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

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

Question

(C) Let $I = \int \frac{dx}{\cos x + \sqrt{3} \sin x}$.Multiply and divide by $2$:$I = \int \frac{dx}{2 \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{dx}{\cos x + \sqrt{3} \sin x}$.Multiply and divide by $2$:$I = \int \frac{dx}{2 \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x ight)}$.Using $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$:$I = \frac{1}{2} \int \frac{dx}{\sin \frac{\pi}{6} \cos x + \cos \frac{\pi}{6} \sin x}$.Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:$I = \frac{1}{2} \int \frac{dx}{\sin(x + \frac{\pi}{6})} = \frac{1}{2} \int \csc(x + \frac{\pi}{6}) dx$.Using the standard integral $\int \csc 	heta d	heta = \log |	an \frac{	heta}{2}| + C$:$I = \frac{1}{2} \log |	an(\frac{x}{2} + \frac{\pi}{12})| + C$.

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

Similar Questions

Find the following integral: $\int(2x^2 - 3\sin x + 5\sqrt{x}) dx$

Integrate the function: $\sqrt{1-4x^2}$

$\int {\frac{{\cot x \tan x}}{{{{\sec }^2}x - 1}}} \;dx = $

$\int \sqrt{1 + \sin \frac{x}{2}} \, dx = $

If $f\left( \frac{3x - 4}{3x + 4} \right) = x + 2, x \ne -\frac{4}{3}$,and $\int f(x) dx = A \log |1 - x| + Bx + C$,then the ordered pair $(A, B)$ is equal to: (where $C$ is a constant of integration)

Vedclass Test Series

Exam Paper Generator

Online Exam Module