The following integral ∫ π 4 π 2 2 c o s e c x 17 d x is equal to

∫ π 4 π 2 2 cosec x 17 d x Let e u + e - u = 2 cosec x , x = π 4 ⇒ u = log 1 + 2 , x = π 2 ⇒ u = 0 ⇒ cosec x + cot x = e u a n d cosec x - cot x = e - u ⇒ cot x = e u - e - u 2 e u - e - u d u = - 2 cosec x cot x d x ∫ π 4 π 2 2 cosec x 17 d x = - 2 ∫ log 1 + 2 0 e u + e - u 16 d u = ∫ 0 l og 1 + 2 2 e u + e - u 16 d u

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDefinite Integration by SubstitutionHard2 minPYQ_2014

MathematicsHardsingle choice

The following integral $\int_{\frac{π}{4}}^{\frac{π}{2}} {(2 c o s e c x)}^{17} d x$ is equal to

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Solution:

$\int_{\frac{π}{4}}^{\frac{π}{2}} {(2 cosec x)}^{17} d x$
Let $e^{u} + e^{- u} = 2 cosec x, x = \frac{π}{4} \Rightarrow u = \log (1 + \sqrt{2}), x = \frac{π}{2} \Rightarrow u = 0$
$\Rightarrow cosec x + \cot x = e^{u} a n d cosec x - \cot x = e^{- u} \Rightarrow \cot x = \frac{e^{u} - e^{- u}}{2}$
$(e^{u} - e^{- u}) d u = - 2$ $cosec x \cot x d x$
$\int_{\frac{π}{4}}^{\frac{π}{2}} {(2 cosec x)}^{17} d x$ $= - 2 \int_{\log (1 + \sqrt{2})}^{0} {(e^{u} + e^{- u})}^{16} d u$
$=$ $\int_{0}^{l og (1 + \sqrt{2})} 2 {(e^{u} + e^{- u})}^{16} d u$

Stream:JEE_ADVSubject:MathematicsTopic:Definite IntegrationSubtopic:Definite Integration by Substitution

⏱ 2mℹ️ Source: PYQ_2014

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