JEE Advanced | 2014Trigonometric EquationHard

Question

The following integral (2 cosec x)17 dx is equal to

Options

A. 2(eu + e-u)16du
B. (eu + e-u)17du
C. (eu - e-u)17du
D. 2(eu - e-u)16du

Solution

I = (2 cosec x)17 dx
= cosec x + cot x + . 2cosec xdx
Let cosecx + cotx = eu
⇒ - cosecxdx = du
⇒ I = - 2 (eu + e-u)16du = 2(eu + e-u)16 du

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