KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.6

Exercise 19.6

Question 1
$\int \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x$
Sol :
$=\int \sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}} d x=\int \frac{\sin x}{\cos x} d x$

$=\int \tan x d x=\log |\sec x|+c$


Question 2
$\int \frac{d x}{\sqrt{1-\cos 2 x}}$
Sol :
$=\int \frac{d x}{\sqrt{2 \sin ^{2} x}}=\int \frac{d x}{\sqrt{2} \sin x}$

$=\frac{1}{\sqrt{2}} \int \operatorname{cosec} x d x$

$=\frac{1}{\sqrt{2}} \log \left|\tan \frac{x}{2}\right|+c$


Question 3
$\int \frac{d x}{\sqrt{1+\cos x}}$
Sol :
$=\int \frac{d x}{\sqrt{2 \cos ^{2} \frac{x}{2}}}=\frac{1}{\sqrt{2}} \int \frac{1}{\cos \frac{x}{2}} d x$

$=\frac{1}{\sqrt{2}} \int \sec \frac{x}{2} d x$

$=\frac{1}{\sqrt{2}} \log \left|\sec \frac{x}{2}+\tan \frac{x}{2}\right|+c$


Question 4
$\int \sec (a x+b) d x$
Sol :
$=\frac{1}{a} \log |\sec (a x+b)+\tan (a x+b)|$


Question 5
$\int \operatorname{cosec}(2 x+3) d x$
Sol :
$=\frac{1}{2} \log \left|\tan \left(\frac{2 x+3}{2}\right)\right|+c$

$=\frac{1}{2} \log \left|\tan \left(x+\frac{3}{2}\right)\right|+c$

Question 6
$=\int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x\right)}$
Sol :
$=\int \frac{d x}{\sqrt{2}\left(\cos x \cdot \cos \frac{\pi}{4}-\sin x \cdot \sin \frac{\pi}{4}\right)}$

$=\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x+\frac{\pi}{4}\right)}$

$=\frac{1}{\sqrt{2}} \int \sec \left(x+\frac{\pi}{4}\right) d x$

$=\frac{1}{\sqrt{2}} \log \left|\sec \left(x+\frac{\pi}{4}\right)+\tan \left(x+\frac{x}{4}\right)\right|+c$

Question 7
$\int \frac{d x}{\cos x \cdot \sin ^{2} x}$
Sol :
$=\int \frac{1}{\cos x \cdot \sin ^{2} x} d x=\int \frac{\left(\cos ^{2} x+\sin ^{2} x\right) d x}{\cos x \cdot \sin ^{2} x}$

$=\int \frac{\cos ^{2} x}{\cos x \cdot \sin ^{2} x} d x+\int \frac{\sin ^{2} x}{\cos x \cdot \sin ^{2} x} d x$

$=\int \frac{\cos x}{\sin x \cdot \sin x} d x+\int \sec x d x$

$=\int \cot x \cdot \operatorname{cosec} x d x+\log |\sec x+\tan x|+c$

=-cosecx+log|secx+tanx|+c

Question 8
$\int \frac{1+\cos x}{\sin x} d x$
Sol :
$=\int \frac{1}{\sin x} d x+\int \frac{\cos x d x}{\sin x}$

$=\int \operatorname{cosec} x d x+\int \cot x d x$

$=\log \left|\tan \frac{x}{2}\right|+\log |\sin x|+c$

Question 9
$\int \frac{d x}{\sqrt{3} \cos x+\sin x}$
Sol :
$=\int \frac{d x}{2\left(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right)}$

$=\frac{1}{2} \int \frac{d x}{\left(\cos \frac{\pi}{6} \cos x+\sin \frac{\pi}{6} \cdot \sin x\right)}$

$=\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\pi}{6}\right)}$

$\frac{1}{\sqrt{2}} \int \sec \left(x-\frac{\pi}{6}\right) d x$

$=\frac{1}{\sqrt{2}} \log \left|\sec \left(x-\frac{\pi}{6}\right)+\tan \left(x-\frac{\pi}{6}\right)\right|+c$


Question 10
$\int \frac{d x}{a \sin x+b \cos x}$
Sol :
Let a=rcosɑ  , b=rsinɑ

∴ asinx+bcosx=rcosɑ.sinx+rsinɑ.cosx

=r(sinx.cosɑ+cosx.sinɑ)
=rsin(x+ɑ)
aisnx+bcosx=rsin(x+ɑ)

Now  , r²sin²α+r²cos²α=a²+b²
r²(sin²α+cos²α)=a²+b²
r²=a²+b²
r=√a²+b²

also , $\frac{b}{a}=\frac{r\sin \alpha}{r\cos a}=\tan \alpha$

$\alpha=\tan ^{-1} \frac{b}{a}$

A.T.Q
$\int \frac{d x}{a \sin x+b \cos x}$

$=\int \frac{d x}{r \sin (x+\alpha)}$

$=\int \frac{d x}{\sqrt{a^{2}+b^{2}} \cdot \sin (x+\alpha)}$

$=\frac{1}{\sqrt{a^{2}+b^{2}}} \int \operatorname{cosec}(x+\alpha) d x$

$=\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \frac{(x+\alpha)}{2}\right|+c$

$=\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \left(\frac{x}{2}+\frac{\alpha}{2}\right)\right|+c$

$=\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \left(\frac{x}{2}+\frac{1}{2}tan^{-1}\frac{b}{\alpha}\right) \right|+c$


Question 11
$\int \frac{d x}{\sin x+\cos x}$
Sol :
$=\int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \alpha+\frac{1}{\sqrt2} \cos x\right)}$

$=\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \frac{\pi}{4} \cdot \sin x+\cos \frac{\pi}{4} \cdot \cos x}$

$=\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\pi}{4}\right)}$

∴cos(A-B)=cosA.cosB+sinA.sinB

$=\frac{1}{\sqrt{2}} \int \sec \left(x-\frac{\pi}{4}\right) d x$

$=\frac{1}{\sqrt{2}} \log \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right|+c$



Question 12
$\int \frac{\sec x}{b+a \tan x}$
Sol :
$=\int \frac{1 d x}{\cos x \cdot\left(b+\frac{a \sin x}{\cos x}\right)}$

$=\int \frac{d x}{\cos x\left(\frac{b \cos x+a \sin x}{\cos x}\right)}$

$=\int \frac{d x}{b \cos x+\operatorname{asin} x}$

$=\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \left(\frac{x}{2}+\frac{1}{2} \tan ^{-1} \frac{b}{a}\right)\right|+c$

Question 13
$\int \tan 2 x \cdot \tan 5 x \cdot \tan 7 x d x$
Sol :
$\because \quad 7 x=2 x+5 x$

$\Rightarrow \tan 7 x=\tan (2 x+5 x)$

$\Rightarrow \frac{\tan 7 x}{1}=\frac{\tan 2 x+\tan 5 x}{1-\tan 2 x \cdot \tan 5 x}$

$\Rightarrow \tan 7 x(1-\tan 2 x \cdot \tan 5 x)=1 \times(\tan 2 x+\tan 5 x)$

$\Rightarrow \tan 7 x-\tan 7 x \cdot \tan 2 x \cdot \tan 5 x=\tan 2 x+\tan 5 x$

$\Rightarrow \tan 7 x-\tan 2 x-\tan 5 x=\tan 2 x \cdot \tan 5 x \cdot \tan 7 x$


A.T.Q $\int \tan 2 x \cdot \tan 5 x \cdot \tan 7 x d x=\int \tan 7 x-\tan 2 x-\tan 5 x d x$

$=\int \tan 7 x d x-\int \tan 2 x d x-\int \tan 5 x d x$

$\frac{1}{7} \log |\sec 7 x|-\frac{1}{2} \log |\sec 2 x|-\frac{1}{5} \log |\sec 5 x|+c$


Question 14
$\int \frac{d x}{\sec x+\operatorname{cosec} x}$
Sol :
$=\int \frac{d x}{\frac{1}{\cos x}+\frac{1}{\sin x}}$


$=\int \frac{d x}{\frac{\sin x+\cos x}{\sin x \cdot \cos x}}$

$=\int \frac{\sin x \cdot \cos x}{\sin x+\cos x} d x$

$=\int \frac{2 \sin x \cdot \cos x}{2(\sin x+\cos x)} d x$

$=\frac{1}{2} \int \frac{1+2 \sin x \cdot \cos x-1}{(\sin x+\cos x)} d x$

$=\frac{1}{2} \int \frac{\sin ^{2} x+\cos ^{2} x+2 \sin x \cdot \cos x-1}{(\sin x+\cos x)}dx$

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

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

$=\frac{1}{2} \int(\sin x+\cos x) d x-\frac{1}{2} \int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\cos x \cdot \frac{1}{\sqrt{2}}\right)}$

$=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2 \sqrt{2}} \int \frac{d x}{\cos \frac{\pi}{4} \cdot \sin x+\cos x \cdot \sin \frac{x}{4}}$

$=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2 \sqrt{2}} \int \frac{d x}{\sin \left(x+\frac{\pi}{4}\right)}$

$=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2 \sqrt{2}} \int \operatorname{cosec}\left(x+\frac{\pi}{4}\right) d x$

$=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2 \sqrt{2}} \log \left|\tan \left(\frac{x+\frac{\pi}{4}}{2}\right)\right|+c$

$=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2 \sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{8}\right)\right|+c$

$=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{\pi}{8}+\frac{x}{2}\right)\right|\right]+c$


Question 15
$\int \frac{d x}{\sin x+\tan x}$
Sol :
$=\int \frac{d x}{\sin x+\frac{\sin x}{\cos x}}$

$=\int \frac{d x}{\frac{\sin x \cdot \cos x+\sin x}{\cos x}}$

$=\int \frac{\cos x d x}{\sin x(1+\cos x)}$

$=\int \frac{\cos \frac{x}{2}-\sin^2 \frac{ x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2} \cdot 2 \cos ^{2} \frac{x}{2}}$

$=\frac{1}{4} \int \frac{\cos ^{2} \frac{x}{2}-\sin \frac{2 x}{2}}{\sin \frac{x}{2} \cdot \cos ^{3} \frac{x}{2}}$

$=\frac{1}{4} \int \frac{\cos ^{2} \frac{x}{2}}{\sin \frac{x}{2} \cdot \cos ^{3} \frac{x}{2}} d x-\frac{1}{4} \int \frac{\sin ^{2} \frac{x}{2}}{\sin \frac{x}{2} \cdot \cos ^{3} \frac{x}{2}} d x$

$=\frac{1}{2} \int \frac{1}{2 \sin \frac{\pi}{2} \cdot \cos \frac{x}{2}} d x-\frac{1}{4} \int \frac{\sin \frac{x}{2}}{\cos \frac{x}{2} \cdot \cos ^{2} \frac{x}{2}} d x$

$=\frac{1}{2} \int \frac{1}{\sin x} d x-\frac{1}{4} \int \tan \frac{x}{2} \cdot \sec ^{2} \frac{x}{2} d x$

$=\frac{1}{2} \log \left|\tan \frac{x}{2}\right|-\frac{1}{4} \int \tan \frac{x}{2} \cdot \sec ^{2} \frac{x}{2} d x$

Let $z=\tan \frac{x}{2}$ then $d z=\frac{1}{2} \sec ^{2} \frac{x}{2} d x$

$2 d z=\sec ^{2} \frac{x}{2} d x$

Now , $\frac{1}{2} \log \left|\tan \frac{x}{2}\right|-\frac{1}{4} \int z \cdot 2 d z$

$=\frac{1}{2} \log \left|\tan \frac{x}{2}\right|-\frac{2}{4} \cdot \frac{z^{2}}{2}+c$

$\frac{1}{2} \log \left|\tan \frac{x}{2}\right|-\frac{1}{4} z^{2}+c$

$=\frac{1}{2} \log \left|\tan \frac{x}{2}\right|-\frac{1}{4} \tan ^{2} \frac{x}{2}+c$


Question 16
$\int \frac{d x}{\sin x \cdot \cos ^{2} x}$
Sol :
$=\int \frac{1}{\sin x \cdot \cos ^{2} x} d x$

$=\int \frac{\left(\sin ^{2} x+\cos ^{2} x\right)}{\sin x \cos ^{2} x} d x$

$=\int \frac{\sin ^{2} x}{\sin x \cdot \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin x \cdot \cos ^{2} x} d x$

$=\int \frac{\sin x}{\cos ^{2} x} d x+\int \frac{1}{\sin x} d x$

$=\int \frac{\sin x}{\cos ^{2} x} d x+\int \operatorname{cosec} x d x$

Let z=cosx then dz=-sinxdx

-dz=sinxdx

Now , $\int \frac{\sin x d x}{\cos ^{2} x}+\log \left|\tan \frac{x}{2}\right|+c$

$=\int \frac{-d z}{z^{2}}+\log \left|\tan \frac{x}{2}\right|+c$

$=-\int z^{-2} d z+\log \left|\tan \frac{x}{2}\right|+c$

$=-\frac{z^{-2+1}}{-2+1}+\log \left|\tan \frac{x}{2}\right|+c$

$=-\frac{z^{-1}}{-1}+\log \left|\tan \frac{x}{2}\right|+c$

$=\frac{1}{z}+\log \left|\tan \frac{x}{2}\right|+c$

$=\frac{1}{\cos x}+\log \left|\tan \frac{x}{2}\right|+c$

$=\sec x+\log \left|\tan \frac{x}{2}\right|+c$


Question 17
$\int \frac{d x}{\sqrt{13+5 \cos x-12 \sin x}}$
Sol :
Let 5=rcosɑ , 12=rsinɑ

∴ 5cosɑ-12sinx=rcosɑ.cosx-rsinɑ.sinx
=r(cosɑ.cosx-sinɑ.sinx)
=r.cos(ɑ+x)
=r.cos(x+ɑ)

r=√a²+b²
r=√5²+12²
r=√25+144
r=√169=13

=13cos(x+ɑ)

∴5cosɑ ⇒$\cos \alpha=\frac{5}{13}$

$\alpha=\cos ^{-1} \frac{5}{13}$

Now , $\int \frac{d x}{\sqrt{13+(5 \cos x-12 \sin x)}}$

$=\int \frac{d x}{\sqrt{13+13 \cos (x+d)}}$

$=\int \frac{d x}{\sqrt{13(1+\cos (x+\alpha)}}$

$=\frac{1}{\sqrt{13}} \int \frac{d x}{\sqrt{1+\cos (x+\alpha)}}$

$=\frac{1}{\sqrt{13}} \int \frac{d x}{\sqrt{2 \cos ^{2}\left(\frac{x+\alpha}{2}\right)}}$

$=\frac{1}{\sqrt{26}} \int \frac{d x}{\cos \left(\frac{x+\alpha}{2}\right)}$

$=\frac{1}{\sqrt{2} 6} \int \sec \left(\frac{x+\alpha}{2}\right) d x$

$\frac{1 \times 2}{\sqrt{2} 6} \log \left|\sec \left(\frac{x+\alpha}{2}\right)+\tan \left(\frac{x+\alpha}{2}\right)\right|+c$

 $\frac{2}{\sqrt{26}} \log \left|\sec \left(\frac{x+\alpha}{2}\right)+\tan \left(\frac{x+\alpha}{2}\right)\right|+c$

where $\alpha=\cos ^{-1} \frac{5}{13}$


Question 18
$\int \frac{\sin x}{\sin (x+\alpha)} d x$
Sol :
$=\int \frac{\sin (x+\alpha-\alpha)}{\sin (x+\alpha)} d x$

$=\int \frac{\sin (x+\alpha) \cdot \cos \alpha-\cos (x+\alpha) \cdot \sin \alpha}{\sin (x+\alpha)} d x$

$=\int \frac{\sin (x+\alpha) \cdot \cos \alpha}{\sin (x+\alpha)} d x-\int \frac{\cos (x+\alpha) \cdot \sin \alpha}{\sin (x+\alpha)} d x$

$=\int \cos \alpha d x-\sin \alpha \int \cot (x+\alpha) d x$

$=\cos \alpha \int d x-\sin \alpha \int \cot (x+\alpha) d x$

=x.cosɑ-sinɑ.log|sin(x+ɑ)|+c


Question 19
$\int \tan (x-\theta) \cdot \tan 2 x \cdot \tan (x+\theta) d x$
Sol :
∴ tan2x=tan[(x+θ)+(x-θ)]

$\frac{\tan 2 x}{1}=\frac{\tan (x+\theta)+\tan (x-\theta)}{1-\tan (x+\theta) \cdot \tan (x-\theta)}$

⇒tan2x(1-tan(x+θ).tan(x-θ)=tan(x+θ)+tan(x-θ))

⇒tan2x-tan2x.tan(x+θ).tan(x-θ)=tan(x+θ)+tan(x-θ)

⇒tan2x-tan(x+θ)-tan(x-θ)=tan2x.tan(x+θ)+tan(x-θ)

⇒tan2x.tan(x+θ).tan(x-θ)=tan2x-tan(x+θ)-tan(x-θ)

⇒tan(x-θ).tan2x.tan(x+θ)=tan2x-tan(x+θ)-tan(x-θ)

A.T.Q $\int \tan (x-\theta) \cdot \tan 2 x \cdot \tan (x+\theta) d x$

$=\int \tan 2 x-\tan (x+\theta)-\tan (x-\theta)dx$

$=\int \tan 2 x d x-\int \tan (x+\alpha) d x-\int \tan (x-\alpha) d x$

$=\frac{1}{2} \log |\sec 2 x|-\log |\sec (x+\theta)|-\log |\sec (x-\theta)|+c$

$\frac{1}{2} \log |\sec 2 x|-\log |\sec (x-\theta)|-\log |\sec (x+\theta)|+c$


Question 20
$\int \frac{\cos x}{\sqrt{1+\cos x}} d x$
Sol :
$=\int \frac{2 \cos ^{2} \frac{x}{2}-1}{\sqrt{2 \cos ^{2} \frac{x}{2}}} d x$

$=\frac{1}{\sqrt{2}} \int \frac{2 \cos^2 \frac{x}{2}-1}{\cos \frac{x}{2}} d x$

$=\frac{2}{\sqrt{2}} \int \frac{\cos ^{2} \frac{x}{2}}{\cos \frac{x}{2}} d x-\frac{1}{\sqrt{2}} \int \frac{1}{\cos \frac{x}{2}} d x$

$=\sqrt{2} \int \cos \frac{x}{2} d x-\frac{1}{\sqrt{2}} \int \sec \frac{x}{2} d x$

$=2 \sqrt{2} \cdot \sin \frac{x}{2}-\frac{1 \times 2}{\sqrt{2}} \log \left|\sec \frac{x}{2}+\tan \frac{x}{2}\right|+c$

$=2 \sqrt{2} \sin \frac{x}{2}-\sqrt{2} \log \left|\sec \frac{x}{2}+\tan \frac{x}{2}\right|+c$




Question 21

Question 22
$\int \frac{\sin 2 x}{\sin (x-\alpha) \cdot \sin (x+\alpha)} d x$
Sol :
$=\int \frac{\sin [(x-\alpha)+(x+\alpha)]}{\sin (x-\alpha) \cdot \sin (x+\alpha)} d x$

$=\int \frac{\sin (x-a) \cdot \cos (x+\alpha)+\cos (x-a) \cdot \sin (x+\alpha)}{\sin (x-\alpha) \cdot \sin (x+a)} d x$

$=\int \frac{\sin (x-\alpha) \cdot \cos (x+\alpha)}{\sin (x-\alpha) \cdot \sin (x+\alpha)} d x+\int \frac{\cos (x-\alpha) \cdot \sin (x+\alpha)}{\sin (x-\alpha) \cdot \sin (x+\alpha)} d x$

$=\int \cot (x+\alpha) d x+\int \cot (x-\alpha) d x$

=log|sin(x+ɑ)|+log|sin(x-ɑ)|+c

=log|sin(x+ɑ).sin(x-ɑ)|+c

∵logm+logn=logmn

=log|sin²x-sin²ɑ|+c


Question 23

$\int \frac{d x}{\cos (x+a) \cdot \cos (x+b)}$

Sol :

$=\int \frac{\sin (b-a)}{\sin (b-a) \cdot \cos (x+a) \cdot \cos (x+b)}$

$=\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\cos (x+a) \cdot \cos (x+b)} d x$

$=\frac{1}{\sin (b-a)} \int \frac{\sin [(x+b)-(x+a)]}{\cos (x+a) \cdot \cos (x+b)} d x$

$=\frac{1}{\sin (b-a)} \int \frac{\sin (x+b) \cdot \cos (x+a)-\cos (x+b) \cdot \sin (x+a)}{\cos (x+a) \cdot \cos (x+b)} d x$

$\left.=\operatorname{cosec}(b-a) \int \frac{\sin (x+b) \cdot \cos (x+a)}{\cos (x+a) \cdot \cos (x+b)}-\frac{\cos (x+b) \cdot \sin (x+a)}{\cos (x+a) \cdot \cos (x+b)}\right) d x$

$=\operatorname{cosec}(b-a) \int \tan (x+b)-\tan (x+a) d x$

$=\operatorname{cosec}(b-a)[\log |\sec (x+b)|-\log |\sec (x+a)|]+c$

$\because \log \frac{m}{m}=\log{m}-log n$

$=\operatorname{cosec}(b-a) \cdot \log \left|\frac{\sec (x+b)}{\sec (x+a)}\right|+c$