Integral

[shadypalm88]

I've been stuck on this for some time now (sexy LaTeX rendering):

The only general techniques that I've been taught (excluding blind guessing) are substitution and integration by parts.  Somewhat similar ones like

are much easier.  With a substitution w = cos(2θ) it solves for:

But this one I seemingly can't crack.  Any advice?

[Yoni]

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).

[rabbit]

That first one is relatively simple.  You can use the S[u du] property on it, and you get S[sin³(z)cos³(z) dz] = -cos³(z) + C
Well..assuming I did it correctly that is.

[shadypalm88]

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).

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.

Let w = sin(z)
dw = cos(z) dz

(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:


Let w = cos(z)
dw = -sin(z) dz


Thoughts?

[Yoni]

That first one is relatively simple. You can use the S[u du] property on it, and you get S[sin³(z)cos³(z) dz] = -cos³(z) + C
Well..assuming I did it correctly that is.

Nope! You didn't. Try again! (Hint: d/dz sin³(z) != cos³(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.)

[shadypalm88]

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.)

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.)

[Yoni]

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

[rabbit]

Bleh!  I suck at integrals -.-