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Sine and cos integration

  1. Oct 14, 2006 #1
    I need to find [tex] \int \cos^2(x)\sin^7(x)dx[/tex]

    I'm not sure what substitution to make
  2. jcsd
  3. Oct 14, 2006 #2


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    If you start by turning [tex] \cos^2 (x)[/tex] into [tex] 1 - \sin^2(x)[/tex], you get an integral that can be done by a reduction formula.
  4. Oct 14, 2006 #3
    Or you can do intergration by parts but the above way is prob quicker
  5. Oct 14, 2006 #4
    ok so now i got

    [tex]\int (1-\sin^2(x))\sin^7(x)dx[/tex]

    and then what?

    [tex] \int \sin^7(x)-\sin^9(x) dx[/tex] ?? and now im stuck :(

    and then [tex] \int (1-\cos^2(x))^3\sin(x) - (1-\cos^2(x))^4\sin(x)) dx [/tex]

    Last edited: Oct 14, 2006
  6. Oct 14, 2006 #5


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    Try doing [tex] \int \sin^n(x)dx[/tex] by integration by parts, to get a formula in terms of an integral of sin to a lower power
  7. Oct 14, 2006 #6
    sorry, im not sure what integration by parts is? :frown:
  8. Oct 14, 2006 #7


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    integration by parts:

    [tex] \int udv = uv - \int vdu[/tex]

    basically, if you have an integral that is the product of a function u, and the derivative of another function v, it equals u*v - the integral of v times the derivative of u. Here's an example:

    [tex] \int x*e^xdx[/tex] Say u=x, and dv=exdx. Then v=ex, and du=dx. So:

    [tex] \int x*e^xdx = x*e^x - \int e^xdx = x*e^x - e^x[/tex]
  9. Oct 14, 2006 #8
    Learning another method to integrate is good, but if meee is required to do this particular problem by just using sub. methods, then this is probably a way.

    Rewrite the integral as...

    [tex] \int{\cos^2{x}\sin^6{x}\sin{x}dx}[/tex]

    Convert the [tex]\sin^6{x}[/tex] in terms of cos, and the do the substitution u = cos(x).
  10. Oct 14, 2006 #9
    ok thanks guys...

    Office_Shredder thanks for that.. but i think integration by parts is beyond my current course.

    neutrino, i tried what you said, but it looks kinda weird...

    [tex] \int{\cos^2{x}\sin^6{x}\sin{x}dx}[/tex]

    [tex]= \int{\cos^2{x}(1-\cos^2{x})^3\sin{x}dx}[/tex]

    let u = [tex]\cos{x}[/tex]
    [tex]\frac{du}{dx} = -\sin{x}[/tex]

    [tex]= -\int{u^2(1-u^2)^3 du}[/tex]
    [tex]= -\int{u^2(1-3u^2+3u^4-u^6) du}[/tex]
    [tex]= -\int{u^2-3u^4+3u^6-u^8)du}[/tex]
    [tex]= -(\frac{1}{3}u^3-\frac{3}{5}u^5+\frac{3}{7}u^7-\frac{1}{9}u^9)[/tex]

    and then sub u=cos(x) back in

    is that right?
    Last edited: Oct 14, 2006
  11. Oct 14, 2006 #10
    that is correct
  12. Oct 14, 2006 #11
    thanks :D cool
  13. Oct 14, 2006 #12

    Btw, I don't think you'll have to wait till university to find more about integration by parts. You'll definitely meet it at the school level.
  14. Oct 14, 2006 #13
    oh reali?
    ive finished my year 12 course (exams in 2 weeks), and havent seen it in school or outside school lectures.

    maybe it will be useful for me to learn it.
  15. Oct 14, 2006 #14
    Okay, shouldn't have generalised. I was under the assumption that every school, in every part of the world taught this. :rolleyes:

    It most certainly will.
  16. Oct 14, 2006 #15
    ok, thanks

    and lol im in australia our schools are probly weird
  17. Oct 27, 2006 #16
    My school which is in Australia are doing it now. Its fairly easy to learn.
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