I've been stuck on this for some time now (sexy LaTeX rendering):
[img]http://misc.ionws.com/math/omg.png[/img]
The only general techniques that I've been taught (excluding blind guessing) are substitution and integration by parts.  Somewhat similar ones like
[img]http://misc.ionws.com/math/easier.png[/img]
are much easier.  With a substitution w = cos(2θ) it solves for:
[img]http://misc.ionws.com/math/easier_solved.png[/img]
But this one I seemingly can't crack.  Any advice?

That LaTeX is hot.

Blindly integrating by parts is not the best way to solve that integral. Try using some trigonometric identities.
Start with: cos^2(z) = 1 - sin^2(z) ...
Continue with integration by parts, then think how to continue (this is a tricky one, so post again if you need more help).

That first one is relatively simple.  You can use the S[u du] property on it, and you get S[sin[sup]3[/sup](z)cos[sup]3[/sup](z) dz] = -cos[sup]3[/sup](z) + C
Well..assuming I did it correctly that is.

[quote author=Yoni link=topic=14434.msg147626#msg147626 date=1141541514]
That LaTeX is hot.

Blindly integrating by parts is not the best way to solve that integral. Try using some trigonometric identities.
Start with: cos^2(z) = 1 - sin^2(z) ...
Continue with integration by parts, then think how to continue (this is a tricky one, so post again if you need more help).
[/quote]I didn't end up using parts at all.  This question comes from my calculus textbook and I have the answer to it, so I was able to check it.  What I don't understand is that I got different answers depending on whether I used the identity for cosine or the identity for sine; and only the sine one matches the answer provided.
[img]http://misc.ionws.com/math/omg2.png[/img]
Let w = sin(z)
dw = cos(z) dz
[img]http://misc.ionws.com/math/omg3.png[/img]
(sorry, seems to be no way to control the image's export size)
Which doesn't seem to be equivavent to the expression that the salopes want:

[img]http://misc.ionws.com/math/omg4.png[/img]
Let w = cos(z)
dw = -sin(z) dz
[img]http://misc.ionws.com/math/omg5.png[/img]

Thoughts?

[quote author=rabbit link=topic=14434.msg147635#msg147635 date=1141570919]
That first one is relatively simple. You can use the S[u du] property on it, and you get S[sin[sup]3[/sup](z)cos[sup]3[/sup](z) dz] = -cos[sup]3[/sup](z) + C
Well..assuming I did it correctly that is.
[/quote]Nope! You didn't. Try again! (Hint: d/dz sin[sup]3[/sup](z) != cos[sup]3[/sup](z) ...)

shadypalm: Good job, it's easier than the way I thought of doing it.

Both your solutions are correct. Don't forget the "+ C" at the end.
The 2 functions you got are not equivalent with the same C - but they are equivalent with different C's. (i.e., their difference is a constant function.)

[quote author=Yoni link=topic=14434.msg147648#msg147648 date=1141591375]
Both your solutions are correct. Don't forget the "+ C" at the end.
The 2 functions you got are not equivalent with the same C - but they are equivalent with different C's. (i.e., their difference is a constant function.)
[/quote]Oh!  I didn't think of that, but yeah, that makes sense.  Cool, thanks for the help.
(PS- I have the + C on paper but forgot to transcribe it into LaTeX.)

Actually you didn't, I see it in your images.

Bleh!  I suck at integrals -.-