Question

integral of sin^3(3x)*cos^5(3x)dx

Answer 1

\[\sin^3(3x)\cos^5(3x)=\sin(3x)\sin^2(3x)\cos^5(3x)=\sin(3x)(1-\cos^2(3x))\cos^5(3x)\]

Answer 2

wow youre amazing!

Answer 3

but there is more to it right?

Answer 4

now use u-substitution

Answer 5

so u =sin(3x)?

Answer 6

no

Answer 7

then it equals the cos?

Answer 8

yes u=cos(3x)

Answer 9

so du= -sin(3x) ! :)

Answer 10

no...but close

Answer 11

chain rule

Answer 12

times 3 also!

Answer 13

yes

Answer 14

and don't forget your dx

Answer 15

and now dv=sin(3x)?

Answer 16

right! youre so amazing!

Answer 17

please stop saying that....

\[du= -\sin(3x)3 dx\]

Answer 18

yes thats what I got

Answer 19

im sorry

Answer 20

now replace everything, simplify (or expand) and then integrate

Answer 21

dont I need to also get dv?

Answer 22

dv?

what is v


are you thinking about integration by parts?

Answer 23

oh yeah that must be it...
so in u subsitution I just use u and then replace it?

Answer 24

just u

Answer 25

find du...replace the stuff...then integrate

Answer 26

how can i replace the sin(3x) though and the 1-cos^2

Answer 27

cuz I dont have those in either u nor du

Answer 28

\[du=-\sin(3x)3dx\]

\[-\frac{1}{3}du=\sin(3x)dx\]

Answer 29

I have to go, but thank you so much for your help. I really appreciate it. I have to look into this u subsitution some more.

Answer 30

oh I see.. so change the du accordingly...thank you a lot

Answer 31

\[1-\cos^2(3x)\]

becomes

\[1-u^2\]