Question

Evaluate ∫ sin(x) cos(x) dx by four methods.

Answer 1

(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)