# Trigonometric Integrals

• Apr 30th 2010, 01:18 AM
Trigonometric Integrals
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!
• Apr 30th 2010, 01:22 AM
Prove It
Quote:

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&#91;x]%29]

Click Show Steps
• Apr 30th 2010, 01:50 AM
Well, thank you, but I already know this site and checked for the integral there - the solution did not involve what I need - I have to use that trigonometric equation somehow.
• Apr 30th 2010, 02:31 AM
Unbeatable0
$\displaystyle \int \frac{dx}{\sqrt{3}\sin x+\cos x} = \int \frac{dx}{\frac{\cos\frac{\pi}{6}}{\sin\frac{\pi}{ 6}}\sin x+\cos x} = \sin\frac{\pi}{6}\int \frac{dx}{\cos\frac{\pi}{6}\sin x + \sin\frac{\pi}{6}\cos x} =$

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

I suppose you can continue from here
• Apr 30th 2010, 03:46 AM
$\displaystyle \int \frac{dx}{\sqrt{3}\sin x+\cos x} = \int \frac{dx}{\frac{\cos\frac{\pi}{6}}{\sin\frac{\pi}{ 6}}\sin x+\cos x} = \sin\frac{\pi}{6}\int \frac{dx}{\cos\frac{\pi}{6}\sin x + \sin\frac{\pi}{6}\cos x} =$
$\displaystyle = \frac{1}{2}\int\frac{dx}{\sin(x+\frac{\pi}{6})}$