int_{ - pi /2}^{pi /2} {sqrt {frac{1}{2}(1 - | vedclass

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

Question

(B) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this into the integral,we get: $\int_{ - \pi /2}^{\pi /2} \sqrt{\frac{1}{2}(2 \sin^2 x)} \,d

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this into the integral,we get: $\int_{ - \pi /2}^{\pi /2} \sqrt{\frac{1}{2}(2 \sin^2 x)} \,dx = \int_{ - \pi /2}^{\pi /2} \sqrt{\sin^2 x} \,dx = \int_{ - \pi /2}^{\pi /2} |\sin x| \,dx$. Since $|\sin x|$ is an even function,we can write: $2 \int_0^{\pi /2} \sin x \,dx$. Evaluating the integral: $2 [-\cos x]_0^{\pi /2} = 2 [-\cos(\pi/2) - (-\cos 0)] = 2 [0 + 1] = 2$.

$\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = $

Similar Questions

$\int_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$ is equal to

$\int_{0}^{\frac{\pi}{4}} e^{\tan^2 \theta} \sin^2 \theta \tan \theta d\theta =$

If $f(x) = \begin{cases} \sqrt{1 - x} & 0 \le x \le 1 \\ (7x - 6)^{-1/3} & 1 < x \le 2 \end{cases}$,then $\int_{0}^{2} f(x) \, dx$ is equal to

Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule,$\int_0^2 f(x) dx$ is equal to

$\int_0^{\pi /2} \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module