Question

$\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\frac{{\tan }^{3}x}{1+\cos 2x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^{3}x}{1+\cos 2x}\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^{3}x}{2{\cos }^{2}x}\mathit{}\mathit{d}x \\ \Rightarrow I=\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\tan }^{3}x{\sec }^{2}xdx \\ \mathrm{Let}\tan x=t.\mathrm{Then},{\sec }^{2}xdx=dt \\ \mathrm{When}x=0,t=0\mathrm{and}x=\frac{\mathrm{\pi }}{4},t=1 \\ \therefore I=\frac{1}{2}{\int }_0^1{t}^{3}dt \\ \Rightarrow I=\frac{1}{2}{\left[\frac{t^4}{4}\right]}_0^1 \\ \Rightarrow I=\frac{1}{2}\left(\frac{1}{4}-0\right) \\ \Rightarrow I=\frac{1}{8} \end{gathered}$$