Question

Prove that ${\int }_0^{\pi /4}\frac{dx}{{\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x}=\frac{\pi }{2}$.

Answer 1

Let $I={\int }_0^{\pi /4}\frac{dx}{{\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x}=\frac{\pi }{2}$
As ${\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x$
$={\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x+2{\cos }^{2}x{\sin }^{2}x-{\cos }^{2}x{\sin }^{2}x$
$={\cos }^{2}2x+\frac{1}{4}(1+\cos 2x)(1-\cos 2x)$
$=\frac{1}{4}(4{\cos }^{2}2x+1-{\cos }^{2}2x)$

$=\frac{1}{4}(1+3{\cos }^{2}2x)$

Therefore
$I={\int }_0^{\pi /4}\frac{4dx}{(1+3{\cos }^{2}2x)}={\int }_0^{\pi /4}\frac{4{\sec }^{2}2x}{{\sec }^{2}2x+3}dx$
Put $\tan 2x=t$

$\Rightarrow 2{\sec }^{2}2xdx=dt$

When $x=0\Rightarrow t=0$

When $x=\frac{\pi }{4}\Rightarrow t=\infty$

Hence
$I={\int }_0^{\infty }\frac{2dt}{t^2+4}$

$={\left[{\tan }^{-1}\frac{t}{2}\right]}_0^{\infty }$

$\Rightarrow I=\frac{\pi }{2}$