Question

Thomas' Calculus 14th Edition, Chapter 5.4 Problem 13E: The question states to evaluate the integral of pi/2 to 0 with the equation of [(1+cos2t)/2][dt]. My question is this: When taking the anti-derivative of "cos2t", why wouldn't we use the trig identity for "cos2x = cos^2x - sin^2x" before we take the anti-derivative of cos2t/2? The book says to take the AD of cos2t/2 = sin2t/4. I'm very confused here. Please help if you can.

Answer 1