x \cdot \cos 6 x d x$
Sol :
$I=\int x \sin 5 x \cdot \cos 6 x d x$
$=\frac{1}{2} \int x \cdot 2 \cos 6 x \cdot \sin 5 x d x$
$I=\frac{1}{2} \int x(\sin 11 x-\sin x) d x$
[2cosA.sinB=sin(A-B)-sin(A-B)]
$I=\frac{1}{2} \int x \cdot \sin 11 x d x-\frac{1}{2} \int x \sin x d x$
$I=\frac{1}{2}\left[x\left(-\frac{\cos 11 x}{11}\right)-\int 1 \times\left(\frac{-\cos 11 x}{11}\right) d x\right]-\frac{1}{2}\left[x(-\cos x)-\int 1 x(-\cos x)\right]$
$I=\frac{1}{2}\left[\frac{-x \cos 11 x}{11}+\frac{\sin 11 x}{121}\right]-\frac{1}{2}[-x \cos x+\sin x]+c$
$I=\frac{1}{2}\left[\frac{-x \cos 11 x}{11}+\frac{\sin 11 x}{121}+x \cos x-\sin x\right]+c$
Question 14
$\int x \cdot \cos 2 x \cdot \cos 3 x d x$
Sol :
$I=\int x \cdot \cos 2 x \cdot \cos 3 x d x$
$=\frac{1}{2} \int x \cdot 2 \cos 3 x \cdot \cos 2 x d x$
$I=\frac{1}{2} \int x(\cos x+\cos 5 x) d x$
$I=\frac{1}{2} \int x \cos x d x+\frac{1}{2} \int x \cos 5 x d x$
$I=\frac{1}{2}\left[x \sin x-\int 1 \times \sin x d x\right]+\frac{1}{2}\left[\frac{x \sin 5 x}{5}-\int 1 \times \sin 5 x d x\right]+c$
$I=\frac{1}{2}[x \sin x+\cos x]+\frac{1}{2}\left[\frac{x \sin 5 x}{5}+\frac{\cos 5 x}{25}\right]+c$
$I=\frac{1}{2}\left[\frac{x \sin 5 x}{5}+\frac{\cos 5 x}{25}+x \sin x+\cos x\right]+c$
Question 15
$\int x \cos a x \cos b x d x$
Sol :
Question 16
$\int x \sin x \cdot \sin 3 x d x$
Sol :
$I=\int x \sin x \cdot \sin 3 x d x$
$=\frac{1}{2} \int x \cdot 2 \sin 3 x \cdot \sin x d x$
$I=\frac{1}{2} \int x(\cos 2 x-\cos 4 x) d x$
$I=\frac{1}{2} \int x \cos 2 x d x-\frac{1}{2} \int x \cos 4 x d x$
$I=\frac{1}{2}\left[x \frac{\sin 2 x}{2}-\int 1 \times \frac{\sin 2 x}{2} d x\right]-\frac{1}{2}\left[x \frac{\sin 4 x}{4}-\int 1 \times \frac{\sin 4 x}{4} d x\right.$
$\left.I=\frac{1}{2}\left[\frac{x \sin 2 x}{2}+\frac{\cos 2 x}{4}\right]-\frac{1}{2} \int \frac{x \sin 4 x}{4}+\frac{\cos 4 x}{16}\right]+c$
$I=\frac{x \sin 2 x}{4}+\frac{\cos 2 x}{8}-\frac{x \sin 4 x}{8}-\frac{1}{32} \cos 4 x+c$
$I=\frac{1}{4}\left[x \sin 2 x+\frac{\cos 2 x}{2}-\frac{x \sin 4 x}{2}-\frac{\cos 4 x}{8}\right]+c$
Question 17
$\int x \cdot \sin 2 x \cdot \sin 3 x d x$
Sol :
$I=\int x \sin 2 x \cdot \sin 3 x d x$
$=\frac{1}{2} \int x \cdot 2 \sin 3 x \cdot \sin 2 x d x$
$I=\frac{1}{2} \int x(\cos x-\cos 5 x) d x$
$I=\frac{1}{2} \int x \cos x d x-\frac{1}{2} \int x \cos 5 x d x$
$I=\frac{1}{2}\left[x \sin x-\int 1 \times \sin x d x\right]-\frac{1}{2}\left[x \frac{\sin 5 x}{5}-\int 1 \times \frac{\sin 5 x}{5} d x\right]$
$I=\frac{1}{2}[x \sin x+\cos x]-\frac{1}{2}\left[\frac{x \sin 5 x}{5}+\frac{\cos 5 x}{25}\right]+c$
$I=\frac{1}{2}\left[x \sin x+\cos x-\frac{x \sin 5 x}{5}-\frac{\cos 5 x}{25}\right]$
Question 18
$\int x \cdot \sec ^{2} x d x$
Sol :
$I=\int x \cdot \sec ^{2} x d x$
$I=x \int \sec ^{2} x d x-\int\left(\frac{d(x)}{d x} \times \int \sec ^{2} x d x\right) d x$
$I=x \tan x-\int \tan x d x$
$I=x \tan x-\log |\sec x|+c$
Question 19
$\int x \cdot \sec ^{2} 2 x d x$
Sol :
$I=\int x \cdot \sec ^{2} 2 x d x$
$\left.I=x \int \sec ^{2} 2 x d x-\int\left(\frac{d(x)}{d x}\right) \times \int \sec ^{2} 2 x d x\right) d x$
$I=\frac{x \tan 2 x}{2}-\int \frac{\tan 2 x}{2} d x$
$I=\frac{x \tan 2 x}{2}-\frac{1}{2} \log |\sec 2 x| \times \frac{1}{2}+c$
$I=\frac{x}{2} \tan 2 x-\frac{1}{4} \log \left|\frac{1}{\cos 2 x}\right|+c$
$\left.I=\frac{x}{2} \tan 2 x-\frac{1}{4}[\log 1-\log | \cos 2 x|\right]+c$
$I=\frac{x}{2} \tan 2 x-\frac{1}{4} \log 1+\frac{1}{4} \log |\cos 2 x|+c$
$I=\frac{x}{2} \tan 2 x+\frac{1}{4} \log |\cos 2 x|+c$
Question 20
$\int \frac{x}{1+\cos x} d x$
Sol :
$I=\int \frac{x}{1+\cos x} d x=\int \frac{x}{2 \cos ^{2} \frac{x}{2}}$
$=\frac{1}{2} \int x \sec ^{2} \frac{x}{2} d x$
$I=\left[x \cdot \int \sec ^{2} \frac{x}{2} d x-\int \frac{d x}{d x} \times \int \sec ^{2} \frac{x}{2} d x\right) d x$
$I=\frac{1}{2}\left[2 x \tan \frac{x}{2}-\int 1 \times \tan \frac{x}{2} \times 2 d x\right.$
$I=x \tan \frac{x}{2}-\log \left|\sec \frac{x}{2}\right| \times 2+c$
Question 21
$\int x \cdot \tan ^{2} x d x$
Sol :
$I=\int x \cdot \tan ^{2} x d x$
$I=\int x \cdot\left(\sec ^{2} x-1\right) d x$
$I=\int x \sec ^{2} x d x-\int x d x$
$I=x \int \sec ^{2} x d x-\int\left(\frac{d x}{d x} \times \int \sec ^{2} x d x\right) d x-\frac{x^{2}}{2}+c$
$I=x \tan x-\int 1 \times \tan x d x-\frac{x^{2}}{2}+c$
$I=x \tan x-\log |\sec x|-\frac{x^{2}}{2}+c$
Question 22
$\int x \cdot \tan x \sec ^{2} x d x$
Sol :
$I=\int x \cdot \tan x \sec ^{2} x d x$
$I=x \int \tan x \cdot \sec ^{2} x d x-\int\left(\frac{d x}{d x} \times \int \tan x \cdot \sec ^{2} x d x\right) d x$
$I=x \int z d z-\int 1 \times z^{2} d x$
$I=x \frac{z^{2}}{2}-\frac{1}{2} \int \tan ^{2 x} d x$
$I=\frac{1}{2} x \tan ^{2} x-\frac{1}{2} \int\left(\sec ^{2} x-1\right) d x$
$I=\frac{1}{2} x \tan ^{2} x-\frac{1}{2} \int \sec ^{2} x d x+\frac{1}{2} \int d x$
$I=\frac{1}{2} x \tan ^{2} x-\frac{1}{2} \tan x+\frac{x}{2}+c$
Question 23
$\int x \cdot \sin ^{2} x d x$
Sol :
$I=\int x \sin ^{2} x d x$
$=\int x\left(\frac{1-\cos 2 x}{2}\right) d x$
$I=\frac{1}{2} \int x(1-\cos 2 x) d x$
$=\frac{1}{2} \int x d x-\frac{1}{2} \int x \cos 2 x d x$
$I=\frac{1}{2} \frac{x^{2}}{2}-\frac{1}{2}\left[x \int \cos 2 x d x-\int\left(\frac{d x}{d x} \times \int \cos 2 x d x\right) d x\right]$
$I=\frac{x^{2}}{4}-\frac{1}{2}\left[x \frac{\sin 2 x}{2}-\int 1 \times \frac{\sin 2 x}{2} d x\right]+c$
$I=\frac{x^{2}}{4}-\frac{1}{2}\left[\frac{x \sin 2 x}{2}+\frac{\cos 2 x}{4}\right]+c$
$I=\frac{x^{2}}{4}-\frac{x \sin 2 x}{4}-\frac{\cos 2 x}{8}+c$
$I=\frac{1}{4}\left(x^{2}-x \sin 2 x-\frac{1}{2} \cos 2 x\right)+c$
Question 24
$\int x \cdot \sin ^{2} x \cdot \cos ^{2} x d x$
Sol :
$I=\int x \sin ^{2} x \cdot \cos ^{2} x d x$
$I=\int x \cdot \frac{1}{4} \sin ^{2} 2 x d x$
$I=\frac{1}{4} \int x \sin ^{2} 2 x d x$
$I=\frac{1}{4} \int x\left(\frac{1-\cos 4 x}{2}\right) d x$
$I=\frac{1}{8} \int x(1-\cos 4 x) d x$
$I=\frac{1}{8} \int x d x-\frac{1}{8} \int x \cos 4 x d x$
$\frac{1}{8} \cdot \frac{x^{2}}{2}-\frac{1}{8}\left[x \int \cos 4 x d x-\int\left(\frac{d x}{d x} \times \int \cos 4 x d x\right) d x\right]$
$I=\frac{x^{2}}{16}-\frac{1}{8}\left[x \cdot \frac{\sin 4 x}{4}-\int 1 \times \frac{\sin 4 x}{4} d x\right]+c$
$I=\frac{x^{2}}{16}-\frac{1}{8}\left[\frac{x \sin 4 x}{4}+\frac{\cos 4 x}{16}\right]+c$
$I=\frac{x^{2}}{16}-\frac{x \sin 4 x}{32}-\frac{1}{128} \cos 4 x+c$
$I=\frac{1}{16}\left(x^{2}-\frac{x \sin 4 x}{2}-\frac{\cos 4 x}{8}\right)+c$
Question 25
$\int x \cdot \cos ^{2} a x d x$
Sol :
$I=\int x \cdot \cos ^{2} a x d x$
$=\int x\left(\frac{1+\cos 2 a x}{2}\right) d x$
$I=\frac{1}{2} \int x(1+\cos 2 a x) d x$
$=\frac{1}{2} \int x d x+\frac{1}{2} \int x \cos 2 a x d x$
$I=\frac{1}{2} \cdot \frac{x^{2}}{2}+\frac{1}{2}\left[x \int \cos 2 a x d x-\int\left(\frac{d x}{d x} \right)\times \cos 2 a x d x\right) d x$
$I=\frac{x^{2}}{4}+\frac{1}{2}\left[\frac{x \sin 2 a x}{2 a}-\int 1 \times \frac{\sin 2 a x}{2 a} d x\right]+c$
$I=\frac{x^{2}}{4}+\frac{1}{2}\left[\frac{x \sin 2 a x}{2 a}-\frac{1}{2 a} \times\left(\frac{-\cos 2 a x}{2 a}\right)\right]+c$
$I=\frac{x^{2}}{4}+\frac{1}{2}\left[\frac{x \sin 2 a x}{2 a}+\frac{\cos 2 a x}{4 a^{2}}\right]+c$
$I=\frac{x^{2}}{4}+\frac{1}{2}\left[\frac{x \sin 2 a x}{2 a}+\frac{\cos 2 a x}{4 a^{2}}\right]+c$
$I=\frac{x^{2}}{4}+\frac{x \sin 2 a x}{4 a}+\frac{\cos 2 a x}{8 a^{2}}+c$
$I=\frac{1}{a^{2}}\left[\frac{a^{2} x^{2}}{4}+\frac{a x \sin 2 a x}{4}+\frac{\cos 2 a x}{8}\right]+c$
Question 26
$\int x \sin ^{3} x d x$
Sol :
$I=\int x \sin ^{3} x d x$
$=\int x\left(\frac{3 \sin x-\sin 2 x}{4}\right)dx$
$I=\frac{3}{4} \int x \sin x d x-\frac{1}{4} \int x \sin 3 x d x$
$I=\frac{3}{4}\left[x(-\cos x)-\int 1 \times(-\cos x) d x\right.-\frac{1}{4} \int x\left(\frac{-\cos 3 x}{3}\right) \left.-\int 1 \times\left(\frac{-\cos 3 x}{3}\right) d x\right]+c$
$I=\frac{3}{4}[-xcosx+-\sin x]-\frac{1}{4}\left[-\frac{x \cos 2 x}{3}+\frac{\sin 3 x}{9}\right]+c$
Question 27
$\int x^{2} \cos x d x$
Sol :
$I=\int x^{2} \cos x d x$
Let u=x2 and v=cosx
then $I=\int x^{2} \cos x d x$
$=x^{2} \int \cos x d x-\int\left(\frac{dx^{2}}{d x} \times \int \cos x d x\right) d x$
$I=x^{2} \sin x-\int 2 x \sin x d x=x^{2} \sin x-2 \int x \sin x d x$
$I=x^{2} \sin x-2\left[x(-\cos x)-\int 1 \times(\cos x) d x\right]+c$
$I=x^{2} \sin x-2\left[-x \cos x+\int \cos x d x\right]+c$
$I=x^{2} \sin x-2 \left[ -x \cos x+\sin x \right]+c$
$I=x^{2} \sin x+2 x \cos x-2 \sin x+c$
Question 28
$\int x^{2} \cos 2 x d x$
Sol :
$I=\int x^{2} \cos 2 x d x$
$I= x^{2} \int \cos 2 x d x-\int\left(\frac{dx^{2}}{d x} \times \int \cos 2 x d x\right) d x$
$I=\left[x^{2} \frac{\sin 2 x}{2}-\int 2 x\times \frac{\sin 2 x}{2} d x\right]$
$I=\left[\frac{x^{2} \sin 2 x}{2}-\int x \sin 2 x d x\right]+c$
$I=\frac{x^{2} \sin 2 x}{2}-\left[x\left(\frac{-\cos 2 x}{2}\right)-\int 1 \times\left(\frac{-\cos 2 x}{2}\right) d x\right]+c$
$I=\frac{x^{2} \sin 2 x}{2}-\left[\frac{-x \cos 2 x}{2}+\frac{1}{2} \times \frac{\sin 2 x}{2}\right]+c$
$I=\frac{x^{2} \sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{\sin 2 x}{4}+c$
Question 29
$\int x^{2} \sin 2 x d x$Sol :
$I=x^{2} \int \sin 2 x d x-\int\left(\frac{dx^{2}}{d x} \times \int \sin 2 x d x\right) d x$
$I=x^{2}\left(\frac{-\cos 2 x}{2}\right)-\int 2 x\left(-\frac{\cos 2 x}{2}\right) d x$
$I=-\frac{-x^{2} \cos 2 x}{2}+\int x \cos 2 x d x$
$I=-\frac{x^{2} \cos 2 x}{2}+\left[x \frac{\sin 2 x}{2}-\int 1 \times \frac{\sin 2 x}{2} d x\right]$
$I=\frac{-x^{2} \cos 2 x}{2}+\left[\frac{x \sin 2 x}{2}-\frac{1}{2}\left(\frac{-\cos 2 x}{2}\right)\right]+c$
$I=\frac{-x^{2} \cos 2 x}{2}+\frac{x \sin 2 x}{2}+\frac{\cos 2 x}{4}+c$
Question 30
$\int x^{2} \sin ^{2} x d x$Sol :
$I=\int x^{2} \sin ^{2} x d x=\int x^{2}\left(\frac{1-\cos 2 x}{2}\right) d x$
$I=\frac{1}{2} \int x^{2} d x-\frac{1}{2} \int x^{2} \cos 2 x d x$
$I=\left.\frac{1}{2} \cdot \frac{x^{3}}{3}-\frac{1}{2}\left[x^{2}\right. \cos 2 x d x-\int\left(\frac{dx^{2}}{d x} \times \int \cos 2 x d x\right) d x\right]$
$I=\frac{x^{3}}{6}-\frac{1}{2}\left[\frac{x^{2} \sin 2 x}{2}-\int 2 x \frac{\sin 2 x}{2} d x\right]+c$
$I=\frac{x^{3}}{6}-\frac{1}{2}\left[\frac{x^{2} \sin 2 x}{2}-\int x \sin 2 x d x\right]+c$
$I=\frac{x^{3}}{6}-\frac{1}{2}\left[\frac{x^{2} \sin 2 x}{2}-\left\{x\left(\frac{-\cos 2 x}{2}\right)-\int 1 \times\left(\frac{-\cos 2 x}{2}\right) d x\right\}\right]$
$I=\frac{x^{3}}{6}-\frac{1}{2}\left[\frac{x^{2} \sin 2 x}{2}-\left\{\frac{-x \cos 2 x}{2}+\frac{1}{2} \frac{\sin 2 x}{2}\right\}\right]+c$
$I=\frac{x^{3}}{6}-\frac{1}{2}\left[\frac{x^{2} \sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{1}{4} \sin 2 x\right] +c$
$I=\frac{x^{3}}{6}-\frac{x^{2} \sin 2 x}{4}-\frac{x \cos 2 x}{4}+\frac{1}{8} \sin 2 x+c$
Question 31
$\int \sin \sqrt{x} d x$Sol :
putting $z=\sqrt{x}$ then
$d z=\frac{1}{2 \sqrt{x}} d x$
$d z=\frac{1}{2 z} d x $
$2 z d z=d x$
Now , $\int \sin \sqrt{x} d x=\int \sin z \cdot 2z d z$
$I=2 \int z \cdot \sin zd z$
$I=2\left[2 \int \sin z d z-\int\left(\frac{d z}{d z} \times \int \sin z d z\right) d z\right.$
$I=2\left[z(-\cos z)-\int 1\times (-\cos z) d z\right.$
$I=2\left[-z \cos z+\int \cos z\right]+c$
$=2[-z \cos z+\sin z]+c$
$I=2(-\sqrt{x} \cos \sqrt{x}+\sin \sqrt{x})+c$
$I=2(\sin \sqrt{x}-\sqrt{x} \cos \sqrt{x})+c$
Question 32
$\int x^{3} \sin x^{2} d x$Sol :
putting $z=x^{2}$ then $d z=2 x d x \Rightarrow \frac{d z}{2 x}=d x$
Now , $\int x^{3} \sin x^{2} d x=\int x^{2} \cdot x \sin x^{2} d x$
$=\int z \cdot x \sin z \frac{d z}{2 x}$
$I=\frac{1}{2} \int z \sin z d z$
$=\frac{1}{2}\left[z(-\cos z)-\int 1 \times(-\cos z) d z\right]+c$
$I=\frac{1}{2}\left[-z \cos z+\int \cos z\right]+c$
$=\frac{1}{2}[-z \cos z+\sin z]+c$
$I=\frac{1}{2}\left[5 \sin x^{2}-x^{2} \cos x^{2}\right]+c$
Question 33
$\int(x+2) \sqrt{2 x+3} d x$Sol :
$I=(x+2) \int(2 x+3)^{\frac{1}{2}} d x-\int\left(\frac{d}{d x}(x+2) \times \int(2 x+3)^{\frac{1}{2}} d x\right) d x$
$I=(x+2) \cdot \frac{(2 x+3)^{\frac{3}{2}}}{2} \times \frac{2}{3}-\int 1 \times \frac{2}{3} \cdot \frac{(2 x+3)^{\frac{3}{2}}}{2} d x$
$I=\frac{1}{3}(x+2)(2 x+3)^{\frac{3}{2}}-\frac{1}{3} \int(2 x+3)^{\frac{3}{2}} d x$
$I=\frac{1}{3}(x+2)(2 x+3)^{\frac{3}{2}}-\frac{1}{3} \cdot \frac{(2 x+3)^{\frac{3}{2}+1}}{\left(-\frac{3}{2}+1\right) \times 2}+c$
$I=\frac{1}{3}(x+2)(2 x+3)^{\frac{3}{2}}-\frac{1}{3} \times \frac{2}{5} \cdot \frac{(2 x+3)^{\frac{5}{2}}}{2}+c$
$I=\frac{1}{3}(x+2)(2 x+3)^{\frac{3}{2}}-\frac{1}{15}(2 x+3)^{\frac{5}{2}}+c$
$I=(2 x+3)^{\frac{3}{2}}\left[\frac{x+2}{3}-\frac{1}{15}(2 x+3)\right]+c$
$I=(2 x+3)^{\frac{3}{2}}\left(\frac{5 x+10-2 x-3}{15}\right)+c$
$I=(2 x+3)^{\frac{3}{2}}\left[\frac{3 x+7}{15}\right]+c$
$I=\frac{1}{15}(2 x+3)^{\frac{3}{2}} \cdot(3 x+7)+c$
Question 34
$\int(a x+b) \sqrt{c x+d} d x$Sol :
$=\int(a x+b)(c x+d)^{\frac{1}{2}} d x$
$I=(a x+b) \int(c x+d)^{\frac{1}{2}} d x-\int \left(\frac{d}{d x} (a x+b) \int(c x+d)^{\frac{1}{2}} d x\right) d x$
$I=\frac{2}{3}(a x+b) \cdot \frac{(c x+d)^{\frac{3}{2}}}{c}-\int a \times \frac{(c x+d)^{\frac{3}{2}}}{c} \times \frac{2}{3} d x$
$I=\frac{2}{3 c}(a x+b)(c x+d)^{\frac{3}{2}}-\frac{2 a}{3 c} \int(c x+d)^{\frac{3}{2}} d x$
$I=\frac{2}{3 c}(a x+b)(c x+d)^{\frac{3}{2}}-\frac{2 a}{3 c} \cdot \frac{(c x+d)^{\frac{5}{2}}}{c} \times \frac{2}{5}+c$
$I=\frac{2}{3 c}(a x+b)(c x+d)^{\frac{3}{2}}-\frac{4 a}{15 c^{2}}(c x+d)^{\frac{5}{2}}+c$
$I=(c x+d)^{\frac{3}{2}}\left[\frac{2(a x+b)}{3 c}-\frac{4 a(c x+d)}{15 c^{2}}\right]+c$
$I=(c x+d)^{\frac{3}{2}}\left[\frac{10 a c x+10 b c-4 a c x-4 a d}{15 c^{2}}\right]+c$
$I=\frac{(c x+d)^{\frac{3}{2}}}{15 c^{2}}[6 a c x+10 b c-4 a d]+c$
$I=\frac{2}{15 c^{2}}(c x+d)^{\frac{3}{2}}[3 a c x+5 b c-2 a d]+c$
Question 35
$\int x^{2} \sqrt{a x+b} d x$
Sol :Question 36
(i) $\int x^{2} \log x d x$Sol :
Let u=logx and v=x2
then $I=\int x^{2} \log x d x$
$=\log x \int x^{2} d x-\int\left(\frac{d}{d x} \log x \times \int x^{2} d x\right)$
$I=\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \times \frac{x^{3}}{3} d x$
$=\frac{x^{3}}{3} \log x-\frac{1}{3} \int x^{2} d x$
$I=\frac{x^{3}}{3} \log x-\frac{1}{3} \frac{x^{3}}{3}+c$
$=\frac{x^{3}}{3}\left(\log x-\frac{1}{3}\right)+c$
(ii) $\int x \log x d x$
Sol :
Let u=logx and v=x
then $I=\int x \log x d x=\log x \int x d x-\int\left(\frac{d}{d x} \log x \times \int x d x \right)d x$
$I={\frac{\log x \times x^{2}}{2}}-\int \frac{1}{x} \times \frac{x^{2}}{2} d x$
$=\frac{x^{2}}{2} \log x-\frac{1}{2} \int x d x$
$I=\frac{x^{2}}{2} \log x-\frac{1}{2} \cdot \frac{x^{2}}{2}+c$
$=\frac{x^{2}}{2} \log x-\frac{x^{2}}{4}+c$
Question 37
$\int \log _{10} x d x$Sol :
$I=\int \log _{10} x d x=\int \log _{e} x \cdot \log _{10} e d x$
$I=\log _{10} e \int \log x d x$
$=\log _{10} e \int \log x \cdot 1 d x$
Let u=logx and v=1dx
then
$I=\int \log x \cdot 1 d x=\log x \int 1 d x-\int\left(\frac{d}{d x} \log x \times \int 1 d x\right) d x$
$I=\log x \cdot x-\int \frac{1}{x} \times x d x$
$=x \log x-\int d x$
$I=x \log x-x$
$=\log_{10} e(x \log x-x)+c$
Question 38
(i) $\int \log (x+1) d x$Sol :
$I=\int \log (x+1) d x=\int \log (x+1) \cdot 1 d x$
Let u=log(x+1) and v=1
then $I=\int \log (x+1) \cdot t d x$
$=\log (x+1) \int t d x-\int\left(\frac{d(\log (x+1)}{d x}\right)(1 d x)dx$
$I=\log (x+1) x-\int \frac{1}{x+1} \times xd x$
$=\log (x+1)-\int \frac{x}{x+1} d x$
$I=x \log (x+1)-\int \frac{x+1-1}{x+1} d x$
$I=x \log (x+1)-\int \frac{x+1}{x+1}+\int \frac{1}{x+1} d x$
$I=x \log (x+1)-\int d x+\int \frac{1}{x+1} d x$
Let x+1=z then dx=dz
∴ $I=x \log (x+1)-x+\int \frac{1}{2} d z$
I=xlog(x+1)-x+logz+c
I=xlog(x+1)-x+log(1+x)+c
I=xlog(x+1)+log(x+1)-x+c
(ii) $\int x \cdot \log \left(1+x^{2}\right) d x$
Sol :
$I=\int x \log \left(1+x^{2}\right) d x$
Let u=log(1+x2) and v=x
then $I=\int x \log \left(1+x^{2}\right) d x$
$=\log \left(1+x^{2}\right) \int x d x-\int\left(\frac{d}{d x} \log \left(1+x^{2}\right)\right. \times \int x d x)$
$I=\log \left(1+x^{2}\right) \cdot \frac{x^{2}}{2}-\int \frac{1 \times 2 x}{1+x^{2}} \times \frac{x^{2}}{2} d x$
$I=\frac{x^{2}}{2} \log \left(1+x^{2}\right)-\int \frac{x^{3}}{1+x^{2}} d x$
$=\frac{x^{2}}{2} \log \left(1+x^{2}\right)-\int \frac{x^{2} \cdot x d x}{1+x^{2}}$
Let $z=1+x^{2}$
$z-1=x^{2}$
then
dz=2xdx
$\frac{d z}{2}=x d x$
Now , $I=\frac{x^{2}}{2} \log \left(1+x^{2}\right)-\frac{1}{2} \cdot \int \frac{(z-1)}{z} \frac{d z}{2}$
$I=\frac{x^{2}}{2} \log \left(1+x^{2}\right)-\frac{1}{2} \int \frac{z}{z} d z+\frac{1}{2} \int \frac{1}{z} d z$
$I=\frac{x^{2}}{2} \log \left(1+x^{2}\right)-\frac{1}{2} z+\frac{1}{2} \log z+c$
$I=\frac{x^{2}}{2} \log \left(1+x^{2}\right)-\frac{1}{2}\left(1+x^{2}\right)+\frac{1}{2} \log \left(1+x^{2}\right)+c$
(iii) $\int x(\log x)^{2} d x$
Sol :
$I=\int x(\log x)^{2} d x$
Let u=(logx)2 and v=x
then $I=\int x(\log x)^{2} d x$
$=(\log x)^{2} \int x d x-\int\left(\frac{d}{d x}(\log x)^{2} \times \int x d x\right) d x$
$I=(\log x)^{2} \cdot \frac{x^{2}}{2}-\int \frac{2 \log x}{x} \times \frac{x^{2}}{2} d x$
$=(\log x)^{2} \cdot \frac{x^{2}}{2}-\int x \log x d x$
$I=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \int x d x-\int\left(\frac{d \log x}{d x} \times(x d x)\right) d x\right.$
I=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \times \frac{x^{2}}{2} d x\right]
$I=\frac{x^{2}}{2}(\log x)^{2}-\left[\frac{x^{2}}{2} \log x-\frac{1}{2} \int x d x \right]$
$I=\frac{x^{2}}{2}(\log x)^{2}-\left[\frac{x^{2}}{2} \log x-\frac{x^{2}}{4}\right]+c$
$I=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+c$
$=\frac{x^{2}}{2}\left[(\log x)^{2}-\log x+\frac{1}{2}\right]$
(iv) $\int\left(x^{2}+1\right) \log x d x$
Sol :
$I=\int\left(x^{2}+1\right) \log x d x$
Let u=logx and $v=\left(x^{2}+1\right)$
then $I=\int\left(x^{2}+1\right) \log x d x$
$\left.=\log x \int\left(x^{2}+1\right) d x-\int\left(\frac{d}{d x} \log x \times \right. \int\left(x^{2}+1\right) d x\right) d x$
$I=\log x\left(\frac{x^{3}}{3}+x\right)-\int \frac{1}{x} \times\left(\frac{x^{3}}{3}+x\right) d x$
$I=\log x\left(\frac{x^{3}}{3}+x\right)-\int \frac{x^{2}+3 x}{3 x} d x$
$I=\log x\left(\frac{x^{3}}{3}+x\right)-\int \frac{x^{3}}{3 x} d x-\int \frac{3 x}{3 x} d x$
$I=\left(\frac{x^{3}}{3}+x\right) \log x-\frac{1}{3} \int x^{2} d x-\int d x$
I=\left(\frac{x^{3}}{3}+x\right) \log x-\frac{x^{3}}{9}-x+c
(v) $\int x \log 2 x d x$
Sol :
$I=\int x \log 2 x d x$
Let u=log2x and v=x
$I=\int x \log 2 x d x=\log 2 x \int x d x-\int\left(\frac{d(\log 2 x)}{d x} \
times \int x d x\right) d x$
$I=\log 2 x \cdot \frac{x^{2}}{2}-\int \frac{1 \times 2}{2 x} \cdot \frac{x^{2}}{2} d x$
$=\log 2 x \frac{x^{2}}{2}-\frac{1}{2} \int x d x$
$I=\log 2 x \cdot \frac{x^{2}}{2}-\frac{1}{2} \cdot \frac{x^{2}}{2}+c$
$=\log 2 x \cdot \frac{x^{2}}{2}-\frac{x^{2}}{4}+c$
$I=\frac{x^{2}}{4}(2 \log 2 x-1)+c$
(vi) $\int \log \left(2+x^{2}\right) d x$
Sol :
$I=\int \log \left(2+x^{2}\right) d x$
$=\int \log \left(2+x^{2}\right) \cdot 1 d x$
let $u=\log \left(2+x^{2}\right)$ and v=L
then $I=\int \log \left(2+x^{2}\right) d x$
$=\log \left(2+x^{2}\right) \int L d x-\int\left(\frac{d}{d x} \log \left(2+x^{2}\right) \times \int 1dx\right)dx$
$I=\log \left(2+x^{2}\right) \cdot x-\int \frac{1 \times 2 x}{2+x^{2}} x \times d x$
$I=x \log \left(2+x^{2}\right)-\int \frac{2 x^{2}}{2+x^{2}} d x$
$=x \log \left(2+x^{2}\right)-2 \int \frac{x^{2}}{2+x^{2}} d x$
$I=x \log \left(2+x^{2}\right)-2 \int \frac{2+x^{2}-2}{2+x^{2}} d x$
$I=x \log \left(2+x^{2}\right)-2 \int \frac{2+x^{2}}{2+x^2} d x-2 \int \frac{-2}{2+x^{2}} d x$
$I=x \log \left(2+x^{2}\right)-2 x+2 \int \frac{2}{2+x^{2}} d x$
$I=x \log \left(2+x^{2}\right)-2 x+2 \tan ^{-1} \frac{x}{\sqrt{2}} \times \sqrt{2}+c$
∵ $ \int \frac{1}{1+\left(\frac{x}{\sqrt{2}}\right)^{2}}=\tan \frac{1}{\sqrt{2}} \times \sqrt{2}$
$I=x \log \left(2+x^{2}\right)-2 x+2 \sqrt{2} \tan^{-1} \frac{x}{\sqrt{2}}+c$
Question 39
$\int \frac{\log (\log x)}{x} d x$Sol :
$I=\int \frac{\log (\log x)}{x} d x$
let logx=z and $d z=\frac{1}{x} d x$
then $I=\int \log z d z=\int \log z \cdot 1 d z$
let u=logz and v=1
then $I=\int \log z \cdot 1 d z=\log z \cdot \int L d z-\int\left(\frac{d}{d z} \log z x \int d z\right) d z$
$I=\log z \cdot z-\int \frac{1}{z} \times z d z$
=zlogz-z+c
I=logx.log(logx)-logx+c
I=logx[log(logx-1)+c]
Question 40
$\int x^{2}(\log x)^{2} d x$Sol :
$I=\int x^{2}(\log x)^{2} d x$
let u=(logx)2 and v=x2
then $I=\int x^{2}(\log x)^{2} d x$
$=(\log x)^{2} \cdot \int x^{2} d x-\int\left(\frac{d{(log x)^{2}} }{d x}\times \int x^2 dx \right)^{2} d x$
$I=(\log x)^{2} \cdot \frac{x^{3}}{3}-\int \frac{2 \log x}{x} \times \frac{x^{3}}{3} d x$
$I=(\log x)^{2} \frac{x^{3}}{3}-\frac{2}{3} \int x^{2} \log x d x$
$I=\frac{x^{3}}{3}(\log x)^{2}-\frac{2}{3}\left[\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \times \frac{x^{3}}{3} d x\right]+C$
$I=\frac{x^{3}}{3}(\log x)^{2}-\frac{2}{3}\left(\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}\right)+c$
$I=\frac{x^{3}}{3}(\log x)^{2}-\frac{2 x^{3}}{9} \log x+\frac{2 x^{3}}{27}$
$I=\frac{x^{3}}{27}\left[9(\log x)^{2}-6 \log x+2\right]+c$
Question 41
$\int \sqrt{x} \cdot(\log x)^{2} d x$Sol :
Let $u=(\log x)^{2}$ and $v=\sqrt{x}=x^{\frac{1}{2}}$
then $I=\int \sqrt{x} \cdot(\log x)^{2} d x$
$=(\log x)^{2} \int \sqrt{x} d x-\int\left(\frac{d}{d x} \log x\right)^{2} \times (\sqrt{x} d x) d x$
$I=(\log x)^{2} \cdot \frac{2}{3} x^{\frac{3}{2}}-\int \frac{2 \log x}{x} \times \frac{2}{3} x^{\frac{3}{2}} d x$
$I=\frac{2}{3} x^{\frac{3}{2}}(\log x)^{2}-\frac{4}{3} \int \log x \cdot x^{\frac{1}{2}} d x$
$I=\frac{2}{3} x^{\frac{3}{2}}(\log x)^{2}-\frac{4}{3}[\log x \int x^{\frac{1}{2}} d x-\int\left(\frac{d \log x}{d x} \times \int x^{\frac{1}{2}} d x\right) d x$
$I=$
$I=\frac{2}{3} x^{\frac{3}{2}}(\log x)-\frac{4}{3}\left[\log x \cdot \frac{2}{3} x^{\frac{3}{2}}-\int \frac{1}{x} \times \frac{2}{3} \cdot x^{\frac{3}{2}} d x\right]$
$I=\frac{2}{3} x^{\frac{3}{2}}(\log x)^{2}-\frac{4}{3}\left[\frac{2}{3} x^{\frac{3}{2}} \log x-\frac{2}{3} \int x^{\frac{1}{2}} d x\right.$
$I=\frac{2}{3} x^{\frac{3}{2}}(\log x)^{2}-\frac{4}{3}\left[\frac{2}{3} x^{\frac{3}{2}} \log x-\frac{2}{3} \times \frac{2}{3} x^{\frac{3}{2}}\right]+c$
$I=\frac{2}{3} x^{\frac{3}{2}}(\log x)^{2}-\frac{8}{9} x^{\frac{3}{2}} \log x+\frac{16}{27} x^{\frac{3}{2}}+c$
$I=\frac{2}{27} x^{\frac{3}{2}}\left[9(\log x)^{2}-12(\log x+8)]+c\right.$
Question 42
$\int \cos ^{-1} x d x$Sol :
$I=\int \cos ^{-1} x d x=\int \cos ^{-1} x \cdot 1 d x$
Let $u=\cos ^{-1} x$ and v=1
Now , $I=\int \cos ^{-1} x \cdot 1 d x$
$=\cos ^{-1} x \int 1 d x-\int\left(\frac{d \cos ^{-1} x}{d x} x \int 1 d x\right) d x$
$I=\cos ^{-1} x \cdot x-\int \frac{-1}{\sqrt{1-x^{2}}} \times x d x$
$I=x \cos ^{-1} x+\int \frac{x}{\sqrt{1-x^{2}}} d x$
Let $z=1-x^{2}$ then $dz=-2 x d x$
$\frac{d z}{-2}=x d x$
Now , $I=x \cos ^{-1} x+\int \frac{x d x}{\sqrt{1-x^{2}}}$
$I=x \cos ^{-1} x+\int \frac{-d z}{2\sqrt{z} }$
$=x \cos ^{-1} x-\frac{1}{2} \int z^{-\frac{1}{2}} d z$
$I=x \cos ^{-1} x-\frac{1}{2} \frac{z^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}+c$
$=x \cos ^{-1} x-\frac{1}{2} \times 2 z^{\frac{1}{2}}+c$\
$I=x \cos ^{-1} x-z^{\frac{1}{2}}+c$
$=x \cos ^{-1} x-\sqrt{1-x^{2}}+c$
Question 43
$\int \tan ^{-1} x d x$Sol :
$I=\int \tan ^{-1} x d x=\int \tan ^{-1} x \cdot 1 d x$
Let $u=\tan ^{-1} x$ and v=1
Now , $I=\int \tan ^{-1} x \cdot 1 d x$
$=\tan ^{-1} x \int 1 d x-\int\left(\frac{d \tan ^{-1} x}{d x} \times \int 1 d x\right) d x$
$I=\tan ^{-1} x \cdot x-\int \frac{1}{1+x^{2}} \times x d x$
$=x \tan ^{-1} x-\int \frac{x}{1+x^{2}} d x$
Let $z=1+x^{2}$ then dz=2xdx $\Rightarrow \frac{d z}{2}=x d x$
Now , $I=x \tan ^{-1} x-\int \frac{d z}{2z}$
$=x \tan ^{-1} x-\frac{1}{2} \int \frac{1}{2} d z$
$I=x \tan ^{-1} x-\frac{1}{2} \log z+c$
$=x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+c$
Question 44
$\int \sec ^{-1} x d x$Sol :
Let $\sec ^{-1} x d x=z \quad \Rightarrow x=\sec z$
dx=secz.tanzdz
Now , $\int z \cdot \sec z \cdot \tan z d z$
$=z \int \sec z \tan z d z-\int\left(\frac{d z}{d z} \times \int \sec z \tan z d\theta\right)dz$
$I=z \cdot \sec z-\int \sec z d x$
$=zsecz-\log (\sec z + \tan z)+c$
$I=\sec ^{-1} x \cdot x-\log (x+\sqrt{x^{2}-1})+c$
$I=x \sec ^{-1} x-\log (x+\sqrt{x^{2}-1})+c$
Question 45
(i) $\int \cos ^{-1} \frac{1-x^{2}}{1+x^{2}} d x$Sol :
$I=\int 2 \tan ^{-1} x d x$
$I=2\left[x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+c\right]$
$I=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+c$
(ii) $\int \sin ^{-1} \frac{2 x}{1+x^{2}} d x$
Sol :
$I=\int \sin ^{-1} \frac{2 x}{1+x^{2}} d x=2 \int \tan ^{-1} x d x$
$I=2 \int \tan ^{-1} x \cdot 1 d x$
$\left.I=2 [ x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+c\right]$
$I=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+c$
Question 46
$\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$Sol :
$I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$
Let $u=\cos ^{-1} x$ and $v=\frac{x}{\sqrt{1-x^{2}}}$
then $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$
$=\cos ^{-1} x \int \frac{x}{\sqrt{1-x^{2}}} d x-\int\left(\frac{d \cos ^{-1} x}{d x} \times \int \frac{x}{\sqrt{1-x^{2}}} d x\right) d x$
$I=\cos ^{-1} x \int \frac{x}{\sqrt{1-x^{2}}} d x-\int\left(\frac{-1}{\sqrt{1-x^{2}}} \times \int \frac{x}{\sqrt{1-x^{2}}}\right) d x$
let $1-x^{2}=z$ then dz=-2xdx $\Rightarrow x d x=\frac{-d z}{2}$
$I=\cos ^{-1} x \int \frac{-d z}{2\sqrt{z} }+\int \frac{1}{\sqrt{z}} x \int \frac{-d z}{\sqrt{2z} } d x$
$I=-\frac{1}{2} \cos ^{-1} x \int z^{-\frac{1}{2}} d z-\int\left(\frac{1}{\sqrt{2}} \times \frac{1}{2} \int z^{-\frac{1}{2}} d z\right) d x$
$I=-\frac{1}{2} \cos ^{-1} x \cdot 2 \sqrt{z}-\int\left(\frac{1}{\sqrt{z}} \times \frac{1}{2} 2 \sqrt{z}\right) d x$
$I=-\cos ^{-1} x \cdot \sqrt{z}-\int d x+c$
$I=-\sqrt{1-x^{2}} \cdot \cos ^{-1} x-x+c$
$I=-x-\sqrt{1-x^{2}} \cdot \cos ^{-1} x+c$
Question 47
$\int x^{2} \sin ^{-1} x d x$Sol :
$I=\int x^{2} \sin ^{-1} x d x$
Let $u=\sin ^{-1} x$ and $v=x^{2}$
then $I=\int x^{2} \sin ^{-1} x d x$
$=\sin ^{-1} x \int x^{2} d x-\int\left(\frac{d \sin ^{-1} x}{d x} \times \int x^{2} d x\right) d x$
$I=\sin ^{-1} x \cdot \frac{x^{3}}{3}-\int \frac{1}{\sqrt{1-x^{2}}} \times \frac{x^{3}}{3} d x$
$I=\frac{x^{3}}{3} \sin ^{-1} x-\frac{1}{3} \int \frac{x^{2} \cdot x}{\sqrt{1-x^{2}}} d x$
let $1-x^{2}=z$ $x^{2}=1-z$ then -2xdx=dz
$\Rightarrow x d x=-\frac{d z}{2}$
Now , $I=\frac{x^{3}}{3} \sin ^{-1} x-\frac{1}{3} \int \frac{(1-z)(-d z)}{2\sqrt{z}}$
$I=\frac{x^{3}}{3} \sin^{-1} x+\frac{1}{6} \int\left(\frac{1}{\sqrt{z}}-\frac{z}{\sqrt{z}}\right) d z$
$I=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6} \int\left(z^{-\frac{1}{2}}-\sqrt{z}\right) d z$
$I=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6} \int z^{-\frac{1}{2}} d z-\frac{1}{6} \int \sqrt{z} d z$
$I=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6} 2 z^{\frac{1}{2}}-\frac{1}{6} \times \frac{2}{3} z^{\frac{3}{2}}+c$
$I=\frac{x^{3} \sin ^{-1} x}{3}+\frac{1}{3} \sqrt{1-x^{2}}-\frac{1}{9}\left(1-x^{2}\right)^{\frac{3}{2}}+c$
Question 48
$\int \operatorname{cosec}^{3} x d x$Sol :
$I=\int \operatorname{cosec}^{3} x d x=\int \operatorname{cosec} x \cdot \operatorname{cosec}^{2} x d x$
Let u=cosecx and $v=\operatorname{cosec}^{2} x$
then $I=\int \operatorname{cosec}^{3} x d x$
$=\operatorname{cosec} x \int \operatorname{cosec}^{2} x d x-\int\left(\frac{d \operatorname{cosec} x}{d x} \times \int cosec^2 x dx\right)dx$
$I=\operatorname{cosec} x(-\cot x)-\int-\operatorname{cosec} x \cdot \cot x(-\cot x) d x$
$I=-\operatorname{cosec} x \cdot \cot c-\int \operatorname{cosec} x \cdot \cot ^{2} x d x$
$I=-\operatorname{cosec} x \cdot \cot x-\int \operatorname{cosec} x\left(\operatorname{cosec}^{2} x-1\right) d x$
$I=-\operatorname{cosec} x \cot x-\int\left(\operatorname{cosec}^{3} x-\operatorname{cosec} x\right) d x$
$I=-\operatorname{cosec} x \cdot \cot x-\int \operatorname{cosec}^{3} x d x+\int \operatorname{cosec} x d x$
$I=-\operatorname{cosec} x \cdot \cot x-I+\log \left|\tan \frac{x}{2}\right|+c$
$2 I=\log \left|\tan \frac{x}{2}\right|-\operatorname{cosec} x \cdot \cot x+c$
$I=\frac{1}{2} \log \left|\tan \frac{x}{2}\right|-\frac{1}{2} \operatorname{cosec} x \cdot \cot x+k$
Question 49
$\int e^{a x} \cdot \cos b x d x$Sol :
$I=\int e^{a x} \cdot \cos b x d x$
Let u=cosbx and $v=e^{a x}$
then $I=\int e^{a x} \cdot \cos b x d x$
$=\cos b x \int e^{a x} d x-\int\left(\frac{d \cos b x}{d x} \times \int e^{a x} d x\right)dx$
$I=\cos b x \cdot \frac{e^{a x}}{a}-\int \frac{-\sin b x\times b}{a} \times \frac{e^{a x}}{a} d x$
$I=\frac{\cos b x \cdot e^{a x}}{a}+\frac{b}{a} \int \sin b x \cdot e^{a x} d x$
$I=\frac{\cos b x \cdot e^{a x}}{a}+\frac{b}{a}[\sin b x] e^{a x} d x-\int\left(\frac{d \sin b x}{d x} \times \int e^{a x} d x\right) d x$
$I=\frac{\cos b x \cdot e^{a x}}{a}+\frac{b}{a}\left[\frac{\sin b x \cdot e^{a x}}{a}-\int \frac{\cos b x \times b}{a} \times \frac{e^{a x}}{a} d x\right]$
$I=\frac{\cos b x \cdot e^{a x}}{a}+\frac{b}{a}\left[\frac{\sin b x \cdot e^{a x}}{a}-\frac{a}{b} \int e^{a x} \cdot \cos b x d x\right]$
$I=\frac{\cos b x \cdot e^{a x}}{a}+\frac{b}{a^{2}} \sin b x \cdot e^{a x}-\frac{b^{2}}{a^{2}} I+c$
$I+I \frac{b^{2}}{a^{2}}=e^{a x}\left(\frac{\cos b x}{a}+\frac{\sin b x \cdot b}{a^{2}}\right)+c$
$I\left(1+\frac{b^{2}}{a^{2}}\right)=e^{a x}\left(\frac{a \cos b x+b \sin b x}{a^{2}}\right)+c$
$I\left(\frac{a^{2}+b^{2}}{a^{2}}\right)=e^{a x} \frac{(a \cos b x+b \sin b x)}{a^{2}}+c$
$I=\frac{e^{a x}}{a^{2}+b^{2}}(a \cos b x+b \sin b x)+c$
Question 50
(i) $\int e^{x} \cos x d x$Sol :
$I=\int e^{x} \cos x d x$
Let u=cos x and $v=e^{x}$
then $I=\int e^{x} \cos x d x$
$=\cos x \int e^{x} d x-\int\left(\frac{d \cos x}{d x} \times \int e^{x} d x\right) d x$
$I=\cos x \cdot e^{x}-\int-\sin x \cdot e^{x} d x=\cos x \cdot e^{x}+\int \sin x \cdot e^{x} d x$
$I=\cos x \cdot e^{x}+\sin x \int e^{x} d x-\int\left(\frac{d \sin x}{d x} \times \int e^{x} d x\right) d x$
$I=\cos x \cdot e^{x}+\sin x \cdot e^{x}-\int \cos x \cdot e^{x} d x$
$I=\cos x \cdot e^{x}+\sin x \cdot e^{x}-I+c$
$2 I=e^{x}(\cos x+\sin x)+c$
$I=\frac{e^{x}}{2}(\sin x+\cos x)+c$
(ii) $\int e^{x} \sin x d x$
Sol :
$I=\int e^{x} \sin x d x$
Let u=sinx and $v=e^{x}$
then $I=\int \sin x \cdot e^{x} d x$
$=\sin x \int e^{x} d x-\int\left(\frac{d \sin x}{d x} \times \int e^{x} d x\right) d x$
$I=\sin x \cdot e^{x}-\int \cos x e^{x} d x$
$I=\sin x \cdot e^{x}-\left[\cos x \cdot \int e^{x} d x-\int\left(\frac{d \cos x}{d x}\times \right.\left. \int e^{x} d x\right) d x\right]$
$I=\sin x \cdot e^{x}-\left[\cos x \cdot e^{x}-\int-\sin x \times e^{x} d x\right.$
$I=\sin x \cdot e^{x}-\cos x \cdot e^{x}-\int \sin x \cdot e^{x} d x$
$I=\sin x \cdot e^{x}-\cos x \cdot e^{x}-I+c$
$2 I=e^{x}(\sin x-\cos x)+c$
$I=\frac{e^{x}}{2}(\sin x-\cos x)+c$
Question 51
(i) $\int e^{-x} \sin x d x$Sol :
$I=\int e^{-x} \sin x d x$
Let u=sinx and $v=e^{-x}$
then $I=\int e^{-x} \sin x d x$
$=\sin x \int e^{-x} d x-\int\left(\frac{d \sin x}{d x} \times \int e^{-x} d x\right) d x$
$I=\sin x \cdot \frac{e^{-x}}{-1}-\int \cos x \times \frac{e^{-x}}{-1} d x$
$=-\sin x \cdot e^{-x}+\int \cos x \cdot e^{-x} d x$
$I=-e^{-x} \sin x+\left[\cos x\int e^{-x} d x-\int\left(\frac{d \cos x}{d x} \times \int e^{-x} d x\right) d x\right]$
$I=-e^{-x} \sin x+\left[\cos x \cdot \frac{e^{-x}}{-1}-\int-\sin x \times \frac{e^{-x}}{-1} d x\right.$
$I=-e^{-x} \sin x-\cos x \cdot e^{-x}-\int \sin x \cdot e^{-x} d x$
$I=-e^{-x} \sin x-\cos x \cdot e^{-x}-I$
$2 I=-e^{-x}(\sin x+\cos x)+c$
$I=\frac{-e^{-x}}{2}(\sin x+\cos x)+c$
(ii) $\int e^{2 x} \cdot \sin x d x$
Sol :
$I=\int e^{2x} \sin x d x$
Let u=sinx and $v=e^{2 x}$
then $I=\int e^{2 x} \sin x d x$
$=\sin x \int e^{2 x} d x-\int\left(\frac{d \sin x}{d x} \times \int e^{2 x} d x\right) d x$
$I=\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x$
$=\frac{e^{2x}}{2} \sin x-\frac{1}{2} \int \cos x \cdot e^{2} d x$
$I=\frac{e^{2x}}{2} \sin x-\frac{1}{2}\left[\cos x \cdot \int e^{2 x} d x-\int\left(\frac{d \cos x}{d x} \times \int e^{2 x} d x\right) d x\right]$
$I=\frac{e^{2 x}}{2} \sin x-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int-\sin x \frac{e^{2x}}{2} d x\right]$
$I=\frac{e^{2 x}}{2} \sin x-\frac{1}{4} \cos x \cdot e^{2 x}-\frac{1}{4} \int \sin x \cdot e^{2 x} d x$
$I=\frac{e^{2 x}}{2} \sin x-\frac{1}{4} \cos x \cdot e^{2 x}-\frac{1}{4} I+c$
$I+\frac{1}{4} I=\frac{e^{2 x}}{2}\left(\sin x-\frac{\cos x}{2}\right)$
$I\left(1+\frac{1}{4}\right)=\frac{e^{2 x}}{2}\left(\frac{2 \sin x-\cos x}{2}\right)$
$I\left(\frac{5}{4}\right)=\frac{e^{2 x}}{2}\left(\frac{2 \sin x-\cos x}{2}\right)+c$
$I=\frac{e^{2 x}}{2} \times \frac{4}{5} \frac{(2 \sin x-\cos x)}{2}+c$
$I=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c$
Question 52
$\int \frac{\cos ^{-1} x}{x^{2}} d x$Sol :
$I=\int \frac{\cos ^{-1} x}{x^{2}} d x$
putting $z=\cos ^{-1} x$ then x=cosz
⇒ dx=-sinzdz
Now , $I=\int \frac{\cos ^{-1} x}{x^{2}} d x$
$=\int \frac{z \cdot(-\sin z) d z}{\cos z \cdot \cos z}$
$=-\int z \cdot \sec z \cdot \tan z d z$
let u=z and v=sexz.tanz
then $I=-\int z \sec z \cdot \tan z d z$
$=-\left[z \int \sec z \cdot \tan z d z-\int\left(\frac{d z}{d z} \times \int \sec z. \tan zdz\right.\right]dz$
$I=-\left[z \cdot \sec z-\int \sec z d z\right]+c$
$I=-z \sec z+\log |\sec z+\tan z|+c$
$I=-\cos^{-1}x \times\frac{1}{x} +\log \left|\frac{1}{x}+\frac{\sqrt{1-x^{2}}}{x}\right|+c$
$I=-\frac{\cos ^{-1} x}{x}+\log \left|\frac{1+\sqrt{1-x^{2}}}{x}\right|+c$
Question 53
$\int x \cdot 2^{x} d x$Sol :
$I=\int x \cdot 2^{x} d x=\int x \cdot e^{x \log 2} d x$
Let u=x and $v=e^{x \log 2}$
then $I=\int x \cdot e^{x \log 2} d x$
$=x \int e^{x \log 2} d x-\int\left(\frac{d x}{d x} \times \int e^{x \log 2} d x\right) d x$
$I=x \cdot \frac{e^{x \log 2}}{\log 2}-\int \frac{e^{x \log 2}}{\log 2} d x$
$=\frac{x \cdot e^{x \log2}}{\log 2}-\frac{e^{x \log 2}}{(\log 2)^{2}}$
$I=\frac{x \cdot e^{x \log 2}}{\log 2}-\frac{e^{x \log 2}}{(\log 2)^{2}}+c$
$I=\frac{x \cdot 2^{x}}{\log 2}-\frac{2^{x}}{(\log 2)^{2}}+c$
Question 54
$\int \frac{\log x}{x^{2}} d x$Sol :
$I=\int \frac{\log x}{x^{2}} d x$
putting z=logx then $e^{z}=x \quad \Rightarrow e^{z} d z=d x$
Now , $I=\int \frac{\log x}{x^{2}} d x$
$=\int \frac{z}{e^{z} \times e^{z}} \times e^zd z=\int z \cdot e^{-z} d z$
$I=z \int e^{-z} d z-\int\left(\frac{d z}{d z} x \int e^{-z} d z\right) d z$
$I=z \cdot \frac{e^{-z}}{-1}-\int \frac{e^{-z}}{-1} d z$
$=-z \cdot e^{-z}+\frac{e^{-z}}{-1}+c$
$I=-e^{-z}(z+1)+c$
$=-\frac{1}{e^{z}}(\log x+1)+c$
$I=-\frac{(1+\log x)}{x}+c$
Question 55
$\int e^{\sin x} \cdot \sin 2 x d x$Sol :
$I=\int e^{\sin x} \cdot \sin 2 x d x$
putting z=sinx then dz=cosxdx
Now ,$I=\int e^{\sin x} \cdot 2 \sin x \cdot \cos x d x$
$I=2 \int e^{2} \cdot z d z=2\left[z \int e^{z} d z-\int\left(\frac{d z}{d z} \times \int e^{2} d z\right) d z \right]$
$I=2\left[z \cdot e^{z}-\int e^{z} dz\right]$
$=2\left[z \cdot e^{z}-e^{z}\right]+c$
$I=2\left(\sin x \cdot e^{\sin x}-e^{\sin x}\right)+c$
$I=2 e^{\sin x}(\sin x-1)+c$
$I=2(\sin x-1) \cdot e^{\sin x}+c$
Question 56
(i) $\int x^{2} \cdot e^{3 x} d x$Sol :
$I=\int x^{2} \cdot e^{3 x} d x$
Let $u=x^{2}$ and $v=e^{3 x}$
then $I=\int x^{2} \cdot e^{3 x} d x$
$=x^{2} \int e^{3 x} d x-\int\left(\frac{d x^{2}}{d x}\times \int e^{3 x} d x\right) d x$
$I=x^{2} \cdot \frac{e^{3 x}}{3}-\int 2 x \times \frac{e^{3 x}}{3} d x$
$=\frac{x^{2}}{3} \cdot e^{3 x}-\frac{2}{3} \int x \cdot e^{3 x} d x$
$I=\frac{x^{2}}{3} \cdot e^{3 x}-\frac{2}{3}\left[x \cdot \int e^{3 x} d x-\int\left(\frac{d x}{d x} \times \int e^{3 x} d x\right) d x\right]$
$I=\frac{x^{2}}{3} \cdot e^{2 x}-\frac{2}{3}\left[x \cdot \frac{e^{3 x}}{3}-\int \frac{e^{3 x}}{3} d x\right]$
$I=\frac{x^{2}}{3} \cdot e^{3 x}-\frac{2}{3}\left[\frac{x}{3} \cdot e^{3 x}-\frac{e^{3 x}}{9}\right]+c$
$I=\frac{x^{2}}{3} \cdot e^{3 x}-\frac{2 x}{9} \cdot e^{3 x}+\frac{2}{27} \cdot e^{3 x}+c$
(ii) $\int x^{2} \cdot e^{x} d x$
Sol :
$I=\int x^{2} \cdot e^{x} d x$
Let $u=x^{2}$ and $v=e^{x}$
then $\left.I=\int x^{2} \cdot e^{x} d x=x^{2} \int e^{x} d x-\int\left(\frac{d x^{2}}{d x} \times \right. e^{x} d x\right) d x$
$1=x^{2} \cdot e^{x}-\int 2 x \times e^{x} d x$
$=x^{2} \cdot e^{x}-2 \int x \cdot e^{x} d x$
$\left.I=x^{2} \cdot e^{x}-2\left[x \int e^{x} d x-\int\left(\frac{d x}{d x} \times \right. \int e^{x} d x\right) d x\right]$
$I=x^{2} \cdot e^{x}-2\left[x \cdot e^{x}-\int e^{x} d x\right]+c$
$I=x^{2} e^{x}-2\left[x \cdot e^{x}-e^{x}\right]+c$
$=x^{2} \cdot e^{x}-2 x \cdot e^{x}+2 e^{x}$
$I=e^{x}\left(x^{2}-2 x+2\right)+c$
Question 57
$\int x \cos ^{-1} x d x$Sol :
$I=\int x \cos ^{-1} x d x$
Let $u=\cos ^{-1} x$ and v=x
$I=\cos ^{-1} x \int x d x-\int\left(\frac{d \cos ^{-1} x}{d x} \times \int x d x\right) d x$
$I=\cos ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{-1}{\sqrt{1-x^{2}}} \times \frac{x^{2}}{2} d x$
$=\frac{x^{2}}{2} \cos ^{-1} x+\int \frac{x^{2}}{\sqrt{1-x^{2}}} d x$
let x=sinš¯›¼ then dx=cosš¯›¼dš¯›¼
$\alpha=\sin ^{-1} x$
then $I=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{2} \int \frac{\sin ^{2} \alpha \cos \alpha d \alpha}{\sqrt{1-\sin ^{2} \alpha}}$
$I=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{2} \int \frac{\sin ^{2} \alpha \cos \alpha d \alpha}{\cos \alpha}$
$=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{2 \times 2} \int 2 \sin ^{2} \alpha d \alpha$
$I=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{4} \int(1-\cos 2 \alpha) d \alpha$
$I=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{4}\left[\alpha-\frac{\sin 2 \alpha}{2}\right]+c$
$I=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{4} \alpha-\frac{1 \times 2 \sin \alpha \cdot \cos \alpha}{4 \times 2}+c$
$I=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{4} \sin ^{-1} x-\frac{1}{4} x \sqrt{1-x^{2}}+c$
Question 58
$\int x \sin^{-1} x d x$Sol :
$I=\int x \sin ^{-1} x d x$
Let $\sin ^{-1} x=u$ and v=x
$I=\int x \sin ^{-1} x d x=\sin ^{-1} x \int x d x-\int\left(\frac{d \sin ^{-1} x}{d x} \times \int x d x\right) d x$
$I=\sin ^{2} x \cdot \frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}} \times \frac{x^{2}}{2} d x$
$=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{2} \int \frac{x^{2}}{\sqrt{1 \cdot x^{2}}} d x$
Let $x=\sin \alpha$ then $d x=\cos \alpha d \alpha$
$I=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{2} \int \frac{\sin ^{2} \alpha}{\sqrt{1-\sin ^{2} \alpha}} \times {\cos \alpha d \alpha}$
$I=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{2} \int \frac{\sin ^{2} \alpha \cdot \cos \alpha d \alpha}{\cos \alpha}$
$=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{2 \times 2} \int 2 \sin ^{2} \alpha d \alpha$
$I=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{4} \int(1-\cos 2 \alpha) d a$
$=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{4}\left[\alpha-\frac{\sin 2 \alpha}{2}\right]+c$
$I=\frac{x^{2}}{2} \sin x-\frac{1}{4} \alpha+\frac{1 \times 2 \sin \alpha \cdot \cos \alpha}{4 \times 2}+c$
$I=\frac{x^{2}}{2} \sin ^{-1} x-\frac{1}{4} \sin ^{-1} x+\frac{1}{4} x \sqrt{1-x^{2}}+c$
Question 59
(i) $\int\left(\sin ^{-1} x\right)^{2} d x$Sol :
$I=\int\left(\sin ^{-1} x\right)^{2} d x=\int\left(\sin ^{-1} x\right)^{2} \cdot 1 d x$
Let $u=\left(\sin ^{-1} x\right)^{2}$ amd v=1
then $\left.I=\int\left(\sin ^{-1} x\right)^{2} \cdot 1 d x=\left(\sin ^{-1} x\right)^{2} \int 1 d x-\int\left(\frac{d}{d x} \sin ^{-1} \times \right)^{2} x \int 1 d x\right) d x$
$I=\left(\sin ^{-1} x\right)^{2} \cdot x-\int \frac{2 \sin ^{2} x \times 1}{\sqrt{1-x^{2}}} x \times d x$
$I=x\left(\sin ^{-1} x\right)^{2}-2 \int \sin ^{-1} x \cdot \frac{x}{\sqrt{1-x^{2}}} d x$
$\left.I=x\left(\sin ^{-1} x\right)^{2}-2[\sin ^{-1} x\int \frac{x}{\sqrt{1-x^{2}}} d x-\int\left(\frac{d \sin^{-1} x}{d x}\right) \cdot \int \frac{x}{\sqrt{1-x^{2}}} d x\right] d x$
Let $z=1-x^{2}$ then dz=-2xdx $\Rightarrow x d x=-\frac{d z}{2}$
Now , $\left.I=x\left(\sin ^{-1} x\right)^{2}-2\left[\sin ^{-1} x\right. \int \frac{-d z}{\sqrt{2} 2}-\int\left(\frac{1}{(\sqrt{1-x^{2}}} \times \int \frac{-d z}{\sqrt{z} 2}\right)\right]+c$
$I=x\left(\sin ^{-1} x\right)^{2}-2\left[-\frac{1}{2} \sin ^{-1} x \int z^{-\frac{1}{2}} d z+\int\left(\frac{1}{\sqrt{z}} \times \frac{1}{2} \int z^{-\frac{1}{2}} d z\right) d x\right.$
$I=x\left(\sin ^{-1} x\right)^{2}-2\left[-\frac{1}{2} \sin ^{-1} 2\sqrt{z}+\int\left(\frac{1}{\sqrt{z}} \times \frac{1}{2} \int z^{-\frac{1}{2}} d z\right) d x\right.$
$\left.I=x\left(\sin ^{-1} x\right)^{2}-2\left[-\sin ^{-1} x\right.\sqrt{1-x^{2}}+x\right]+c$
$I=x\left(\sin ^{-1} x\right)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+c$
(ii) $\int \sin ^{-1}\left(3 x-4 x^{3}\right) d x$
Sol :
$I=\int \sin ^{-1}\left(3 x-4 x^{3}\right) d x=\int 3 \sin ^{-1} x d x=3 \int \sin ^{-1} x d x$
$I=3\left[\sin ^{-1} x \int 1 d x-\int\left(\frac{d}{d x} \sin ^{-1} x \times \int 1 d x\right) d x\right]$
$I=3\left[\sin ^{-1} x \cdot x-\int \frac{x}{\sqrt{1-x^{2}}} d x\right]$
Let $z=1-x^{2}$ then dz=-2xdx $\Rightarrow-\frac{d z}{2}=x d x$
$I=3\left[x \sin ^{-1} x-\int \frac{-d z}{\sqrt{z} 2}\right]+C$
$I=3\left[x \sin ^{-1} x+\frac{1}{2} \int z^{-\frac{1}{2}} d z\right]+c$
$=3\left[x \sin ^{-1} x+\frac{1}{2} \times 2 \sqrt{z}\right]+c$
$I=3 x \sin ^{-1} x+3 \sqrt{1-x^{2}}+c$
Question 60
$\left.\int \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right.}{x^{4}}\right] d x$Sol :
$I=\left.\int \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right.}{x^{4}}\right] d x$
$I=\int \sqrt{1+\frac{1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right) \cdot \frac{1}{x^{3}} d x$
putting $z=1+\frac{1}{x^{2}}$ then $d z=-2 x^{-3} d x$
$d z=-\frac{2}{x^{3}} d x$
$-\frac{d z}{2}=\frac{1}{x^{2}} d x$
Now , $I=\int \sqrt{z} \log z \cdot\left(-\frac{d z}{2}\right)=-\frac{1}{2} \int \sqrt{z} \log z d z$
$I=-\frac{1}{2}\left[\log z \int \sqrt{z} d z-\int\left(\frac{d}{d z} \log z x \int \sqrt{z} d z\right) d z\right]$
$I=-\frac{1}{2}\left[\log 2 \cdot \frac{2}{3} z^{\frac{3}{2}}-\int\left(\frac{1}{2} \times \frac{2}{3} \cdot z^{\frac{3}{2}}\right) d z\right]$
$I=-\frac{1}{2}\left[\frac{2}{3} z^{\frac{3}{2}} \log z-\frac{2}{3} \int z^{\frac{1}{2}} d z\right]+c$
$I=-\frac{1}{2}\left[\frac{2}{3} z^{\frac{3}{2}} \log z-\frac{2}{3} \times \frac{2}{3} \cdot z^{\frac{3}{2}}\right]+c$
$I=-\frac{1}{3} z^{\frac{3}{2}} \log z+\frac{2}{9} z^{\frac{3}{2}}+c$
$I=\frac{1}{9} z^{\frac{3}{2}}\left[-3 \log z+2\right]+c$
$=\frac{1}{9} z^{\frac{3}{2}}[2-3 \log z]+c$
$I=\frac{1}{9}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}\left[2-3 \log \left(1+\frac{1}{x^{2}}\right)\right]+c$
Question 61
$\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$Sol :
$I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$
putting $x=\cos 2 \theta$ $\begin{aligned} d x &=-\sin 2 \theta \cdot 2 d \theta \\ d x &=-2 \sin 2 \theta d \theta \end{aligned}$
Now , $I=\int \tan ^{-1} \sqrt\frac{1-\cos 2 \theta}{1+\cos 2 \theta}(-2 \sin 2 \theta) d \theta$
$I=\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}} \times(-2 \sin 2 \theta)$
$I=\int \tan ^{-1} \tan \theta \times(-2 \sin 2 \theta) d \theta$
$I=-2 \int \theta \cdot \sin 2 \theta d \theta$
$=-2\left(\theta \int \sin 2 \theta-\int\left(\frac{d \theta}{d \theta} \times \int \sin 2 \theta d \theta\right) d \theta\right.$
$I=-2\left[\theta\left(\frac{-\cos 2 \theta}{2}\right)-\int \frac{-\cos 2 \theta}{2} d \theta\right]$
$=-2\left[-\frac{\theta \cos 2 \theta}{2}+\frac{\sin 2 \theta}{4}\right]+c$
$I=\theta \cos 2 \theta-\frac{1}{2} \sin 2 \theta+c$
$I=x \cdot \frac{1}{2} \cos ^{2} x-\frac{1}{2} \sqrt{1-x^{2}}+c$
$=\frac{x}{2} \cos ^{-1} x-\frac{1}{2} \sqrt{1-x^{2}}+c$
Question 62
$\int \frac{x-\sin x}{1-\cos x} d x$Sol :
$I=\int \frac{x-\sin x}{1-\cos x} d x$
$=\int \frac{x-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}} d x$
$I=\int \frac{x}{2 \sin ^{2} \frac{x}{2}} d x-\int \frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} d x$
$I=\frac{1}{2} \int x \cdot \operatorname{cosec}^{2} \frac{x}{2}-\int \cot \frac{x}{2} d x$
$I=\frac{1}{2}\left[x \int \operatorname{cosec}^{2} \frac{x}{2} d x-\int\left(\frac{d x}{d x} \times \int \operatorname{cosec}^{2} x d x\right) d x\right]-\int \cot \frac{x}{2} d x$
$I=\frac{1}{2}\left[2 x\left(-\cot \frac{x}{2}\right)-\int-2 \cot \frac{x}{2} d x\right]-\int \cot \frac{x}{2} d x$
$I=\frac{1}{2} \times \left(-2 x \cot \frac{x}{2}\right)+\frac{1}{2} \times 2 \int \cot \frac{x}{2}-\int \cot \frac{x}{2} d x$
$I=-x \cot \frac{x}{2}+\int \cot \frac{x}{2} d x-\int \cot \frac{x}{2} d x+c$
$I=-x \cot \frac{x}{2}+c$
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