Evaluate the integral using any method..?

Question

integral of 48sinxcosx dx from 0 to pi/6

thanks for the help :)

Answer 1

Use the identity 2 sin x cos x = sin 2x.

∫ (48 sin x cos x) dx = ∫ (24 sin 2x) dx = -12 cos 2x.

Evaluating from 0 to pi/6, we get

-12 cos (pi/3) + 12 cos (0) = -12 * (1/2) + 12 * (1) = 6.

Answer 2

factor out a 24 so you get
24 * integral(2sinxcosx dx)
using trig identities, you get
24 * integral(sin2x dx)
substitute, u=2x du=2
24 * integral(sinu (1/2)du) now from 0 to pi/3
12 * integral(sinu du)
12 *( -cos(pi/3) + cos(0))
12 * 0.5

6