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Another integral getting a bunch of conflicting info, extremely confused.

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    This is perhaps the most confused I've ever been.

    [itex]\int sin^{3}7x dx[/itex]
    2. Relevant equations

    3. The attempt at a solution

    First, I let u = 7x. du = 7 dx

    Here's where the trouble begins.

    [itex]\frac{1}{7}\int sin^{3}U dU[/itex]

    Now my book claims the following:

    [itex]\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx[/itex]

    Working that gives me:

    [itex]\frac{1}{7}[-\frac{1}{3}sin^{2}u cos u - \frac{2}{3} cos u][/itex]

    Here's the problem. Not only does this simplify to a completely incorrect answer, but my application of:

    [itex]\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx[/itex]

    Doesn't even lead to a correct integral of (sin x)^3 alone, according to wolfram. Is this formula BS?
  2. jcsd
  3. Sep 17, 2011 #2


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    Gold Member

    Why don't you start by writing

    [tex]\sin^3(7x) = \sin(7x)(1 - cos^2(7x)[/tex]

    You don't need a reduction formula at all.
  4. Sep 17, 2011 #3
    That's all well and good, but I really need to understand why what I did does not work.
  5. Sep 17, 2011 #4


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    Integrating by parts mean that

    [tex]\int{u'vdx}=uv-\int{uv' dx}[/tex]

    [tex]\int{sin^n(x)}dx =\int{sin(x)sin^{n-1}(x)dx}[/tex],

    u'=sin(x) and v=sin(n-1)(x).


    Edit: Replacing cos^2(x) by 1-sin^2(x) and rearranging, it leads to your formula.

    [tex]\int sin^{n}x dx = -\frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n}\int sin^{n-2}xdx[/tex]

    Last edited: Sep 17, 2011
  6. Sep 17, 2011 #5
    Your formula is correct. My book also lists the formula
    [tex] \int{sin^3udu}=-\frac{1}{3}(2+sin^2u)cosu+C [/tex]
    Which is what I got when I used the generic [itex]n[/itex] formula.

    Perhaps you simplified incorrectly. (your work posted here appears correct) And perhaps what wolfram returned was equivalent to the known answer.
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