I've been stuck on this for some time now (sexy LaTeX rendering):
(http://misc.ionws.com/math/omg.png)
The only general techniques that I've been taught (excluding blind guessing) are substitution and integration by parts. Somewhat similar ones like
(http://misc.ionws.com/math/easier.png)
are much easier. With a substitution w = cos(2θ) it solves for:
(http://misc.ionws.com/math/easier_solved.png)
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[sin3(z)cos3(z) dz] = -cos3(z) + C
Well..assuming I did it correctly that is.
Quote from: Yoni on March 05, 2006, 12:51 AM
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.
(http://misc.ionws.com/math/omg2.png)
Let w = sin(z)
dw = cos(z) dz(http://misc.ionws.com/math/omg3.png)
(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:
(http://misc.ionws.com/math/omg4.png)
Let w = cos(z)
dw = -sin(z) dz(http://misc.ionws.com/math/omg5.png)
Thoughts?
Quote from: rabbit on March 05, 2006, 09:01 AM
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 sin
3(z) != cos
3(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 from: Yoni on March 05, 2006, 02:42 PM
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.)
Actually you didn't, I see it in your images.
Bleh! I suck at integrals -.-