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Math Help - integration question

  1. #1
    Super Member angel.white's Avatar
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    integration question

    Had a quiz today, one of the questions was

    \int sin^n(x) ~dx

    :/ I could do it if it were a number for n. I could probably write an algorithm for it if I knew whether it were even or odd, but I don't know how I'm supposed to approach it.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    I would go about saying that

    If it is even you rewrite it as \int[\frac{1-cos(x)}{2}]^{n-2} and keep decomposing it using trig identities...this is one where you cant get a definite answer youll get part of an answer followed by another indefinite integral...if I am not mistaken
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    If it is even you rewrite it as \int[\frac{1-cos(x)}{2}]^{n-2} and keep decomposing it using trig identities...this is one where you cant get a definite answer youll get part of an answer followed by another indefinite integral...if I am not mistaken
    But I don't know the value of n, I don't know if it is even or odd.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Well I

    dont know what to tell you..there is ways of doing it but it involves hypergeometric functions...
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    Had a quiz today, one of the questions was

    \int sin^n(x) ~dx

    :/ I could do it if it were a number for n. I could probably write an algorithm for it if I knew whether it were even or odd, but I don't know how I'm supposed to approach it.
    try by parts. you should be able to discern a pattern after doing it two or three times

    Let u = \sin^n x and dv = dx
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Man

    I was just about to say what Jhevon said lol \int sin(x)^{n},dx= xsin(x)^{n}-\int \frac{nxsin(x)^{n}}{tan(x)},dx\ ad infinitum
    Last edited by Mathstud28; April 3rd 2008 at 08:13 PM.
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  7. #7
    Super Member angel.white's Avatar
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    Quote Originally Posted by Jhevon View Post
    try by parts. you should be able to discern a pattern after doing it two or three times

    Let u = \sin^n x and dv = dx
    x~\sin^nx -\frac n2 x^2~\cos^{n-1}x +  \frac {n(n-1)}{1*2*3}x^3\sin^{n-2}x - ... \pm \frac{n(n-1)(n-2)*...*(1)}{1*2*3*...*n}(\sin~ or~ \cos)~x + \frac 1n x^{n-1}+ C

    It feels like there are too many unknowns to say for sure.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    x~\sin^nx -\frac n2 x^2~\cos^{n-1}x +  \frac {n(n-1)}{1*2*3}x^3\sin^{n-2}x - ... \pm \frac{n(n-1)(n-2)*...*(1)}{1*2*3*...*n}(\sin~ or~ \cos)~x + \frac 1n x^{n-1}+ C

    It feels like there are too many unknowns to say for sure.
    indeed. perhaps using the reduction formula would be more fruitful. (it is derived the same way, by the way, using u = \sin^{n - 1} x and dv = \sin x)

    you could also do two cases, n is odd, or n is even, then you'd know if the last function should be cosine or sine
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  9. #9
    Super Member angel.white's Avatar
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    Quote Originally Posted by Jhevon View Post
    you could also do two cases, n is odd, or n is even, then you'd know if the last function should be cosine or sine
    Yeah, thats what I started to do, but I just got a little overwhelmed (there was huge time crunch on this test too)
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    I dont know if this is right but..

    since sin(x)≡ \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} woulding sin(x)^{m} or [sin(x)]^{m}
    be the same as \bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\bigg]^{m} which is equal to \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}} so \int\sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}},dx= \sum_{n=0}^{\infty}\int\frac{(-1)^{mn}x^{2mn+mm}}{[(2n+1)!]^{m}},dx which is equal to \sum_{n=0}^{\infty}\frac{(-1)^{mn}}{[(2n+1)!]^{m}}\int[x^{2nm+m}],dx= \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m+1}}{[(2n+1)!]^{m}(2mn+m+1)}
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  11. #11
    Super Member angel.white's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    since sin(x)≡ \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} woulding sin(x)^{m} or [sin(x)]^{m}
    be the same as \bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\bigg]^{m} which is equal to \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}} so \int\sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}},dx= \sum_{n=0}^{\infty}\int\frac{(-1)^{mn}x^{2mn+mm}}{[(2n+1)!]^{m}},dx which is equal to \sum_{n=0}^{\infty}\frac{(-1)^{mn}}{[(2n+1)!]^{m}}\int[x^{2nm+m}],dx= \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m+1}}{[(2n+1)!]^{m}(2mn+m+1)}
    Wow, thank you for putting all that time and effort in. Unfortunately that is a bit beyond what I have learned, I did not know sin was equal to that sum.

    Wouldn't surprise me if we learn that soon, though, we are into sequences and series' now.
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  12. #12
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    Quote Originally Posted by angel.white View Post
    Yeah, thats what I started to do, but I just got a little overwhelmed (there was huge time crunch on this test too)
    Read this. The sine case would be done similarly.
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Hey!

    no problem
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