Originally Posted by

**elocin** I have: 64 integral from (pi/4) to (pi/3) (1+2cos(2theta)+cos(2theta)^2) dtheta. Then I split those up into the integrals (still multiplying everything by 64) of 1, 2cos(2theta) and cos(2theta)^2. I have a lot of work shown, but I ended up getting the answer 64(3pi/32+sin(pi/6)+1/8sin(pi/3). That answer isn't right and I simplified it to the best of my ability and I don't know where I went wrong! But for the integral of 1, I got (pi/3-pi/4) and for the integral of 2cos(2theta) I did u substitution and got sin(2theta) and used the limits, and then for cos(2theta)^2 I used the half angle formula and got (1+cos(4theta) over 2, since its double that of a single half angle formula. Using the formula, I got (1/8) integral of 1 + integral sin(4theta). SO that was for the last integral when I first split everything up.