Trigonometric Integrals

**adam63** · April 30th 2010, 01:18 AM

How can I solve this problem:

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

Using the fact that $\cos({\frac{\pi}{6}})=\frac{\sqrt{3}}{2}$ ?

Thank you!

**Prove It** · April 30th 2010, 01:22 AM

Originally Posted by adam63

How can I solve this problem:

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

Using the fact that $\cos({\frac{\pi}{6}})=\frac{\sqrt{3}}{2}$ ?

Thank you!

http://www.wolframalpha.com/input/?i...Cos[x]%29]

Click Show Steps

**adam63** · April 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.

**Unbeatable0** · April 30th 2010, 02:31 AM

$ \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} = $

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

I suppose you can continue from here

**adam63** · April 30th 2010, 03:46 AM

Originally Posted by Unbeatable0

$ \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} = $

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

I suppose you can continue from here

Wow, beautiful, thank you very much

!!!

Thread: Trigonometric Integrals

LinkBack

Thread Tools

Display

Trigonometric Integrals

Similar Math Help Forum Discussions

trigonometric integrals

trigonometric integrals

trigonometric integrals

trigonometric integrals

Trigonometric integrals

Search Tags