Question

${\int }_0^{\pi /2}\frac{ta{n}^{7}x}{co{t}^{7}x+ta{n}^{7}x}dx$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$I={\int }_0^{\pi /2}\frac{{\tan }^{7}x}{{\cot }^{7}x+{\tan }^{7}x}dx⟶1$
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I={\int }_0^{\pi /2}\frac{{\tan }^7(\frac{\pi }{2}-x)}{{\cot }^7(\frac{\pi }{2}-x)+{\tan }^7(\frac{\pi }{2}-x)}$
$I={\int }_0^{\pi /2}\frac{{\cot }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}dx⟶2$
Add equations 1 and 2
$⟹2I={\int }_0^{\pi /2}(\frac{{\tan }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}+\frac{{\cot }^{7}x}{{\tan }^{7}x+{\cot }^{7}x})$
$⟹2I={\int }_0^{\pi /2}1\cdot dx$
$⟹2I={\left|x\right|}_0^{\pi /2}$
$⟹2I=\frac{\pi }{2}$
$I=\frac{\pi }{4}$