int_{-pi / 2}^{pi / 2} sin |x| , dx is equal | vedclass

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

Question

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$. Since $f(x) = \sin |x|$ is an even function because $f(-x) = \sin |-x| = \sin |x| = f(x)$,we c

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$. Since $f(x) = \sin |x|$ is an even function because $f(-x) = \sin |-x| = \sin |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$. Thus,$I = 2 \int_{0}^{\pi / 2} \sin |x| \, dx$. For $x \in [0, \pi / 2]$,$|x| = x$,so $I = 2 \int_{0}^{\pi / 2} \sin x \, dx$. Evaluating the integral,$I = 2 [-\cos x]_{0}^{\pi / 2}$. $I = 2 [-\cos(\pi / 2) - (-\cos(0))]$. $I = 2 [0 - (-1)] = 2(1) = 2$.

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

Similar Questions

If $a > 0$,then $\int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ is equal to

By using the properties of definite integrals,evaluate the integral $\int_{-5}^{5}|x+2| d x$.

The value of $\int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{\left(1+\cos ^{2} x\right)\left(e^{\cos x}+e^{-\cos x}\right)} d x$ is equal to

$\int_0^\pi x \log(\sin x) \, dx = $

$\int\limits_0^\pi {{e^{{{\cos }^4}x}}} \cdot \cos^5(2n + 1)x \,dx, (n \in I)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module