$ \text { The integral } \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x \text { is…
$
\text { The integral } \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x \text { is equal to: }
$
-
$\frac{1}{\left(1+\cot ^3 x\right.}+c$
-
$\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$
-
$\frac{\sin ^3 x}{\left(1+\cos ^3 x\right.}+c$
-
$-\frac{\cos ^3 x}{3\left(1+\sin ^3 x\right.}+c$
Solution
Let $I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$
$
\begin{aligned}
&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\sin ^3 x+\cos ^3 x}\right)^2 d x \\
&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\cos ^3 x\left(1+\tan ^3 x\right)}\right)^2 d x \\
&=\int\left(\frac{\sin x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)}\right)^2 d x \\
&\text { Put } 1+\tan ^3 x=\mathrm{t} \\
&d t=3 \tan ^2 x \sec ^2 x d x \text { or } d x \\
&=\frac{d t}{3 \tan ^2 x \sec ^2 x} \\
&\therefore \mathrm{I}=\int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{3 \tan ^2 x \sec ^2 x}
\end{aligned}
$
$
\begin{aligned}
&\mathrm{I}=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\frac{\sin ^2 x}{\cos ^2 x} \times \sec ^2 x} \\
&=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\sin ^2 x \sec ^4 x} \\
&\therefore \mathrm{I}=\frac{1}{3} \int \frac{d t}{t^2}=\frac{1}{3} \int t^{-2} d t \\
&\mathrm{I}=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c=\frac{-1}{3}\left[\frac{1}{t}\right]+c \\
&\text { or } \mathrm{I}=-\frac{1}{3\left(1+\tan ^3 x\right)}+c
\end{aligned}
$
Asked in: JEE Main 2014 (12 Apr Online)
Practice more Indefinite Integration questions on Aicharya