Question

Evaluate : $\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}dx$

Answer 1

$I=\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}dx$
$\text{Divide numerator and denominator by}$ ${\cos }^{4}x$

$=\int \frac{{\sec }^{4}x}{{\tan }^{4}x+{\tan }^{2}x+1}dx$,

$[Put\tan x=t,{\tan }^{2}x=t^2,{\sec }^{2}x=1+t^2,{\sec }^{2}xdx=dt]$

= $\int \frac{(1+t^2)dt}{t^4+t^2+1}$, $\text{[Divide numerator and denominator by}{t}^2\text{]}$

= $\int \frac{(\frac{1}{t^2}+1)}{t^2+\frac{1}{t^2}+1}dt,[Putt-\frac{1}{t}=u,(1+\frac{1}{t^2})dt=du,t^2+\frac{1}{t^2}-2=u^2\Rightarrow t^2+\frac{1}{t^2}=u^2+2]$


= $\int \frac{du}{u^2+2+1}$

= $\int \frac{du}{u^2+\left({\sqrt{3}}^2\right)}=\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{u}{\sqrt{3}}$

=$\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t^2-1}{t\sqrt{3}}\right)$

= $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{{\tan }^{2}x-1}{\left(\tan x\right)\sqrt{3}}\right)+c$