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

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(B) To solve the integral $\int \frac{dx}{\sin x + \sqrt{3} \cos x}$,we multiply and divide by $2$: $\int \frac{dx}{\sin x + \sqrt{3} \cos x} = \frac{

Vedclass · Accepted Answer

$\frac{1}{2} \log 	an \left( \frac{x}{2} + \frac{\pi}{6} ight) + c$

Vedclass · Answer

(B) To solve the integral $\int \frac{dx}{\sin x + \sqrt{3} \cos x}$,we multiply and divide by $2$: $\int \frac{dx}{\sin x + \sqrt{3} \cos x} = \frac{1}{2} \int \frac{dx}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}$ Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write $\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x = \sin(x + \frac{\pi}{3})$: $= \frac{1}{2} \int \frac{dx}{\sin(x + \frac{\pi}{3})} = \frac{1}{2} \int \csc(x + \frac{\pi}{3}) dx$ Using the standard integral $\int \csc 	heta d	heta = \log 	an(\frac{	heta}{2}) + c$: $= \frac{1}{2} \log 	an \left( \frac{x + \frac{\pi}{3}}{2} ight) + c = \frac{1}{2} \log 	an \left( \frac{x}{2} + \frac{\pi}{6} ight) + c$.

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

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