**can someone do the integral of cos^9xdx, this is the location of the problem and it gives no explanation ?**

**Answer:**

1/9)sin(9x)+C

Let U=9x

DU=9dx. In other words, (1/9)DU=dx [via division].

Now, factor the constant (1/9) out in front of the integral.

You are now left with the integral of cos(u)DU.

That equals sin(u)+C *** don’t forget the constant.

It becomes (1/9)sin(u)+C after you carry the one ninth.

Now: Resubstitute in for U

(1/9)sin(9x) +C

Now, let’s double check. The derivative of (1/9)sin(9x)+C.

The derivative of sine is cosine

and then using the chain rule, we take the derivative of 9x.

And finally the C goes to zero.

We should get (9/9)cos(9x)+0

Which then equals cos(9x).