int frac{1}{cos x+sqrt{3} sin x} dx = | vedclass

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

Question

(A) To evaluate the integral $I = \int \frac{1}{\cos x+\sqrt{3} \sin x} dx$,we multiply and divide the denominator by $2$: $I = \int \frac{1}{2(\frac{

Vedclass · Accepted Answer

$\frac{1}{2} \log \left|	an \left(\frac{x}{2}+\frac{\pi}{12}ight)ight|+c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{1}{\cos x+\sqrt{3} \sin x} dx$,we multiply and divide the denominator by $2$: $I = \int \frac{1}{2(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x)} dx$ Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write $\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \sin(x + \frac{\pi}{6})$: $I = \frac{1}{2} \int \frac{1}{\sin(x + \frac{\pi}{6})} dx$ $I = \frac{1}{2} \int \csc(x + \frac{\pi}{6}) dx$ Using the standard integral formula $\int \csc 	heta d	heta = \log |	an(\frac{	heta}{2})| + C$: $I = \frac{1}{2} \log |	an(\frac{x}{2} + \frac{\pi}{12})| + C$

$\int \frac{1}{\cos x+\sqrt{3} \sin x} dx =$

Similar Questions

$\int {\frac{{{x^2}dx}}{{{{(a + bx)}^2}}}} = $

$\int \sin^3 x \, dx$ is equal to

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

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

$\int \sin 2x \cos 3x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module