Tolong bantu jawab!!!

intergral tertentu

(0...π/2)∫ 2 sin 4x cos 2x dx =
..
2 sin A cos B = sin (A+B) + sin (A - B)
2 sin 4x cos 2x  = sin (4x+2x) + sin (4x-2x)
2 sin 4x cos 2x = sin 6x + sin 2x
..
(0..π/2) ∫ 2 sin 4x cos 2x dx = (0..π/2) ∫ sin 6x + sin 2x dx
= { -1/6 cos 6x -  1/2 cos 2x](π/2...0)
= -1/6 ( cos 3π - cos 0) - 1/2 cos(π - cos 0)
= -1/6 (-1 - 1) - 1/2 (-1 -1)
= -1/6(-2) - 1/2 (-2)
= 1/3+ 1
=  4/3