Question

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

Answer 1

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

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

$=\frac{{\sec }^{2}x.{\sec }^{2}xdx}{1+{\tan }^{4}x}$

$=\frac{(1+{\tan }^{2}x){\sec }^{2}xdx}{1+{\tan }^{4}x}$

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

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

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

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

Put $t-\frac{1}{t}=z\Rightarrow (1+\frac{1}{t^2})dt=dz$

$=\frac{dz}{z^2+2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{z}{\sqrt{2}}+C$

$\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}(t-\frac{1}{t})+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}(\tan x-\frac{1}{\tan x})+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}(\tan x-\cot x)+C.$