Question

If $I_n={\int }_0^{\pi /4}{\tan }^n\theta d\theta$ for $n=1,2,3,...$ then $I_{n-1}+I_{n+1}=...$

• A. $0$
• B. $1$
• C. $\frac{1}{n}$
• D. $\frac{1}{n+1}$

Answer 1

Answer: (C)

$\frac{1}{n}$

$I_{n-1}+I_{n+1}={\int }_0^{\pi /4}{\tan }^{n-1}\theta d\theta +{\int }_0^{\pi /4}{\tan }^{n+1}\theta d\theta$

$={\int }_0^{\pi /4}{\tan }^{n-1}\theta (1+{\tan }^2\theta )d\theta$

$={\int }_0^{\pi /4}{\tan }^{n-1}\theta {\sec }^2\theta d\theta$

Substitute $\tan \theta =t\Rightarrow {\sec }^2\theta d\theta =dt$

$\therefore I_{n-1}+I_{n+1}={\int }_0^1{t}^{n-1}dt=\frac{t^n}{n}{|}_0^1=\frac{1}{n}$