Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\sin x+\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx \\ =\int \frac{\sin x+\cos x}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}dx \\ =\int \frac{\sin x+\cos x}{1-2{\sin }^{2}x{\cos }^{2}x}dx \\ =\int \frac{\sin x+\cos x}{1-{\displaystyle \frac{1}{2}}{\left(2\sin x\cos x\right)}^2}dx \\ =\int \frac{\sin x+\cos x}{1-{\displaystyle \frac{1}{2}}{\sin }^{2}2x}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}\sin x-\cos x=t.....\left(1\right) \\ \Rightarrow {\left(\sin x-\cos x\right)}^2=t^2 \\ \Rightarrow {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=t^2 \\ \Rightarrow 1-2\sin x\cos x=t^2 \\ \Rightarrow \sin 2x=1-t^2 \\ \mathrm{Differentiating}\left(1\right),\mathrm{we}\mathrm{get} \\ \left(\cos x+\sin x\right)dx=dt \\ \therefore I=\int \frac{1}{1-{\displaystyle \frac{1}{2}}{\left(1-t^2\right)}^2}dt \\ =\int \frac{2}{2-{\left(1-t^2\right)}^2}dt \\ =\int \frac{2}{{\left(\sqrt{2}\right)}^2-{\left(1-t^2\right)}^2}dt \\ =2\int \frac{1}{\left(\sqrt{2}+1-t^2\right)\left(\sqrt{2}-1+t^2\right)}dt \end{gathered}$$

$$\begin{gathered} =\frac{2}{2\sqrt{2}}\int \left[\frac{1}{\sqrt{2}+1-t^2}+\frac{1}{\sqrt{2}-1+t^2}\right]dt \\ =\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{2}+1-t^2}dt+\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{2}-1+t^2}dt \\ =\frac{1}{\sqrt{2}}\int \frac{1}{{\left(\sqrt{\sqrt{2}+1}\right)}^2-t^2}dt+\frac{1}{\sqrt{2}}\int \frac{1}{{\left(\sqrt{\sqrt{2}-1}\right)}^2+t^2}dt \\ =\frac{1}{\sqrt{2}}\times \frac{1}{2\sqrt{\sqrt{2}+1}}\log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{\sqrt{2}+1}}{\tan }^{-1}\frac{t}{\sqrt{\sqrt{2}+1}}+C \\ =\frac{1}{\sqrt{2}}\left[\frac{1}{2\sqrt{\sqrt{2}+1}}\log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{\sqrt{2}+1}}{\tan }^{-1}\frac{t}{\sqrt{\sqrt{2}+1}}\right]+C,\mathrm{where}t=\sin x-\cos x \end{gathered}$$