The following integral int_{frac{pi}{4}}^{fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \operatorname{cosec} x)^{17} dx$.Substitute $	an(\frac{x}{2}) = e^u$,then $\frac{1}{2} \sec^2(\f

Vedclass · Accepted Answer

$\int_0^{\log (1+\sqrt{2})} 2(e^u+e^{-u})^{16} du$

Vedclass · Answer

(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \operatorname{cosec} x)^{17} dx$.Substitute $	an(\frac{x}{2}) = e^u$,then $\frac{1}{2} \sec^2(\frac{x}{2}) dx = e^u du$,so $dx = \frac{2 e^u}{1+e^{2u}} du$.Also,$\operatorname{cosec} x = \frac{1}{\sin x} = \frac{1+e^{2u}}{2e^u} = \frac{e^u + e^{-u}}{2}$.When $x = \frac{\pi}{4}$,$e^u = 	an(\frac{\pi}{8}) = \sqrt{2}-1$,so $u = \ln(\sqrt{2}-1) = -\ln(\sqrt{2}+1)$.When $x = \frac{\pi}{2}$,$e^u = 	an(\frac{\pi}{4}) = 1$,so $u = 0$.Substituting these into the integral:$I = \int_{-\ln(\sqrt{2}+1)}^{0} (e^u + e^{-u})^{17} \cdot \frac{2}{e^u + e^{-u}} du = \int_{-\ln(\sqrt{2}+1)}^{0} 2(e^u + e^{-u})^{16} du$.Since $f(u) = (e^u + e^{-u})^{16}$ is an even function,$\int_{-a}^{0} f(u) du = \int_{0}^{a} f(u) du$.Thus,$I = \int_{0}^{\ln(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$.

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

Similar Questions

$\int_0^{\pi /4} \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \,dx = $

$\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6} \Rightarrow x=$

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$ is equal to

$\int_0^3 \frac{dx}{(x+2) \sqrt{x+1}} = $

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module