Question

If $I=\int \frac{dx}{{\sin }^{4}x+{\cos }^{4}x}$, then Find out $I$

Answer 1

$=\int \frac{dx}{{\sin }^{4}x+{\cos }^{4}x}=\int \frac{{\sin }^{2}x(1+{\tan }^{2}x)}{1+{\tan }^{4}x}dx$

$=\int \frac{1+u^2}{1+u^2}du(whereu=\tan x)$

$=\int \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}}du=\int \frac{d(u-\frac{1}{u})}{{(u-\frac{1}{u})}^2+2}du$

$=\frac{1}{\sqrt{2}}\tan x\left(\frac{u-\frac{1}{u}}{\sqrt{2}}\right)+c$

$\frac{1}{\sqrt{2}}{\tan x}^{\perp }\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+c$

$\Rightarrow \frac{25+9t^2+30t+25+16t^2-40t+25}{25t^2}=1$

$X=\frac{-5+3t}{5t}$