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!

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- Apr 30th 2010, 01:18 AMadam63Trigonometric 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 AMProve It
http://www.wolframalpha.com/input/?i...Cos[x]%29]

Click Show Steps - Apr 30th 2010, 01:50 AMadam63
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 AMUnbeatable0
$\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 AMadam63