int_0^{pi /4} {frac{{sec x}}{{1 + 2{{sin }^2} | vedclass

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

Question

(A) Let $I = \int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx = \int_0^{\pi /4} {\frac{{\cos x}}{{{{\cos }^2}x(1 + 2{{\sin }^2}x)}}} dx$$= \i

Vedclass · Accepted Answer

$\frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$

Vedclass · Answer

(A) Let $I = \int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx = \int_0^{\pi /4} {\frac{{\cos x}}{{{{\cos }^2}x(1 + 2{{\sin }^2}x)}}} dx$$= \int_0^{\pi /4} {\frac{{\cos x}}{{(1 - {{\sin }^2}x)(1 + 2{{\sin }^2}x)}}} dx$Let $t = \sin x$,then $dt = \cos x dx$. When $x=0, t=0$ and when $x=\pi/4, t=1/\sqrt{2}$.$I = \int_0^{1/\sqrt{2}} \frac{dt}{(1-t^2)(1+2t^2)}$.Using partial fractions: $\frac{1}{(1-t^2)(1+2t^2)} = \frac{1}{3} \left( \frac{1}{1-t^2} + \frac{2}{1+2t^2} ight)$.$I = \frac{1}{3} \left[ \int_0^{1/\sqrt{2}} \frac{dt}{1-t^2} + 2 \int_0^{1/\sqrt{2}} \frac{dt}{1+2t^2} ight]$$I = \frac{1}{3} \left[ \frac{1}{2} \log \left| \frac{1+t}{1-t} ight| + 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1}(t\sqrt{2}) ight]_0^{1/\sqrt{2}}$$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{1+1/\sqrt{2}}{1-1/\sqrt{2}} ight) + \sqrt{2} 	an^{-1}(1) ight]$$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} ight) + \sqrt{2} \cdot \frac{\pi}{4} ight]$Since $\frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2$,we have $\frac{1}{2} \log (\sqrt{2}+1)^2 = \log(\sqrt{2}+1)$.$I = \frac{1}{3} \left[ \log(\sqrt{2}+1) + \frac{\pi}{2\sqrt{2}} ight]$.

$\int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx$ is equal to

Similar Questions

$\int_0^\pi \frac{dx}{1 + \sin x} = $

$\int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} = $

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

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

Let $[x]$ denote the greatest integer less than or equal to $x$,then the value of the integral $\int_{-1}^{1}(|x|-2[x]) \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module