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Math Help - Trig Substitution

  1. #1
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    Trig Substitution

    Thanks!!
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  2. #2
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    \int\sin^4x\cdot\cos^2x\,dx=\int\sin^4x\,dx-\int\sin^6x\,dx

    From here you can use the identity \cos2x=\cos^2x-\sin^2x

    It's messy, I hate this integral.
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  3. #3
    MHF Contributor red_dog's Avatar
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    \displaystyle I_n=\int\sin^nxdx=\int\sin^{n-1}x\sin xdx=\int\sin^{n-1}x(-\cos x)'dx=
    \displaystyle=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}\cos^2dx=
    \displaystyle=-\sin^{n-1}x\cos x+(n-1)\int(\sin^{n-2}x-\sin^nx)dx=
    \displaystyle=-\sin^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_n

    Then I_n=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}
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  4. #4
    Math Engineering Student
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    Yeah, but we actually don't know if qbkr21 may use reduction formulas.
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