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

integral-of-cos9xd

#1

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).