Mathematics - Definite Integration Question with Solution | TestHub

Using the property ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, the given integral

$$I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+{\tan }^{5}x}={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+{\tan }^5\left(\frac{\pi }{3}+\frac{\pi }{6}-x\right)}={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+{\cot }^{5}x}={\int }_{\pi /6}^{\pi /3}\frac{{\tan }^{5}dx}{1+{\tan }^{5}x}$$

Hence $2I={\int }_{\pi /6}^{\pi /3}dx\Rightarrow 1=\frac{1}{2}\left(\frac{\pi }{3}-\frac{\pi }{6}\right)=\frac{\pi }{12}$. Hence (C) is the correct answer.