Question

integrate sin^3(6x)cos^2(6x) dx

Answer 1

Hmm so we have an `odd power`, that's really good :)

Answer 2

yeah, you can pull off a sinx

Answer 3

\[\Large \int\limits \sin6x \color{royalblue}{\sin^26x}\cos^26x\;dx\]Yah that sounds right peg.
We'll want to use our `Square Identity` to convert this blue part to cosines.

Answer 4

\[\int\limits_{?}^{?} (1-\cos ^{2}6x)\cos ^{2}6xsin6x\]

Answer 5

yah looks good.
NOW you can do a u-sub ^^

u=cos6x

Answer 6

this is where I'm getting confused \[\int\limits_{}^{} (1-u ^{2}u ^{2} du ?\]

Answer 7

woops, forgot the 2nd parenthesis

Answer 8

I think you get a negative on your du right?\[\large u=\cos6x \qquad\to\qquad du=-6\sin6x\;dx \qquad\to\qquad \left(-\frac{1}{6}du\right)=\sin6x\;dx\]

Answer 9

yeah, that's right

Answer 10

That wasn't the part you were confused on I guess, I just wanted to make sure :)\[\Large \int\limits(1-u^2)u^2\left(-\frac{1}{6}du\right)\]

Answer 11

Next, just distribute the u^2 to each term in the brackets.
Then you can apply the power rule as normal.

Answer 12

excellent, just got the answer. I see what I was doing wrong.

Answer 13

cool c:

Answer 14

thank you!