Please help me the way through this sum Integrate Cos squared(3x) btwn x=pie/2 andx=0
Grrr! It's pi, not pie!
∫[0,π/2] cos^2(3x) dx
Now, you know that cos(2u) = 2cos^2(u)-1, so
cos^2(u) = (1+cos(2u))/2
and the integral becomes
∫[0,π/2] (1+cos6x)/2 dx
= x/2 + 1/12 sin6x [0,π/2]
= π/4