Question

The value of the integral $\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{1+{\tan }^{3}x}dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{1+{\tan }^{3}x}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{1+{\displaystyle \frac{{\sin }^{3}x}{{\cos }^{3}x}}}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{{\displaystyle \frac{{\cos }^{3}x+{\sin }^{3}x}{{\cos }^{3}x}}}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\cos }^{3}x}{{\displaystyle {\cos }^{3}x+{\sin }^{3}x}}dx...\left(1\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(a-x\right)dx \\ \mathrm{Thus}, \\ I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\cos }^3\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{{\displaystyle {\cos }^3\left(\frac{\mathrm{\pi }}{2}-x\right)+{\sin }^3\left(\frac{\mathrm{\pi }}{2}-x\right)}}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\sin }^{3}x}{{\displaystyle {\sin }^{3}x+{\cos }^{3}x}}dx...\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\cos }^{3}x}{{\displaystyle {\cos }^{3}x+{\sin }^{3}x}}dx+\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\sin }^{3}x}{{\displaystyle {\sin }^{3}x+{\cos }^{3}x}}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{{\cos }^{3}x+{\sin }^{3}x}{{\displaystyle {\sin }^{3}x+{\cos }^{3}x}}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}1dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$​

​Hence, the value of the integral $\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{1}{1+{\tan }^{3}x}dx$ is $\underline{\frac{\mathrm{\pi }}{4}}$.