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

  • Thread starter 1MileCrash
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  • #1
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Homework Statement



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


[itex]\int sin^{3}7x dx[/itex]

Homework Equations





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?
 

Answers and Replies

  • #2
LCKurtz
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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.
 
  • #3
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That's all well and good, but I really need to understand why what I did does not work.
 
  • #4
ehild
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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).

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

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]

ehild
 
Last edited:
  • #5
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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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