The following integral _{π/4}^{π/2} (2 x)¹7}dx is equal to [JEE Adv 2014 P2]

Correct Answer is : ∫0log⁡(1+2)2(eu+e−u)16₀^{(1+{2})} 2(e^u + e^{-u})¹6}0∫log(1+2)2(eu+e−u)16 du

₀^{(1+{2})} 2(e^u + e^{-u})¹6} du

₀^{(1+{2})} (e^u + e^{-u})¹7} du

₀^{(1+{2})} (e^u - e^{-u})¹7} du

₀^{(1+{2})} 2(e^u - e^{-u})¹6} du

Correct Answer is : ∫0log⁡(1+2)2(eu+e−u)16₀^{(1+{2})} 2(e^u + e^{-u})¹6}0∫log(1+2)2(eu+e−u)16 du

Test Index

Definite Integration

Section: Mathematics

Question : 31 of 46

Marks: +1, -0

The following integral

\int\limits_{\pi/4}^{\pi/2} (2\csc x)^{17}

Solution:

\int\limits_{\pi/4}^{\pi/2} (2\csc x)^{17}

Let

e^u - e^{-u}

= 2 cosec x, x =

\frac{\pi}{4}

⇒ u = ln (1 +

\sqrt{2}

), x =

\frac{\pi}{2}

⇒ u = 0

⇒ cosec x + cot x =

e^u

and cosec x - cot x =

e^{-u}

⇒ cot x =

\frac{e^u - e^{-u}}{2}

(e^u - e^{-u})

dx = -2 cosec x cot x dx

⇒ -

\int (e^u + e^{-u})^{17}

\frac{e^u - e^{-u}}{2\csc x \cot x}

= -2

\int\limits_{\ln(1+\sqrt{2})}^{0} (e^u + e^{-u})^{16}

\int\limits_{0}^{\ln(1+\sqrt{2})} (e^u + e^{-u})^{16}

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