Question

Prove that:
$\int \frac{dx}{\sqrt{3}\sin x+\cos x}$

Answer 1

$I=\int \frac{dx}{\sqrt{3}sinx+cosx}$

$=\int \frac{dx}{2(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx)}$

$=\frac{1}{2}\int \frac{dx}{sin(x+\frac{\pi }{6})}$ $(\because sin(A+B)=sinA.cosB+cosA.sinB)$

$=\frac{1}{2}\int cosec(x+\frac{\pi }{6})dx$

$=\frac{1}{2}\times log\left[tan\right(\frac{\pi }{2}+\frac{\pi }{12}\left)\right]+c$

$\because \int cosecxdx=log\left(tan\right(\frac{\pi }{2}\left)\right)+c$