(a) Since u = cos(x) gives du = -sin(x) dx, we have
∫ sin(x) cos(x) dx = - ∫ (-sin(x)) cos(x) dx
= - ∫ cos(x) d(cos(x))
= - ∫ u du
= - 1/2 u² + C
= -1/2 cos²(x) + C
(b) Now u = sin(x) gives du = cos(x) dx, so
∫ sin(x) cos(x) dx = ∫ sin(x) d(sin(x))
= ∫ u du
= 1/2 u² + C
= 1/2 sin²(x) + C
(c) Since sin(2x) = 2 sin(x) cos(x), we have
∫ sin(x) cos(x) dx = 1/2 ∫ sin(2x) dx
Substitute u = 2x, so that du = 2 dx, and
∫ sin(x) cos(x) dx = 1/2 ∫ sin(2x) dx
= 1/4 ∫ 2 sin(2x) dx
= 1/4 ∫ sin(u) du
= -1/4 cos(u) + C
= -1/4 cos(2x) + C
(d) Integrate by parts, setting
u = sin(x) ==> du = cos(x) dx
dv = cos(x) dx ==> v = sin(x)
Then
∫ sin(x) cos(x) dx = sin²(x) - ∫ sin(x) cos(x) dx
2 ∫ sin(x) cos(x) dx = sin²(x) + C
∫ sin(x) cos(x) dx = 1/2 sin²(x) + C
The solutions in (b) and (d) are identical, but all 4 are equivalent, and this follows from the identities,
sin²(x) + cos²(x) = 1
cos(2x) = cos²(x) - sin²(x)
