Question

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

Answer 1

integration by parts?

Answer 2

no need :)

Answer 3

u sub

Answer 4

still trying to learn intergrals better... I can derive all day long :) but turning around and going back is killin' me :)

Answer 5

Yea same here. Deriving is easy for me, going back is whats hurting me

Answer 6

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 :)

Answer 7

or integration by parts if you want, but that'll be complicated ^_^

Answer 8

orect me if I'm wrong please.

Answer 9

correct*

Answer 10

so, basicaly, I could "reverse" these calculations, if i use the antiderivative, right?

Answer 11

yep ^_^, by using the antiderivative, you get the original function

Answer 12

i knew it

Answer 13

lol

Answer 14

Oh! Hmm is \[ \sin ^{2}2u=\sin (2u)^{2}\]

Answer 15

why sin^2 2u? it should be sin^2 (2x)/2

I jumped a step ahead lol

Answer 16

yes, that is right

Answer 17

becareful, put the brackets all over the thing (sin2u)^2 ^_^

Answer 18

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

Answer 19

LOL! I don't like that feeling! >_<

Answer 20

yeah, it really sucks.... debuging can be bloody annoying

Answer 21

._. tell me abt it

Answer 22

haha! im doing this calculus to be a doctor...i honestly dont know when im going to integrate anything while being one though lol

Answer 23

sorius did you get your answer? ^_^

Answer 24

lol, mathematics is found everywhere

Answer 25

working on it right now lol

Answer 26

you use it to compute the area of something

Answer 27

or anything really, not just area

Answer 28

there is alot more :)