Answer:
Option A
Explanation:
Plan This type of question can be done using appropriate subsitiution
Given, $I= \int_{\pi/4}^{\pi/2} (2 cosec$ $x)^{17}$ dx
$2^{17}(cosec$ $x)^{16}$cosec$ x$
$=\int_{\pi/4}^{\pi/2} \frac{(cosec x+cot x)}{(cosec x+cot x)}dx$
Let $cosec$ $x +cot$ $x= t$
$\Rightarrow $ $ (-cosec$ $x.cot$ $x-cosec^{2}$ $x)$dx= dt
and $cosec$ $x-cot$ x=1/t
$\Rightarrow 2cosec$ $x=t+\frac{1}{t}$
$\therefore I= -\int_{\sqrt{2}+1}^{1} 2^{17}\left(\frac{t+1/t}{2}\right)^{16}\frac{dt}{t}$
Let t= eu $\Rightarrow$ dt = eu du. when t=1
eu =1
$\Rightarrow$
u=0
and when $t=\sqrt{2}+1,e^{u}=\sqrt{2}+1$
$\Rightarrow$ $u= ln(\sqrt{2}+1)$
$\Rightarrow$ $I=-\int_{ln(\sqrt{2}+1)}^{0} 2(e^{u}+e^{-u})^{16}\frac{e^{u}du}{e^{u}}$
= $2\int_{0}^{ln(\sqrt{2}+1)}(e^{u}+e^{-u})^{16}du$
