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Math Help - Trigonometric Integrals

  1. #1
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    Trigonometric Integrals

    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!
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  2. #2
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    Quote Originally Posted by adam63 View Post
    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
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  3. #3
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    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.
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  4. #4
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    <br />
\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} =<br />

    <br />
= \frac{1}{2}\int\frac{dx}{\sin(x+\frac{\pi}{6})}<br />

    I suppose you can continue from here
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  5. #5
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    Quote Originally Posted by Unbeatable0 View Post
    <br />
\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} =<br />

    <br />
= \frac{1}{2}\int\frac{dx}{\sin(x+\frac{\pi}{6})}<br />

    I suppose you can continue from here
    Wow, beautiful, thank you very much !!!
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