Q. The value of $I = \int \limits^ \frac{\pi}{4}_{0}\left(tan^{n+1} x\right)dx + \frac{1}{2} \int \limits^ \frac{\pi}{2}_{0} tan^{n-1} \left(x / 2\right)dx$ is equal to

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A

$\frac{1}{n}$

0%

B

$\frac{n+2}{2n+1}$

100%

C

$\frac{2n-1}{n}$

0%

D

$\frac{2n-3}{3n -2}$

0%

Solution:

Given,
$I=\int_ \limits{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\frac{1}{2} \int_{0}^{\pi / 2} \tan ^{n-1}\left(\frac{x}{2}\right) d x$
In second integral, put $t=\frac{x}{2}$
$ \Rightarrow d x=2 d t$
$\Rightarrow $ Also, when $x=0$ then $t=0$,
When $ x=\pi / 2, \text { then } t=\pi / 4$
Then, $I=\int_\limits{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\int_{0}^{\pi / 4} \tan ^{n-1} t d t $
$ I=\int_\limits{0}^{\pi / 4} \tan ^{n+1} x \cdot d x+\int_\limits{0}^{\pi / 4} \tan ^{n-1} x d x$
$\left\{\because \int_\limits{a}^{b} f(x) d x=\int_{a}^{b} f(y) d y\right\}$
$\Rightarrow I=\int_\limits{0}^{\pi / 4}\left(\tan ^{n+1} x+\tan ^{n-1} x\right) d x$
$\Rightarrow I=\int_\limits{0}^{\pi / 4} \tan ^{n-1} x \cdot\left(\tan ^{2} x+1\right) d x$
$\Rightarrow I=\int_\limits{0}^{\pi / 4} \tan ^{n-1} x\left(\sec ^{2} x\right) d x$
Put $t=\tan x$
$ \Rightarrow d t=\sec ^{2} x d x$
Also, when $x=0$, then $t=0$
when $x=\pi / 4$, then $t=1$
$I=\int_{0}^{1} t^{n-1} d t=\left[\frac{t^{n}}{n}\right]_{0}^{1}=\frac{1}{n}$

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