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Math Help - Trig integration

  1. #1
    Newbie SuumEorum's Avatar
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    Trig integration

    I need some help with this integral.

    \int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta

    I've tried using integration by parts:

    u=\sin ^{2}2\theta

    \frac{dv}{d\theta }=\cos 2\theta

    And got the integral as:

    \frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right).

    I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks .
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by SuumEorum View Post
    I need some help with this integral.

    \int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta

    I've tried using integration by parts:

    u=\sin ^{2}2\theta

    \frac{dv}{d\theta }=\cos 2\theta

    And got the integral as:

    \frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right).

    I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks .
    Let \varphi=\sin(2\theta)\Rightarrow{d\varphi=2\cos(2\  theta)d\theta}\Rightarrow\frac{d\varphi}{2}=\cos(2  \theta)d\theta

    So we have

    \frac{1}{8}\int{\varphi^2d\varphi}

    You can take it from theree.
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  3. #3
    Newbie SuumEorum's Avatar
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    Yay I have it now, thanks a lot . So what I did was completely wrong? D:
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by SuumEorum View Post
    Yay I have it now, thanks a lot . So what I did was completely wrong? D:
    I think you had the right idea...but when you apply integration by parts to non integration by parts friendly integrals, the result can be ugly. So right idea for the method, but the wrong method.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SuumEorum View Post
    I need some help with this integral.

    \int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta

    I've tried using integration by parts:

    u=\sin ^{2}2\theta

    \frac{dv}{d\theta }=\cos 2\theta

    And got the integral as:

    \frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right).

    I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks .
    This question has already been answered satisfactorally, but I'd like to make the observation that if we first put y = 2 \theta the rest of the logic is easier to see, since we don't have those pesky 2's cropping up.

    -Dan
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