Question

If $u_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^n\theta d\theta$ then $u_n+u_{n-2}$ is:

• A. $\frac{1}{n-1}$
• B. $\frac{1}{n+1}$
• C. $\frac{1}{2n-1}$
• D. $\frac{1}{2n+1}$

Answer 1

Answer: (A)

Given $:u_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^n\theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}{\tan }^2\theta {\tan }^{n-2}\theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}\left({\sec }^2\theta -1\right){\tan }^{n-2}\theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}{\sec }^2\theta {\tan }^{n-2}\theta d\theta$
$-\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}\theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}{\sec }^2\theta {\tan }^{n-2}\theta d\theta -u_{n-2}$
$\Rightarrow u_n+u_{n-2}=\underset{0}{\overset{\pi /4}{\int }}{\sec }^2\theta t{\tan }^{n-2}\theta d\theta$
$={\frac{{\tan }^{n-1}\theta }{n-1}|}_0^{\pi /4}=\frac{1}{n-1}$