Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{{\sin }^{4}x+{\cos }^{4}x} \end{gathered}$$
Dividing numerator and denominator by cos⁴x
$$\begin{gathered} I=\int \frac{{\sec }^{4}xdx}{{\tan }^{4}x+1} \\ =\int \frac{{\sec }^{2}x{\sec }^{2}xdx}{{\tan }^{4}x+1} \\ =\int \frac{\left(1+{\tan }^{2}x\right){\sec }^{2}xdx}{{\tan }^{4}x+1} \\ \mathrm{Putting}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{\left(1+t^2\right)dt}{t^4+1} \\ =\int \frac{\left({\displaystyle \frac{1}{t^2}+1}\right)dt}{t^2+{\displaystyle \frac{1}{t^2}}} \\ =\int \frac{\left(1+{\displaystyle \frac{1}{t^2}}\right)}{{\left(t-{\displaystyle \frac{1}{t}}\right)}^2+2}dt \\ \mathrm{Putting}t-\frac{1}{t}=p \\ \Rightarrow \left(1+\frac{1}{t^2}\right)dt=dp \\ \therefore I=\int \frac{1}{p^2+{\left(\sqrt{2}\right)}^2}dp \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{p}{\sqrt{2}}\right)+C \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-{\displaystyle \frac{1}{t}}}{\sqrt{2}}\right)+C \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+C \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{{\tan }^{2}x-1}{\sqrt{2}\tan x}\right)+C \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(-\sqrt{2}\times \frac{1-{\tan }^{2}x}{2\tan x}\right)+C \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{-\sqrt{2}}{\tan 2x}\right)+C \\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(-\sqrt{2}\cot 2x\right)+C \end{gathered}$$