The following integral is equal to :

Engineering

Mathematics

Properties of Definite Integral

Question

The following integral $\int_{\frac{π}{4}}^{\frac{π}{2}} {(2 cosecx)}^{17} dx$ is equal to :

$\int_{0}^{\log (1 + \sqrt{2})} 2 {(e^{u} - e^{- u})}^{16} du$

$\int_{0}^{\log (1 + \sqrt{2})} 2 {(e^{u} + e^{- u})}^{16} du$

$\int_{0}^{\log (1 + \sqrt{2})} {(e^{u} + e^{- u})}^{17} du$

$\int_{0}^{\log (1 + \sqrt{2})} {(e^{u} - e^{- u})}^{17} du$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$I = \int_{\frac{π}{4}}^{\frac{π}{2}} {(2 cosecx)}^{17} dx$

Let cosec x + cot x = t

$∴$ cosec x – cot x $= \frac{1}{t}$

Hence 2 cosec x $= t + \frac{1}{t}$

Also (– cosec x cot x – cosec²x) dx = dt

$∴$ – cosec x (cot x + cosec x)dx = dt

$∴ dx= \frac{- dt}{cosecx (cotx + cosecx)} = \frac{- 2 dt}{(t + \frac{1}{t}) (t)}$

$Now I = - 2 \int_{\sqrt{2} + 1}^{1} {(t + \frac{1}{t})}^{17} \frac{dt}{t (t + \frac{1}{t})} = - 2 \int_{\sqrt{2} + 1}^{1} {(t + \frac{1}{t})}^{16} \frac{dt}{t}$

Put t = e^u

$∴ I = \int_{0}^{\log (1 + \sqrt{2})} 2 {(e^{u} + e^{- u})}^{16} du$