The following integral \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \operatorname{cosec} x)^{17} d x \] is equal to

(A) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{u}+e^{-u}\right)^{16} d u$

(B) $\int_{0}^{\log (1+\sqrt{2})}\left(e^{u}+e^{-u}\right)^{17} d u$

(C) $\int_{0}^{\log (1+\sqrt{2})}\left(e^{u}-e^{-u}\right)^{17} d u$

(D) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{u}-e^{-u}\right)^{16} d u$

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Definite Integrals

The following integral

\int_{\frac{π}{4}}^{\frac{π}{2}} (2 cosec x)^{17} d x

is equal to

Answer: (A)

Let

2 cosec (x) = e^{u} + e^{- u}

, so

cosec (x) = \frac{e ^{u} + e ^{- u}}{2}

.

Differentiating:

- cosec x cot x d x = (\frac{e ^{u} - e ^{- u}}{2}) d u

.

Since

cot x = cosec^{2} x - 1

, we get

d x = - \frac{2 d u}{e ^{u} + e ^{- u}}

.

When

x = π /4

cosec (π /4) = 2

, so

e^{u} + e^{- u} = 22

, giving

u = lo g (1 + 2)

.

When

x = π /2

u = 0

.

Substituting:

I = \int_{l o g (1 + 2)}^{0} - \frac{2}{e ^{u} + e ^{- u}} (e^{u} + e^{- u})^{17} d u = \int_{0}^{l o g (1 + 2)} 2 (e^{u} + e^{- u})^{16} d u

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option