int_0^{400 pi} sqrt{1-cos 2 x} , dx = (in sqr | vedclass

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

Question

(D) We know that $1 - \cos 2x = 2 \sin^2 x$. Therefore,$\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$. The integral becomes $I = \int_0^

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(D) We know that $1 - \cos 2x = 2 \sin^2 x$. Therefore,$\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$. The integral becomes $I = \int_0^{400 \pi} \sqrt{2} |\sin x| \, dx$. Since $|\sin x|$ is periodic with period $\pi$,we can write $I = \sqrt{2} 	imes 400 \int_0^{\pi} |\sin x| \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$. Thus,$I = 400 \sqrt{2} \int_0^{\pi} \sin x \, dx$. Evaluating the integral: $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. Finally,$I = 400 \sqrt{2} 	imes 2 = 800 \sqrt{2}$.

$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)

Similar Questions

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$,where $[x]$ represents the greatest integer function,equals:

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ where $\alpha, \beta > 0$,then $\alpha^4-\beta^4$ is equal to:

If $F(x) = f(x) + f\left(\frac{1}{x}\right)$,where $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$,then $F(e) = $

The absolute value of $\int_{10}^{19} \frac{\sin x}{1 + x^8} dx$ is less than:

Vedclass Test Series

Exam Paper Generator

Online Exam Module