The value of int_0^{2 pi} sqrt{1+sin left(fra | vedclass

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

Question

(A) We know that $1 + \sin 	heta = \cos^2 \frac{	heta}{2} + \sin^2 \frac{	heta}{2} + 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2} = (\cos \frac{

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) We know that $1 + \sin 	heta = \cos^2 \frac{	heta}{2} + \sin^2 \frac{	heta}{2} + 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2} = (\cos \frac{	heta}{2} + \sin \frac{	heta}{2})^2$. Substituting $	heta = \frac{x}{2}$,we get $\sqrt{1 + \sin \frac{x}{2}} = |\cos \frac{x}{4} + \sin \frac{x}{4}|$. 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 positive. Thus,the integral becomes $\int_0^{2\pi} (\cos \frac{x}{4} + \sin \frac{x}{4}) dx$. Evaluating the integral: $[4 \sin \frac{x}{4} - 4 \cos \frac{x}{4}]_0^{2\pi}$. $= (4 \sin \frac{\pi}{2} - 4 \cos \frac{\pi}{2}) - (4 \sin 0 - 4 \cos 0)$. $= (4(1) - 4(0)) - (4(0) - 4(1)) = 4 - (-4) = 8$.

The value of $\int_0^{2 \pi} \sqrt{1+\sin \left(\frac{x}{2}\right)} d x$ is

Similar Questions

The value of $\int_{-1}^{2} \frac{|x|}{x} d x$ is

Let $f(x) = \max \{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$. Then $\int_{-6}^{0} f(x) \, dx$ is equal to

Evaluate the definite integral $\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$.

$\int_2^3 \frac{d x}{x^2-x}$ is equal to

Prove that $\int_{0}^{\frac{\pi}{4}} 2 \tan^{3} x \, dx = 1 - \log 2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module