Question

求∫[0-->π](1-cos³x ）dx

Answer 1

∫[0-->π](1-cos³x ）dx

=∫[0-->π]dx-∫[0-->π]cos^2x cosxdx

=x|(0-->π)-∫[0-->π](1-sin^2x)dsinx

=π-∫[0-->π]dsinx+∫[0-->π]sin^2xdsinx

=π-sinx|(0-->π)+1/3*sin^3x|(0-->π)

=π