Question

If $I_n=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n}xdx$ where $n$ is positive integer then $I_{10}+I_8$ is equal to

• A. 9
• B. $\frac{1}{7}$
• C. $\frac{1}{8}$
• D. $\frac{1}{9}$

Answer 1

Answer: (D)

$I_n+I_{n-2}=\frac{1}{n-1}$