Question

Evaluate the following integrals:

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\tan }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}dx$

Answer 1

Let I = ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\tan }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}dx$ .....(1)
Then,
$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\tan }^7\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{{\tan }^7\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+{\cot }^7\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cot }^7\mathrm{x}}{{\cot }^7\mathrm{x}+{\tan }^{7}x}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\tan }^{7}x+{\cot }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ \Rightarrow 2I={\overline{)x}}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{2}-0=\frac{\mathrm{\pi }}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$