How do you evaluate the integral (1-cos(x)) ^2 dx?

Answer 1

This is the inverse of the chain rule for differentiation:

d/dx g ( f(x) ) = g' (f(x) * f'(x)

For g, integrate 1-cos(x) to get x - sin(x) + C

For f, it is simply the integral of g^2, which is g^3 / 3 + D

So, the complete integral is ( x - sin(x) + C )^3 ) / 3 + D

Differentiate this to confirm you get the original equation