Indefinite integral of Sin(2x)^2(cos(2x))^2dx

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Indefinite integral of Sin(2x)^2(cos(2x))^2dx

Mathematics
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integration by parts?
no need :)
u sub

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still trying to learn intergrals better... I can derive all day long :) but turning around and going back is killin' me :)
Yea same here. Deriving is easy for me, going back is whats hurting me
you can take :\[u = \sin^2 2u\]\[du = 4\cos^2 2u\] so : \[=1/4 \int\limits_{}^{} u du\] = 1/4 [ sin^2 2x/2] I guess :)
or integration by parts if you want, but that'll be complicated ^_^
orect me if I'm wrong please.
correct*
so, basicaly, I could "reverse" these calculations, if i use the antiderivative, right?
yep ^_^, by using the antiderivative, you get the original function
i knew it
lol
Oh! Hmm is \[ \sin ^{2}2u=\sin (2u)^{2}\]
why sin^2 2u? it should be sin^2 (2x)/2 I jumped a step ahead lol
yes, that is right
becareful, put the brackets all over the thing (sin2u)^2 ^_^
yes, brackets are LIFESAVERS, i remember how i spent 2 sleeples days trying to find a bug in my program code just to realise a pair of brackets was missing and hence the result was rong
LOL! I don't like that feeling! >_<
yeah, it really sucks.... debuging can be bloody annoying
._. tell me abt it
haha! im doing this calculus to be a doctor...i honestly dont know when im going to integrate anything while being one though lol
sorius did you get your answer? ^_^
lol, mathematics is found everywhere
working on it right now lol
you use it to compute the area of something
or anything really, not just area
there is alot more :)

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