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Math Help - messy integral (?)

  1. #1
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    messy integral (?)

    could somebody help me integrate the following please:

    (sin x)^4 (cos x)^5 dx

    (the limits are pi/2 and 0 but i can solve that once the initial integration has been done)

    im not sure how to split it up to make it easier, because when i do use trig identities it becomes very messy....

    cheers
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  2. #2
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    \int_0^{\pi /2} {\sin ^4 x\cos ^5 x\,dx}  = \int_0^{\pi /2} {\sin ^4 x\left( {\cos ^2 x} \right)^2 \cos x\,dx} .

    Since you know that \cos^2x=1-\sin^2x, you just replace it and make the obvious substitution. (Don't forget to get the new integration limits for your new variable.)
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  3. #3
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    Quote Originally Posted by nmanik90 View Post
    could somebody help me integrate the following please:

    (sin x)^4 (cos x)^5 dx

    (the limits are pi/2 and 0 but i can solve that once the initial integration has been done)

    im not sure how to split it up to make it easier, because when i do use trig identities it becomes very messy....

    cheers
    Put \cos^5 x = \cos^4 x \cos x = (1 - \sin^2 x)^2 \cos x. Now make the substitution u = \cos x.
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  4. #4
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    There is a general way for \int_0^{\pi/2} \sin^n x \cos^m x ~ dx. Here
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