Question

If $I_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^n\theta d\theta$, then for any positive integer $n$, the value of $I_{n-1}+I_{n+1}$ is

• A. $1$
• B. $\frac{1}{n}$
• C. $\frac{\pi }{n}$
• D. $\frac{1}{n-1}$

Answer 1

The correct option is B $\frac{1}{n}$

If $I_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n}xdx$,
then $I_n+I_{n-2}=\frac{1}{n-1}$
Replacing $n$ by $n+1,$ we get
$I_{n+1}+I_{n-1}=\frac{1}{n}$