The value of intlimits_{ - pi /2}^{pi /2} {fr | vedclass

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

Question

(D) Let $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$Using the property $\int\limits_a^b f(x)dx = \int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) Let $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$Using the property $\int\limits_a^b f(x)dx = \int\limits_a^b f(a+b-x)dx$,we replace $x$ with $-\frac{\pi}{2} + \frac{\pi}{2} - x = -x$:$I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}(-x)}}{{1 + {2^{-x}}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + \frac{1}{2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}{{\sin }^2}x}}{{2^x + 1}}} dx \quad ......(2)$Adding $(1)$ and $(2)$:$2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x + {2^x}{{\sin }^2}x}}{{1 + {2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x dx}$Since $\sin^2 x$ is an even function,$2I = 2 \int\limits_0^{\pi /2} {\sin^2 x dx} = 2 \int\limits_0^{\pi /2} {\frac{1 - \cos 2x}{2} dx}$$2I = \int\limits_0^{\pi /2} {(1 - \cos 2x) dx} = [x - \frac{\sin 2x}{2}]_0^{\pi /2} = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$Therefore,$I = \frac{\pi}{4}$.

The value of $\int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ is

Similar Questions

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$

The value of the definite integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)}$ is equal to:

$ \int_{-3}^{3} \cot^{-1} x \, dx = $

Let $f(x) = \frac{\sin x}{x}$,then $\int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module