求∫[0-->π](1-cos³x ）dx

求∫[0-->π](1-cos³x ）dx

∫[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

∫[0,π](1-cos³x ）dx
=x[0,π]-∫[0,π]cos^2x dsinx
=π-∫[0,π](1-sin^2x) dsinx
=π-(sinx-sin^3x/3)[0,π]
=π