int sqrt{1+cos x} , dx is equal to | vedclass

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

Question

(B) We know that $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$.Substituting this into the integral, we get:$\int \sqrt{2 \cos^2 \left(\frac{x}{2}\r

Vedclass · Accepted Answer

$2 \sqrt{2} \sin \frac{x}{2} + C$

Vedclass · Answer

(B) We know that $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$.Substituting this into the integral, we get:$\int \sqrt{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \int \sqrt{2} \left| \cos \left(\frac{x}{2}\right) \right| \, dx$.Assuming $\cos \left(\frac{x}{2}\right) > 0$, we have:$\sqrt{2} \int \cos \left(\frac{x}{2}\right) \, dx$.Integrating with respect to $x$:$\sqrt{2} \cdot \frac{\sin \left(\frac{x}{2}\right)}{1/2} + C = 2 \sqrt{2} \sin \left(\frac{x}{2}\right) + C$.

$\int \sqrt{1+\cos x} \, dx$ is equal to

Similar Questions

$\int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} d x=$

$\int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}} } \;dx = $

$\int \frac{1 - x^7}{x(1 + x^7)} dx$ equals:

$\int \frac{1}{\sqrt{4x-x^2}} dx = $ . . . . . . $+ c$.

$\int (\tan x - \cot x)^2 \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module