solve the differential eq
dr/d0 = (sin(pi0))^4
0 = theata
lol integrals then suddenly diff eq lol =)) your curriculum makes me wonder
IT makes EVERYONE wonder.....
It's like they want to add them in to get us ready for diff eq later...
\[\frac{dr}{d\theta} = \sin^4 (\pi \theta)?\]
yessir.
\[dr = (\sin(\pi \theta))^4 d \theta\]
now take the integral of both sides
so it's time for half angle formula!
half angle?
i think this is just some trig sub thingy
Unless just do it normally...
\[\int \sin^4 (\pi \theta) d\theta\] treat pii like a variable
i mean treat pi like a cosntant
I could say u = pi theta du = pi dtheta
\[1/\pi \int\limits \sin^4(u) du\]
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
yup you're doing well
you cann integrate that though...
the integration of that is pretty long haha
I could do a half angle formula now :P
what half lol
\[1/2\pi \int\limits (1-\cos2u)^2 du\]
WE!
\[1/2\pi \int\limits (1-\cos2u) (1-\cos2u) du\]
\[1/2\pi \int\limits) (1-2\cos2u +( \cos2u)^2)du\]
:p
ok I give up hahha.
lol =)) like i said this integration is long
just look at this :( http://www.wolframalpha.com/input/?i=int+sin%5E4+%28x%29dx
long isnt it
Power reduction formula, maybe?
maybe :P
\(\sin^4(x)=\frac{1}{8}\left(3-4\cos(2x)+\cos(4x)\right)\)
Wolfram is wrong! :P
Did you misunderstand? I said your statement is false for all possibilities.
I'll try your answer, it said wolfram's was incorrect.
Cool I win.
@KonradZuse, I apologize if I made a comment earlier that offended you. Many of my rather crazy-sounding or outlandish statements are completely jokes. I had actually meant that original comment in a positive light--I was glad to see that you have taken interest in mathematics and enjoy it more than most people. I had said this to other users on OS and complimented you during conversations with them. Nonetheless, I apologize if my message came off completely contrary to how it was intended. I hope you'll forgive me. P.S. My original comment was also a joke on Paul Erdos. Once again, nothing mean was meant. Thanks, Limitless
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