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} \dfrac{ 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 + \dfrac{1}{t} $
$ \therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, I = - \int^1_{ \sqrt 2 + 1 } \, 2^{17} \bigg ( \dfrac{ t + \dfrac{1}{t}}{2}\bigg )^{16} \dfrac{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} \dfrac{ e^u \, du }{ e^u}$
$\Rightarrow =2 \int^0_{In ( \sqrt 2 + 1 )} ( e^u + e^{-u} )^{16} \, du $

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