How can I solve this problem:
$\displaystyle \int \frac{dx}{\sqrt{3}\sin(x)+\cos(x)}$
Using the fact that $\displaystyle \cos({\frac{\pi}{6}})=\frac{\sqrt{3}}{2}$ ?
Thank you!
http://www.wolframalpha.com/input/?i...Cos[x]%29]
Click Show Steps
$\displaystyle
\int \frac{dx}{\sqrt{3}\sin x+\cos x} = \int \frac{dx}{\frac{\cos\frac{\pi}{6}}{\sin\frac{\pi}{ 6}}\sin x+\cos x} = \sin\frac{\pi}{6}\int \frac{dx}{\cos\frac{\pi}{6}\sin x + \sin\frac{\pi}{6}\cos x} =
$
$\displaystyle
= \frac{1}{2}\int\frac{dx}{\sin(x+\frac{\pi}{6})}
$
I suppose you can continue from here