Question

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

Answer 1

Let $I=\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}dx=\int \frac{{\sec }^{4}xdx}{{\tan }^{4}x+{\tan }^{2}x+1}$

$I=\int \frac{(1+{\tan }^{2}x)\left({\sec }^{2}x\right)dx}{{\tan }^{4}x+{\tan }^{2}x+1}$

Let $\tan x=t$

So, ${\sec }^{2}xdx=dt$

So, $I=\int \frac{1+t^2}{t^4+t^2+1}dt=\int \frac{1+\frac{1}{t^2}}{t^2+1+\frac{1}{t^4}}dt$

$I=\int \frac{(1+\frac{1}{t^2})dt}{{(t-\frac{1}{t})}^2+3}$

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

So, $(1+\frac{1}{t^2})dt=dz$

So, $I=\int \frac{dz}{z^2+3}=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{z}{\sqrt{3}}\right)+C$

$I=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t^2-1}{\sqrt{3}t}\right)+CI=\frac{1}{\sqrt{3}}\left(\frac{{\tan }^{2}x-1}{\sqrt{3}\tan x}\right)+C$