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[tex]\int\frac {k(\sin^2\theta-\cos^2\theta-k\sin^4\theta)} {(1-k\sin^2\theta)^{\frac {3}{2}}} d\theta [/tex]

Thanks in advance!

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- Thread starter melknin
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- #1

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[tex]\int\frac {k(\sin^2\theta-\cos^2\theta-k\sin^4\theta)} {(1-k\sin^2\theta)^{\frac {3}{2}}} d\theta [/tex]

Thanks in advance!

- #2

James R

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[tex]-\frac{k \sin 2\theta}{\sqrt{4 - 2k + 2k \cos 2\theta}}[/tex]

That doesn't give you a method, though.

- #3

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It may or may not help to simplify it to something like

[tex] k\int {\frac{{\cos {2\theta } - k\sin ^4 \theta }}

{{\left( {1 - k\sin ^2 \theta } \right)^{\frac{3}{2}} }}d\theta } [/tex]

or

[tex] k\int {\frac{{\sin ^2 \theta \left( {1 - k\sin ^2 \theta } \right) - \cos ^2 \theta }}{{\left( {1 - k\sin ^2 \theta } \right)^{\frac{3}{2}} }}d\theta } [/tex]

*Well...it still feels like I'm missing something obvious ~

[tex] k\int {\frac{{\cos {2\theta } - k\sin ^4 \theta }}

{{\left( {1 - k\sin ^2 \theta } \right)^{\frac{3}{2}} }}d\theta } [/tex]

or

[tex] k\int {\frac{{\sin ^2 \theta \left( {1 - k\sin ^2 \theta } \right) - \cos ^2 \theta }}{{\left( {1 - k\sin ^2 \theta } \right)^{\frac{3}{2}} }}d\theta } [/tex]

*Well...it still feels like I'm missing something obvious ~

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This looks like a prime candidate for contour integration on the complex plane...

- #5

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James Jackson said:This looks like a prime candidate for contour integration on the complex plane...

Ahh--no wonder my CalcII couldn't do it! :tongue2:

And I thought it was a basic CalcII integral! :grumpy:

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Thing is, contour integrals deal with definite integration. The integral stated has no limits...

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Hmm, don't think I've learnt about contour integration yet. I'll look that up. I didn't mention it before, but the integral I'm working with actually does have limits from 0 to [tex]\phi[/tex] if that helps.

- #8

arildno

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Now, just one piece here:

By the identity [tex]\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}[/tex]

the denominator of the integrand may be rewritten as:

[tex](1-k\sin^{2}\theta)^{\frac{3}{2}}=\frac{(4-2k+2k\cos(2\theta))^{\frac{3}{2}}}{8}[/tex]

since [tex]4^{\frac{3}{2}}=8[/tex]

Perhaps I'll try to do something further later on.

By the identity [tex]\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}[/tex]

the denominator of the integrand may be rewritten as:

[tex](1-k\sin^{2}\theta)^{\frac{3}{2}}=\frac{(4-2k+2k\cos(2\theta))^{\frac{3}{2}}}{8}[/tex]

since [tex]4^{\frac{3}{2}}=8[/tex]

Perhaps I'll try to do something further later on.

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- #9

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melknin said:Hmm, don't think I've learnt about contour integration yet. I'll look that up. I didn't mention it before, but the integral I'm working with actually does have limits from 0 to [tex]\phi[/tex] if that helps.

Does phi have an upper limit? You've got to be careful with contour integration with the values of functions, what with functions such as log z being multi-valued at certain points (leading to lovely things such as branch points, Reimann surfaces, that kind of jazz).

I suggest Boas (Mathematical Methods in the Physical Sciences) for an introduction to Contour Integration (look at Cauchy Residue Theorem - an integral and beautiful part of solving things like this).

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- #11

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Hmm, that seems to make sense, but it gets me to an integral I'm not having much more luck with. The only thing I can think of is some other subsitution that I can't figure out.

[tex] - 2k\int {\frac{{t^8 - 4t^6 + 2(8k - 5)t^4 - 4t^2 + 1}}{{(1 + t^2 )^3 (1 - \frac{{4kt^2 }}{{(1 + t^2 )^2 }})^\frac {3}{2} }}} [/tex]

Making [tex] u=1 - \frac{{4kt^2 }}{{(1 + t^2 )^2 }} [/tex] almost seems to work, but not quite. Any advice on what to do from here?

[tex] - 2k\int {\frac{{t^8 - 4t^6 + 2(8k - 5)t^4 - 4t^2 + 1}}{{(1 + t^2 )^3 (1 - \frac{{4kt^2 }}{{(1 + t^2 )^2 }})^\frac {3}{2} }}} [/tex]

Making [tex] u=1 - \frac{{4kt^2 }}{{(1 + t^2 )^2 }} [/tex] almost seems to work, but not quite. Any advice on what to do from here?

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- #12

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