Question

${\int }_0^{\frac{\pi }{4}}\frac{{\sin }^{2}x.{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx=\frac{m}{6}$.Find $m$

Answer 1

$I={\int }^{}\frac{{\sin }^{2}x.{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx$
$={\int }^{}\frac{{\sin }^{2}x.{\cos }^{2}xdx}{{{\cos }^{6}x({\tan }^{3}x+1)}^2}$
$I={\int }^{}\frac{{\tan }^{2}x.{\sec }^{2}xdx}{{(1+{\tan }^{3}x)}^2}$
Substitute $1+{\tan }^{3}x=Z$
$dZ=3{\tan }^{2}x.{\sec }^{2}xdx$
Now,

$I=\frac{1}{3}{\int }^{}\frac{3{\tan }^{2}x.{\sec }^{2}xdx}{{(1+{\tan }^{3}x)}^2}$

$I=\frac{1}{3}\int \frac{dZ}{Z^2}=-\frac{1}{3Z}+c$

$=-\frac{1}{3(1+{\tan }^{3}x)}+c\dots \left(1\right)$

Now,
$I={\int }_0^{\frac{\pi }{4}}\frac{{\sin }^{2}x.{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx$
$={[-\frac{1}{3}\times \frac{1}{(1+{\tan }^{3}x)}]}_0^{\frac{\pi }{4}}\dots \left(from\right(1\left)\right)$
$=-\frac{1}{3}\times \frac{1}{1+1}+\frac{1}{3}\frac{1}{1+0}$
$=\frac{1}{3}-\frac{1}{6}$

$I=\frac{1}{6}$

$\therefore m=1$