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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