Q. The integral $ \int^{\pi/2}_{\pi/4} ( 2 \ cosec \ x )^{17} \ dx $ is equal to

Solution:

PLAN This type of question can be done using appropriate substitution
Given, $ I = \int^{\pi/2}_{\pi/4} ( 2 \ cosec \ x )^{17} \ dx $
$ \ \ \ \ \ \ \ \ \ = \int^{\pi/2}_{\pi/4} \frac{ 2^{17} ( cosec\ x )^{16} \ cosec \ x ( 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 2 cosec x = t + \frac{1}{t} $
$ \therefore \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = - \int^1_{ \sqrt 2 + 1 } \ 2^{17} \bigg ( \frac{ t + \frac{1}{t}}{2}\bigg )^{16} \frac{dt}{t} $
let $\ t = e^u \Rightarrow \ dt = e^u \ d \ u , when \ t = 1 , e^u = 1 \Rightarrow u = 0 $ and when $ t = \sqrt 2 + 1 , e^u = \sqrt 2 + 1 $
$ \Rightarrow u =$ In $Q( \sqrt 2 + 1 ) $
$ \Rightarrow I = -\int^0_{In ( \sqrt 2 + 1 )} 2( e^u + e^{-u} )^{16} \frac{ e^u \ du }{ e^u}$
$\Rightarrow =2 \int^0_{In ( \sqrt 2 + 1 )} ( e^u + e^{-u} )^{16} \ du $

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