Use integration by parts TWICE and solve for the integral.

It goes like this.

1) Original Integral with sine (A) <== This is where we started.

2) Expression with cosine (B) plus an integral with cosine (C). <== This is where one might give up, realizing that cosine and sine will just keep changing back and forth. Don't quit. Do it again.

3) Expression with cosine (B) + expression with sine (D) + integral with sine (E). - Are you discouraged yet. You will be discouraged until you notice one thing, that "Integral with sine" is just "Original integral with sine" multiplied by a constant. I'll use a constant of (1/9).

A = B + D + E

A = B + D + (1/9)A

(8/9)A = B + D

A = (9/8)(B+D) and you are done.

Definitely put this process in your pocket. It can be very useful. It should be interesting to you that the otherwise difficult calculus problem was solved by some rather ancient algebra. Don't forget your algebra!!