The value of the integral int_{-pi / 2}^{pi / | vedclass

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

Question

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\sin^2 x}{1+e^x} \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\sin^2 x}{1+e^x} \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx$,we get: $I = \int_{0}^{\pi / 2} \left( \frac{\sin^2 x}{1+e^x} + \frac{\sin^2(-x)}{1+e^{-x}} \right) \, dx$ Since $\sin^2(-x) = \sin^2 x$ and $\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1}$,we have: $I = \int_{0}^{\pi / 2} \left( \frac{\sin^2 x}{1+e^x} + \frac{e^x \sin^2 x}{1+e^x} \right) \, dx$ $I = \int_{0}^{\pi / 2} \frac{\sin^2 x (1+e^x)}{1+e^x} \, dx = \int_{0}^{\pi / 2} \sin^2 x \, dx$ Using $\sin^2 x = \frac{1-\cos 2x}{2}$: $I = \int_{0}^{\pi / 2} \frac{1-\cos 2x}{2} \, dx = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\pi / 2}$ $I = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$.

The value of the integral $\int_{-\pi / 2}^{\pi / 2} \frac{\sin^2 x}{1+e^x} \, dx$ is

Similar Questions

Let $f(x) = \int\limits_1^x \frac{\tan^{-1} t}{t} dt$ for $x > 0$. Then $f(e^2) - f\left(\frac{1}{e^2}\right)$ is

$\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Then,the location of the roots of $ax^2 + bx + c = 0$ is:

$\tan ^{-1}\left[\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\right]=$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

The value of $\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2}{1 + \tan x + \sqrt{1 + \tan^2 x}} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module