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Thread: trigonometric integral

  1. #1
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    trigonometric integral

    hello

    Anyone know how to solve this ??

    $\displaystyle \int \dfrac{ sin\theta - cos\theta }{ \sqrt{sin 2\theta} } d\theta$
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  2. #2
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    $\displaystyle sin(2\theta)=1+sin(2\theta)-1=sin^2(\theta)+cos^2(\theta)+sin(2\theta)-1=(sin(\theta)+cos(\theta))^2-1$

    so your integral = $\displaystyle \displaystyle \int \dfrac{ sin(\theta) - cos(\theta) } { \sqrt{ (sin(\theta)+cos(\theta))^2-1 } } \, d\theta $

    Put $\displaystyle t=sin(\theta)+cos(\theta)$ so that $\displaystyle - ( sin(\theta)-cos(\theta) ) d\theta = dt $

    your integral becomes: $\displaystyle \displaystyle - \int \dfrac{dt}{\sqrt{t^2-1}} \, dt$

    A quick trigonometric substitution will finish it.
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  3. #3
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    Quote Originally Posted by Liverpool View Post
    hello

    Anyone know how to solve this ??

    $\displaystyle \int \dfrac{ sin\theta - cos\theta }{ \sqrt{sin 2\theta} } d\theta$
    Wolfram says this ... (won't show the steps to get there)

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  4. #4
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    Thanks.
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