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Math Help - integrating trigs

  1. #1
    eep
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    integrating trigs

    hi guys

    i just recently joined this forum in hope that some of you may help me with my maths problems?

    i'm having trouble integrating this problem:

    I(n) = sin^n(x)dx in terms of I(n-2)

    ps. i didn't put the integral sign at the front

    i've tried for countless hours =\ hope someone can help. thank you
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by eep View Post
    hi guys

    i just recently joined this forum in hope that some of you may help me with my maths problems?

    i'm having trouble integrating this problem:

    I(n) = sin^n(x)dx in terms of I(n-2)

    ps. i didn't put the integral sign at the front

    i've tried for countless hours =\ hope someone can help. thank you
    I dont understand yoru notation? you have I(n)=\int{sin^{n}(x)dx} is that right?
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  3. #3
    eep
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    Quote Originally Posted by Mathstud28 View Post
    I dont understand yoru notation? you have I(n)=\int{sin^{n}(x)dx} is that right?

    yea, that's right .. i didn't know how to type it up =\
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by eep View Post
    yea, that's right .. i didn't know how to type it up =\
    In that case..use integration by parts to get it into the desired form..let u=sin^{n}(x)
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  5. #5
    eep
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    Quote Originally Posted by Mathstud28 View Post
    In that case..use integration by parts to get it into the desired form..let u=sin^{n}(x)
    oo ic ic, thanks a bunch
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by eep View Post
    hi guys

    i just recently joined this forum in hope that some of you may help me with my maths problems?

    i'm having trouble integrating this problem:

    I(n) = sin^n(x)dx in terms of I(n-2)

    ps. i didn't put the integral sign at the front

    i've tried for countless hours =\ hope someone can help. thank you
    this is just asking you for the reduction formula for sine. they derive this all over the internet. just write it as \int \sin x \sin^{n - 1}x~dx and use integration by parts. let u = \sin^{n - 1}x and dv = \sin x
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