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Math Help - Integration

  1. #1
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    Integration

    hi I'm having trouble integrating this

    \int sin^3\theta\cdot 3sin\theta cos\theta

    I don't really have a method for integrating trig it's normally pretty simple to work out but i don't know where to start with this one :/
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    Quote Originally Posted by djmccabie View Post
    hi I'm having trouble integrating this

    \int sin^3\theta\cdot 3sin\theta cos\theta

    I don't really have a method for integrating trig it's normally pretty simple to work out but i don't know where to start with this one :/
    3\int \sin^4{\theta} \cos{\theta} \, d\theta

    integration by substitution ...

    let u = \sin{\theta}
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    ah thanks! how to you know what to let u equal?
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    Hello,
    Quote Originally Posted by djmccabie View Post
    ah thanks! how to you know what to let u equal?
    By noting that cos is the derivative of sin.
    So you'll have something in the form u^4*u', and then, it's just recognizing known antiderivatives
    And they're more easily spotted by making a substitution
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    yes i realised while i was answering the question
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    ok so this question actually has limits \int_{0}^{\frac{\pi}{2}}


    now i have the mark scheme for this question which shows that the limits stay the same. But i thought when you did integration by substitution you had to manipulate the limits :/
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    Quote Originally Posted by djmccabie View Post
    ok so this question actually has limits \int_{0}^{\frac{\pi}{2}}


    now i have the mark scheme for this question which shows that the limits stay the same. But i thought when you did integration by substitution you had to manipulate the limits :/
    You have two possibilities, while making a substitution :

    - you substitute, and you change the boundaries
    - you substitute, and you don't change the boundaries. In which case, you have to «back substitute», that is : after you got an antiderivative in terms of the new variable, substitute it with the old variable, and keep the limits.


    But I think the first one is better, because for the second one, it would be more correct to write :

    \int_{\theta=0}^{\theta=\pi/2} u^4 ~du, so that you know that it's the boundaries for theta.
    While for the first situation, you can just say \int_0^1 u^4 ~du

    Got it ?
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    so why limits \int_{0}^{1} ??

    i was under the impression you had to make theta = 0 to work out the bottom limit, then theta = pi/2 for the top limit :/
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    Quote Originally Posted by djmccabie View Post
    so why limits \int_{0}^{1} ??

    i was under the impression you had to make theta = 0 to work out the bottom limit, then theta = pi/2 for the top limit :/
    \theta=0 \Rightarrow u=\sin\theta=0
    \theta=\pi/2 \Rightarrow u=\sin\theta=1

    Be careful of the order of the boundaries.
    In the case where you let y=\cos\theta, you would have \int_1^0 =-\int_0^1
    But that's another story ^^
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    Thanks!

    would i integrate this by substitution?

    \frac{1}{2} \int_{0}^{\frac{\pi}{2}} sin^22\theta d\theta


    or change into compound angle i.e.

    sin2\theta = 2sin\theta cos\theta and then square it?

    or does the square prevent this?
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  11. #11
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    \sin ^2 2\theta  = 1 - \cos 4\theta
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