Question

What is definite integration of sin(2x) dx / (sin⁴(x) + cos^4​(x)) from 0 to pi/2?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{I}={\int }_0^{\mathrm{\pi }/2}\frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx \\ ={\int }_0^{\mathrm{\pi }/2}\frac{\sin 2x}{{\left({\sin }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^2}dx \\ ={\int }_0^{\mathrm{\pi }/2}\frac{\sin 2x}{{\left({\sin }^{2}x\right)}^2+{\left(1-{\sin }^{2}x\right)}^2}dx \\ \mathrm{put}{\sin }^{2}x=t \\ \Rightarrow 2\sin x.\cos xdx=dt \\ \Rightarrow \sin 2xdx=dt \\ \mathrm{as}x\to 0,t\to 0 \\ \mathrm{as}x\to \frac{\mathrm{\pi }}{2},t\to 1 \\ \mathrm{Now},\mathrm{I}={\int }_0^1\frac{dt}{t^2+{\left(1-t\right)}^2} \\ ={\int }_0^1\frac{dt}{2t^2-2t+1} \\ =\frac{1}{2}{\int }_0^1\frac{dt}{t^2-t+{\displaystyle \frac{1}{2}}} \\ =\frac{1}{2}{\int }_0^1\frac{dt}{t^2-t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}} \\ =\frac{1}{2}{\int }_0^1\frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\frac{1}{2}\times 2{\left[{\tan }^{-1}\left\{\frac{\left(t-1/2\right)}{1/2}\right\}\right]}_0^1 \\ ={\left[{\tan }^{-1}\left(2t-1\right)\right]}_0^1 \\ ={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(-1\right) \\ =\frac{\mathrm{\pi }}{4}-\left(-\frac{\mathrm{\pi }}{4}\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$