Question

Integrate:

$\int \frac{1}{{\sin }^{4}x+{\cos }^{4}x}dx$

Answer 1

$I=\int \frac{1}{{\sin }^{4}x+{\cos }^{4}x}dx$

$=\int \frac{{\sec }^{4}x}{1+{\tan }^{4}x}dx$

Let $\tan x=t$

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

$I=\int \frac{(1+t^2)}{1+t^4}dt$

$=\int \frac{t^2(1+\frac{1}{t^2})}{t^2(t^2+\frac{1}{t^2})}dt$

$=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2}dt$

Let $t-\frac{1}{t}=v$

$\Rightarrow (1+\frac{1}{t^2})dt=dv$

$I=\int \frac{dv}{(\sqrt{2}{)}^2+v^2}$

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

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left[\frac{{\tan }^{2}x+1}{\sqrt{2}\tan x}\right]+c$.