posted by Chiquesta .
Express sin^6theta in multiples of costheta and hence evaluate the integral of sin^6theta from 0 to pi/2.
Thanks in advance! :)
sin^6θ = (sin^2θ)^3
= (1 - cos^2θ)^3
= 1 - 3cos^2θ + 3cos^4θ + cos^6θ
Now you can use your half-angle formula
cos^2θ = (1 + cos 2θ)/2
to get no exponents and multiples of θ.
You will end up with
1/192 (60θ - 45sin2θ + 9sin4θ - sin6θ)
from 0 to pi/2 yields 30pi/192