Question

If $f\left(n\right)=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n}xdx$ then $f\left(30\right)+f\left(28\right)$ is equal to

• A. $\frac{1}{28}$
• B. $\frac{1}{30}$
• C. $\frac{1}{29}$
• D. $\frac{1}{58}$

Answer 1

The correct option is C $\frac{1}{29}$

$f(n-2)=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n-2}xdx$
$\Rightarrow f\left(n\right)+f(n-2)=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n-2}x(1+{\tan }^{2}x)dx$
Let $\tan x=y\Rightarrow {\sec }^{2}xdx=dy$
$\Rightarrow f\left(n\right)+f(n-2)=\underset{0}{\overset{1}{\int }}{y}^{n-2}dy$
$\Rightarrow {\left[\frac{y^{n-1}}{n-1}\right]}_0^1=\frac{1}{n-1}$
Put $n=30$
$f\left(30\right)+f\left(28\right)=\frac{1}{29}$