int frac{1}{sqrt{1 + sin x}} , dx = | vedclass

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

Question

(C) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.Alternatively,$1 + \sin x = 1 + \cos(\pi/2

Vedclass · Accepted Answer

$\sqrt{2} \log 	an \left( \frac{\pi}{8} + \frac{x}{4} ight) + c$

Vedclass · Answer

(C) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.Alternatively,$1 + \sin x = 1 + \cos(\pi/2 - x) = 2\cos^2(\pi/4 - x/2)$.Thus,$\sqrt{1 + \sin x} = \sqrt{2} \cos(\pi/4 - x/2) = \sqrt{2} \sin(\pi/4 + x/2)$.The integral becomes $\int \frac{1}{\sqrt{2} \sin(\pi/4 + x/2)} \, dx$.This is equal to $\frac{1}{\sqrt{2}} \int \csc(\pi/4 + x/2) \, dx$.Using the formula $\int \csc 	heta \, d	heta = \log 	an(	heta/2) + c$,we get:$\frac{1}{\sqrt{2}} \cdot 2 \log 	an \left( \frac{\pi/4 + x/2}{2} ight) + c = \sqrt{2} \log 	an \left( \frac{\pi}{8} + \frac{x}{4} ight) + c$.

$\int \frac{1}{\sqrt{1 + \sin x}} \, dx = $

Similar Questions

The value of the integral $\int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx$ is equal to...

$\int \frac{\sin (x-a)}{\sin (x-b)} d x = A x + B \log |\sin (x-b)| + C \Rightarrow (A, B) = $

Integrate the function: $\frac{x+2}{\sqrt{4x-x^2}}$

Evaluate the integral: $\int \frac{dx}{2+\cos x}$

Integrate the function: $\frac{x+2}{\sqrt{x^{2}-1}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module