Question

Solve : $\int \frac{dx}{{\cos }^{6}x+{\sin }^{6}x}$

Answer 1

$\int \frac{1}{{\sin }^{6}x+{\cos }^{6}x}dx$

$\int \frac{\frac{1}{{\cos }^{6}x}}{\frac{{\sin }^{6}x+{\cos }^{6}x}{{\cos }^{6}x}}$

$\int \frac{{\sec }^{6}x}{1+{\tan }^{6}x}dx$

$\int \frac{{\sec }^{4}x\times {\sec }^{2}x}{1+{\tan }^{6}x}dx$

$\int \frac{(1+{\tan }^{2}x{)}^2\times {\sec }^{2}x}{1+{\tan }^{6}x}dx$

substitute $\tan x=t\to {\sec }^{2}xdx=dt$

$\int \frac{(1+t^2{)}^2}{1+t^6}dt$

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

$\int \frac{1+t^2}{t^4-t^2+1}dt$

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

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

substitute $t-\frac{1}{t}=u\to (1+\frac{1}{t^2})dt=du$

$\frac{du}{u^2+1}={\tan }^{-1}u=ta{n}^{-1}(t-\frac{1}{t})=ta{n}^{-1}\left(\frac{t^2-1}{t}\right)$

$=ta{n}^{-1}\frac{{\tan }^{2}x-1}{\tan x}+C$