# 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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