The following integral ∫ _ (π)/(4) ^ (π)/(2) (2cosecx) ^ 17 dx is equal to

Correct answer: A. ∫ _ 0 ^ (1 + 2 ) 2 ( e ^ u + e ^ - u ) ^ 16 du ∫ _ π4 ^ π2 (2cosecx)17dx Let e^ u + e^ - u = 2cosecx, x = (π)/(4) ⇒ u = (1 + 2 ), x = (π)/(2) ⇒ u = 0 ⇒ cosecx + x = e^ u andcosecx - x = e^ - u ⇒ x = e^ u - e^ - u 2 (e^ u - e^ - u )du = - 2cosecxcotxdx ∫ _ π4 ^ π2 (2cosecx)17dx= - 2 ∫ _ (1 + 2 ) ^ 0 (e^ u + e^ - u ) ^ 16 du = ∫ _ 0 ^ (1 + 2 ) 2 (e^ u + e^ - u ) ^ 16 du

JEE Advanced 2014Paper 2

Exam: JEE Advanced
Year: 2014
Paper: Paper 2
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2014 · Paper 2 · Mathematics · Definite Integration

The following integral ∫ _ (π)/(4) ^ (π)/(2) (2cosecx) ^ 17 dx is...

The following integral $\mathop{∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}{\left(2cosecx\right)}^{17}dx is equal to

Options

A $\mathop{∫}_{0}^{\log \left(1 + \sqrt{2}\right)}2{\left({e}^{u} + {e}^{- u}\right)}^{16}du$
B $\mathop{∫}_{0}^{\log ( 1 + \sqrt{2} )}{\left({e}^{u} + {e}^{- u}\right)}^{17}du$
C $\mathop{∫}_{0}^{\log ( 1 + \sqrt{2} )}{\left({e}^{u} - {e}^{- u}\right)}^{17}du$
D $\mathop{∫}_{0}^{\log ( 1 + \sqrt{2} )}2{\left({e}^{u} - {e}^{- u}\right)}^{16}du$

Correct answer

A. $\mathop{∫}_{0}^{\log \left(1 + \sqrt{2}\right)}2{\left({e}^{u} + {e}^{- u}\right)}^{16}du$

Step-by-step solution

\mathop{∫}_{\pi4}^{\pi2}\left(2cosecx\right)17dx

Let

e^{u} + e^{- u} = 2cosecx, x = \dfrac{π}{4} ⇒ u = \log \left(1 + \sqrt{2}\right), x = \dfrac{π}{2} ⇒ u = 0

⇒ cosecx + \cot x = e^{u}andcosecx - \cot x = e^{- u} ⇒ \cot x = \dfrac{e^{u} - e^{- u}}{2}

\left(e^{u} - e^{- u}\right)du = - 2cosecxcotxdx

<br />

\mathop{∫}_{\pi4}^{\pi2}\left(2cosecx\right)17dx= - 2\mathop{∫}_{\log \left(1 + \sqrt{2}\right)}^{0}{\left(e^{u} + e^{- u}\right)}^{16}du

<br />

=\mathop{∫}_{0}^{\log \left(1 + \sqrt{2}\right)}2{\left(e^{u} + e^{- u}\right)}^{16}du

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