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Integrating Powers of sin 
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#1
Oct2907, 04:09 PM

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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 usub [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 [tex]=\int\sin^4(3x)(1sin^23x)\cos3xdx[/tex] How the hell does that make this problem ANY easier?! Is there a formula for ^^^that! Casey 


#3
Oct2907, 04:53 PM

P: 436

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 usubstitution, as you already have, and almost certainly some integration by parts.



#4
Oct2907, 05:21 PM

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Integrating Powers of sin
I would suggest you use the trig identity sin(x)^2=1cos(x)^2, followed by u=cos(x).



#5
Oct2907, 05:22 PM

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[tex]=\int\sin^4(3x)(1sin^23x)\cos3xdx[/tex]
If you let u=sin3x does that make it any easier? 


#6
Oct2907, 05:24 PM

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#7
Oct2907, 05:31 PM

P: 687

[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. 


#8
Oct2907, 05:49 PM

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Casey 


#9
Oct2907, 06:05 PM

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#10
Oct2907, 06:10 PM

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Casey 


#11
Oct2907, 09:32 PM

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No, it's positive. (1cos^(x)^2)*sin(x)*dx. u=cos(x), du=sin(x)*dx. The u^3/3 term comes out with a plus sign.



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