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Integrating Powers of sin

  1. Oct 29, 2007 #1
    Okay so I am suposed to evaluate 2 of these:

    1.) [tex]\int\sin^3(a\theta)d\theta[/tex] the solution manual looks

    like it used a trig ID to do this. Why??

    Doesn't [tex]\int\sin^3u du=\frac{1}{3}\cos^3u-(\cos u)+C[/tex] ?

    So just use the u-sub [tex]u=a\theta \Rightarrow du/a=d\theta[/tex]

    So it should just be [tex]\frac{1}{3a}\cos^3a\theta-\frac{1}{a}\cosa\theta+C[/tex] ?

    But they got: [tex]-\frac{1}{a}\cos a\theta-\frac{1}{3a}\cos^3a\theta+C[/tex]

    Why the negative -1/3a ? What did I miss? Is it my formula?

    2.) For [tex]\int\sin^4(3x)\cos^3xdx[/tex] they went from <---that to


    How the hell does that make this problem ANY easier?! Is there a formula for ^^^that!

    Last edited: Oct 29, 2007
  2. jcsd
  3. Oct 29, 2007 #2
  4. Oct 29, 2007 #3
    You're using sin(u) like a variable, and integrating, which will certainly get you nowhere. One thing is that [tex] \int sin(u)^{3}[/tex] does not equal [tex]\frac{1}{3}\cos^3u[/tex]. You're going to have to use u-substitution, as you already have, and almost certainly some integration by parts.
    Last edited: Oct 29, 2007
  5. Oct 29, 2007 #4


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    I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
  6. Oct 29, 2007 #5


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    If you let u=sin3x does that make it any easier?
  7. Oct 29, 2007 #6
    I'll try that now. But why is my formula not producing the correct result if I am using u=a*theta ? Shouldn't it be 1/a *{Pug it in} ?
  8. Oct 29, 2007 #7
    [tex]\int \sin^3(a\theta)d\theta = \int \sin(a\theta)d\theta - \int \sin(a\theta)\cos^2(a\theta)d\theta[/tex]

    The first term should be easy for you to integrate, for the second term, try a substitution of

    [tex]u = \cos(a\theta) \Rightarrow \frac{du}{d\theta} = -a\sin(a\theta) \Rightarrow \sin(a\theta)d\theta = -\frac{1}{a}du[/tex]

    After which the equation simplifies a little.

    Use a similar idea for the second one.
  9. Oct 29, 2007 #8
    So I got it using this. But I am still trying to figure out where I went wrong with the formula.

  10. Oct 29, 2007 #9


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    Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
  11. Oct 29, 2007 #10
    Not if you use the trig substitution; it is negative. It is definitely a mistake on my part.

  12. Oct 29, 2007 #11


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    No, it's positive. (1-cos^(x)^2)*sin(x)*dx. u=cos(x), du=-sin(x)*dx. The u^3/3 term comes out with a plus sign.
  13. Oct 29, 2007 #12
    did you mean [tex]\cos^{3}3xdx[/tex]?
  14. Oct 29, 2007 #13
    I don't know what I meant?! I figured it out though.

  15. Oct 29, 2007 #14
    i was trying to solve it for myself but then it saw that ... lol, is all good though.
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