int_0^{2pi } {sqrt {1 + sin frac{x}{2}} ,dx = | vedclass

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

Question

(C) We know that $1 + \sin 	heta = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$. Thus,$\sqrt{1 + \sin \frac{x}{2}} = \sqrt{(\sin \frac{x}{4} +

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) We know that $1 + \sin 	heta = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$. Thus,$\sqrt{1 + \sin \frac{x}{2}} = \sqrt{(\sin \frac{x}{4} + \cos \frac{x}{4})^2} = |\sin \frac{x}{4} + \cos \frac{x}{4}|$. Since $x \in [0, 2\pi]$,$\frac{x}{4} \in [0, \frac{\pi}{2}]$,where both $\sin \frac{x}{4}$ and $\cos \frac{x}{4}$ are non-negative. Therefore,the integral becomes $\int_0^{2\pi} (\sin \frac{x}{4} + \cos \frac{x}{4}) dx$. Evaluating the integral: $= [-4 \cos \frac{x}{4} + 4 \sin \frac{x}{4}]_0^{2\pi}$ $= (-4 \cos \frac{\pi}{2} + 4 \sin \frac{\pi}{2}) - (-4 \cos 0 + 4 \sin 0)$ $= (-4(0) + 4(1)) - (-4(1) + 4(0))$ $= 4 - (-4) = 8$.

$\int_0^{2\pi } {\sqrt {1 + \sin \frac{x}{2}} \,dx = } $

Similar Questions

The value of the definite integral $\int_{0}^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx$ is equal to:

$\int_0^{\pi /4} \tan^2 x \, dx = $

If $f(x) = \operatorname{Max}\{x^3-4, x^4-4\}$ and $g(x) = \operatorname{Min}\{x^2, x^3\}$,then $\int_{-1}^1 (f(x) - g(x)) \, dx =$

$\int\limits_1^{\sqrt 2 } {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx$ is equal to:

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \tan x \,dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module