Question

If $I_n=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n}xdx,$ then $\frac{1}{I_3+I_5}$ is equal to

Answer 1

$I_n=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n}xdx$
For $n\ge 2,n\in I$, we can write
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}({\sec }^{2}x-1){\tan }^{n-2}xdx\Rightarrow I={\left[\frac{{\tan }^{n-1}x}{n-1}\right]}_0^{\frac{\pi }{4}}-\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{n-2}xdx\Rightarrow I_n=\frac{1}{n-1}-I_{n-2}$

Now, putting $n=5$, we get
$I_5=\frac{1}{4}-I_3\Rightarrow I_3+I_5=\frac{1}{4}\therefore \frac{1}{I_3+I_5}=4$