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Math Help - Trig Integrals

  1. #1
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    Question Trig Integrals

    Could use some help with these:

    1. The integral sin^6 x cos^3 x dx
    2. The integral sin^2 x cos^4 x dx

    Thanks!
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  2. #2
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    Hello, coolio!

    The first one is quite straight forward . . .


    1)\;\;\int \sin^6\!x\cos^3\!x\, dx

    We have: . \sin^6\!x\cos^2\!x(\cos x) \;=\;\sin^6\!x(1-\sin^2\!x)(\cos x) \;=\;(\sin^6\!x - \sin^8\!x)(\cos x)

    The integral becomes: . \int(\sin^6\!x - \sin^8\!x)(\cos x\,dx)

    Let: u \:=\: \sin x\quad\Rightarrow\quad du \:=\:\cos x\,dx

    Substitute: . \int(u^6 - u^8)\,du

    . . . Got it?

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  3. #3
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    Thanks Soroban, that makes a lot more sense. The case where sin and cos both have even powers seems to be giving me the most trouble still. I know it has something to do with the half angle formulas.
    Last edited by coolio; January 30th 2008 at 01:55 PM. Reason: sp
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by coolio View Post
    Thanks Soroban, that makes a lot more sense. The case where sin and cos both have even powers seems to be giving me the most trouble still. I know it has something to do with the half angle formulas.
    yes. you would replace them with formulas derived from the half angle formula: \sin^2 x = \frac {1 - \cos 2x}2 and \cos^2 x = \frac {1 + \cos 2x}2
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