Question

I have another integral question.. >_<

Evaluate the definite integral.

sin^2(9x)cos^2(9x) dx

*From 6 to 13

Thanks! :D

Answer 1

I believe you may have to integrate this by parts, do you know how to do that?

Answer 2

Is it the one with the equation S udv = uv - S vdu?

Answer 3

\[\int \sin^2 (9x) (1 - \sin^2 (9x) ) dx\] \[\int \sin ^2 (9x)dx - \int \sin^3 (9x)dx\] \[\int \sin^2 (9x) dx - \int \sin (9x) (1 - \cos^2 (9x) ) dx\] \[\int \sin^2 (9x)dx + \cos(9x) - \frac{\cos^3 (9x)}{3}\]

Answer 4

for the last step i'll give you a hint \[\int \sin ^2 u = \int \frac{1 - cos(2x)}{2}\]

Answer 5

ugh another typo =_= \[\int sin^2 u = \int \frac{1 - \cos(2u)}{2}\]

Answer 6

do you get it? @donnac10

Answer 7

Yeah.. I'm trying it right now.. :D Thanks so much @lgbasallote :D

Answer 8

great! ^_^ do check the last steps of my integration though...i dont have a paper so i might have done something wrong with the + or - signs


well i gotta get to class now so goodluck :D

Answer 9

I will.. Thanks soooo much! Goodluck and have fun on your class! :D

Answer 10

I just saw @lgbasallote that you put \[\int\limits\sin^2(9x) - \int\limits \sin^3(9x)\] I'm just wondering where you got those... It's confusing me..

Answer 11

@donnac10 yeah sorry i realized that too later on...sorry for that....to compensate i shall write the correct one let a = 9x since it's confusing; da = 9dx

\[\int \sin^2 (9x) \cos^2 (9x)dx\] \[\frac{1}{9} \int \sin^2 (a) \cos^2 (a) da\]
\[\frac{1}{9} \int \cos a(\sin^2a \cos a)da\] let u = cos a and dv = sin^2 a cosa da

du = -sin a and v = sin^3 a/3

\[\frac{1}{9} \int \sin^2 (a) \cos^2 (a) da\ = \frac{1}{9} [\frac{\cos a \sin^3 a}{3} + \int \frac{\sin^4 a}{3}da]\] \[\frac{1}{9} \int \sin^2 (a) \cos^2 (a) da\ = \frac{1}{27} [\cos a \sin^3 a + \int \sin^2 a(1 - \cos^2 a)da]\] \[\frac{1}{9} \int \sin^2 (a) \cos^2 (a) da\ = \frac{1}{27} [\cos a \sin^3 a + \int \sin^2 ada - \int \sin^2 a\cos^2 ada]\]
notice how both sides have sin^ a cos^ a so i'll transpose the sin^a cos^a onthe right side to the left side \[\frac{1}{9} \int \sin^2 (a) \cos^2 (a) da + \frac{1}{27} \int \sin^2 a \cos ^2 a da = \frac{1}{27} [\cos a \sin^3 a + \int \sin^2 ada ]\]
\[\frac{4}{27} \int \sin^2 a \cos ^2 a da = \frac{1}{27} [\cos a \sin^3 a + \int \sin^2 ada ]\] now i multiply both sides by 27/4 \[\int \sin^2 a \cos ^2 a da = \frac{1}{4} [\cos a \sin^3 a + \int \sin^2 ada ]\] now that should be right...sorry for the mistake

Answer 12

dont forget to sub back to 9x later on

Answer 13

@lgbasallote Thanks you so much! That was really helpful! ;D

Answer 14

one more thing....

Answer 15

BEFORE you sub back to 9x divide both sides by 9...the left side should have a 1/9 i accidentally removed it :| hehe

Answer 16

so it should be \[\frac{1}{9} \int \sin^2 a \cos^2 a da = \frac{1}{36} [\cos a \sin^3 a + \int \sin^2 ada]\]

Answer 17

solve for the right side then sub back to 9x