Mathematics (2018) | JEE 2018 | Discussion Page

The Integral

$\int_{}^{} \frac{\sin^{2}x cos^{2}x}{(\sin^{5}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x)^{2}}dx$

is equal to (where C is constant of integration)

$\frac{1}{3( 1+tan^{3}x)}+C$

$\frac{-1}{3( 1+tan^{3}x)}+C$

$\frac{1}{ 1+\cot^{3}x}+C$

$\frac{-1}{ 1+\cot^{3}x}+C$

Option B

We have

I= $\int_{}^{} \frac{\sin^{2}x cos^{2}x}{(\sin^{5}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x)^{2}}dx$

$\int_{}^{}\frac{\sin^{2}x cos^{2}x}{( \sin^{3}x( \sin^{2}x+\cos^{2}x)+\cos^{3}x( \sin^{2}x+\cos^{2}x))^{2}}dx$

= $\int_{}^{}\frac{\sin^{2}x cos^{2}x}{( \sin^{3}x+\cos^{3}x)^{2}}dx$

= $\int_{}^{}\frac{\sin^{2}x cos^{2}x}{\cos^{6}x( 1+\tan^{3}x)^{2}}dx$

= $\int_{}^{} \frac{\tan^{2}x\sec^{2}x}{( 1+\tan^{3}x)^{2}}dx$

Put $\tan^{2}x=1\Rightarrow 3\tan^{3}x \sec^{2}x dx=dt$

$\therefore$ $I=\frac{1}{3}\int_{}^{} \frac{dt}{( 1+t)^{2}}$

$\Rightarrow$ I= $\frac{-1}{3( 1+t)}+C$

$\Rightarrow$ $I= \frac{-1}{3( 1+\tan^{3}x)}+C$

Discussion Forum