Compute the following integrals

The integral from negative to positive pi of:
a) 1/2pi (cos(5t))(cos(10t)) dt
b) 1/2pi (cos(3t))(sin(11t)) dt
c) 1/pi (cos^2 (2t)) dt
d) 1/pi (sin^2 (3t)) dt

Remember the following:
Integral from negative to positive pi of an even function is 2 times integral from zero to pi.
Integral from negative to positive pi of an odd function is zero.
2cosAcosB = cos(A+B) + cos(A-B)
2sinAcosB = sin(A+B) + sin(A-B)
2cos^2(t) = 1+cos2t
2sin^2(t) = 1-cos(2t)
Now come to the solution.
a) int -pi to pi 1/2pi cos5tcos10t dt = int 0 to pi 1/2pi (cos15t+cos5t)dt = 1/2pi 0 to pi [(sin15t)/15 + (sin5t)/5] = 1/pi(0+0) = 0.
b) int -pi to pi 1/2pi cos3tsin11t dt = 1/2pi(0) = 0.
c) int -pi to pi 1/pi cos^2(2t)dt = 1/pi int 0 to pi [1+cos4t]dt = 1/pi 0 to pi [t+(sin4t)/4] = 1/pi(pi+0) = 1.
d) int -pi to pi 1/pi sin^2 (2t)dt = 1/pi int 0 to pi [1-cis6t]dt = 1/pi 0 to pi [t-sin6t] = 1.