Integrating Powers of sin


by Saladsamurai
Tags: integrating, powers
Saladsamurai
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#1
Oct29-07, 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 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

[tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex]

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


Casey
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Saladsamurai
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#2
Oct29-07, 04:50 PM
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: mad::m ad::ma d::mad :
hotcommodity
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Oct29-07, 04:53 PM
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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.

Dick
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#4
Oct29-07, 05:21 PM
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Integrating Powers of sin


I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
rock.freak667
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Oct29-07, 05:22 PM
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[tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex]


If you let u=sin3x does that make it any easier?
Saladsamurai
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#6
Oct29-07, 05:24 PM
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Quote Quote by Dick View Post
I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
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} ?
NeoDevin
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Oct29-07, 05:31 PM
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[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.
Saladsamurai
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#8
Oct29-07, 05:49 PM
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Quote Quote by Dick View Post
I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
So I got it using this. But I am still trying to figure out where I went wrong with the formula.

Casey
nrqed
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Oct29-07, 06:05 PM
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Quote Quote by Saladsamurai View Post
So I got it using this. But I am still trying to figure out where I went wrong with the formula.

Casey
Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
Saladsamurai
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#10
Oct29-07, 06:10 PM
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Quote Quote by nrqed View Post
Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
Not if you use the trig substitution; it is negative. It is definitely a mistake on my part.

Casey
Dick
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Oct29-07, 09:32 PM
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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.
rocomath
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#12
Oct29-07, 09:42 PM
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Quote Quote by Saladsamurai
2.) For [tex]\int\sin^4(3x)\cos^3xdx[/tex] they went from <---that to

[tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex]
did you mean [tex]\cos^{3}3xdx[/tex]?
Saladsamurai
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#13
Oct29-07, 09:45 PM
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Quote Quote by rocophysics View Post
did you mean [tex]\cos^{3}3xdx[/tex]?
I don't know what I meant?! I figured it out though.

Thanks!
Casey
rocomath
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#14
Oct29-07, 09:49 PM
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i was trying to solve it for myself but then it saw that ... lol, is all good though.


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