Question

The value of $I={\int }_0^{\frac{\pi }{4}}\left({\tan }^{n+1}x\right)dx+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left({\tan }^{n+1}\left(\frac{x}{2}\right)\right)dx$ is

• A. $\frac{1}{n}$
• B. $\frac{n+2}{2n+1}$
• C. $\frac{2n-1}{n}$
• D. $\frac{2n-3}{3n-2}$

Answer 1

Answer: (A)

$\frac{1}{n}$

Given $I={\int }_0^{\frac{\pi }{4}}\left({\tan }^{n+1}x\right)dx+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left({\tan }^{n+1}\left(\frac{x}{2}\right)\right)dx$

In second integral, put $t=\frac{x}{2}\Rightarrow$ $dx=2dt$
$\Rightarrow$ Also, when $x=0$ then $t=0$

When $x=\frac{\pi }{2}$ then $t=\frac{\pi }{4}$

Then $I={\int }_0^{\pi /4}\left({\tan }^{n+1}x\right)dx+{\int }_0^{\pi /4}\left({\tan }^{n-1}t\right)dt$

$I={\int }_0^{\pi /4}\left({\tan }^{n+1}x\right)dx+{\int }_0^{\pi /4}\left({\tan }^{n-1}x\right)dx(\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(y\right)dy)$

$\Rightarrow I={\int }_0^{\pi /4}({\tan }^{n+1}x+{\tan }^{n-1}x)dx$

$\Rightarrow I={\int }_0^{\pi /4}\left({\tan }^{n-1}x\right).\left({\tan }^{2}x+1\right)dx$

$\Rightarrow I={\int }_0^{\pi /4}\left({\tan }^{n-1}x\right).\left({\sec }^{2}x\right)dx$

Put $t=\tan x$
$\Rightarrow dt={\sec }^{2}xdx$

Also when $x=0$ then $t=0$
when $x=\pi /4$, then $t=1$

$I={\int }_0^1{t}^{n-1}dt$

$={\left[\frac{t^n}{n}\right]}_0^1=\frac{1}{n}$