## anonymous 5 years ago Indefinite integral of Sin(2x)^2(cos(2x))^2dx

1. amistre64

integration by parts?

2. anonymous

no need :)

3. anonymous

u sub

4. amistre64

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

5. anonymous

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

6. anonymous

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 :)

7. anonymous

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

8. anonymous

orect me if I'm wrong please.

9. anonymous

correct*

10. anonymous

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

11. anonymous

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

12. anonymous

i knew it

13. anonymous

lol

14. anonymous

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

15. anonymous

why sin^2 2u? it should be sin^2 (2x)/2 I jumped a step ahead lol

16. anonymous

yes, that is right

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

._. tell me abt it

22. anonymous

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

23. anonymous

24. anonymous

lol, mathematics is found everywhere

25. anonymous

working on it right now lol

26. anonymous

you use it to compute the area of something

27. anonymous

or anything really, not just area

28. anonymous

there is alot more :)