Question

$\int \frac{dx}{si{n}^{4}x+co{s}^{4}x}$

Answer 1

$\begin{array}{c}{\int }^{}\frac{1}{{\sin }^4\left(x\right)+{\cos }^4\left(x\right)}dx={\int }^{}{\sec }^2\left(x\right)\cdot \frac{{\tan }^2\left(x\right)+1}{{\tan }^4\left(x\right)+1}dx\\ substitituteu=\tan x\\ ={\int }^{}\frac{u^2+1}{u^4+1}du={\int }^{}\frac{u^2+1}{\left(u^2-\sqrt{2}u+1\right)\left(u^2+\sqrt{2}u+1\right)}du\\ ={\int }^{}\left(\frac{1}{2\left(u^2+\sqrt{2}u+1\right)}+\frac{1}{2\left(u^2-\sqrt{2}u+1\right)}\right)du\\ =\frac{1}{2}{\int }^{}\frac{1}{u^2+\sqrt{2}u+1}du+\frac{1}{2}{\int }^{}\frac{1}{u^2-\sqrt{2}u+1}du\\ =\frac{{\tan }^{-1}\left(\sqrt{2}\tan \left(x\right)+1\right)}{\sqrt{2}}+\frac{{\tan }^{-1}\left(\sqrt{2}\tan \left(x\right)-1\right)}{\sqrt{2}}+C\end{array}$