Question

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

Answer 1

$$\begin{gathered} \mathrm{Let},I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\tan }^{3}x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\tan }^3\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+co{t}^{3}x}dx....\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{1+{\tan }^{3}x}+\frac{1}{1+co{t}^{3}x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+{\tan }^{3}x+co{t}^{3}x}{\left(1+{\tan }^{3}x\right)\left(1+co{t}^{3}x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+{\tan }^{3}x+co{t}^{3}x}{2+{\tan }^{3}x+co{t}^{3}x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left(x\right)}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence},I=\frac{\mathrm{\pi }}{4} \end{gathered}$$