Question

$\underset{0}{\overset{\pi /2}{\int }}\frac{{\tan }^{7}x}{{\cot }^{7}x+{\tan }^{7}x}dx$ is equal to

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

Answer 1

Answer: (B)

Let $I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\tan }^{7}x}{{\cot }^{7}x+{\tan }^{7}x}dx\dots (i)$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\tan }^7(\pi /2-x)}{{\cot }^7(\pi /2-x)+{\tan }^7(\pi /2-x)}dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\cot }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}dx\dots (ii)$
On adding Eqs. (i) and (ii), we get
$2I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\tan }^{7}x+{\cot }^{7}x}{{\tan }^{7}x+{\cot }^{7}x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}dx=[x{]}_0^{\pi /2}=\frac{\pi }{2}$
$\therefore I=\frac{\pi }{4}$