batas atas phi/2 batas bawah 0 integral (1 - cos x) sin x dx

∫(1 - cos x) sin x dx [π/2 ... 0]
= -∫(1 - cos x) dcos x
= - ∫dcos x + ∫cos x dcos x
= -cos x + 1/2 cos² x
= (-cos π/2 + 1/2 cos² π/2) - (-cos 0 + 1/2 cos² 0)
= (0 + 0) - (-1 + 1/2 . 1²)
= 0 - (-1/2)
= 1/2