tolong bantu integral ya

INTEGRAL TRIGONOMETRI

∫(4cos 2x - 3 sin 3x) dx [π/2_π/3]
= 4 ∫cos 2x dx - 3 ∫sin 3x dx
= 4 . 1/2 ∫dsin 2x - 3 (- 1/3 ∫dcos 3x)
= 2 sin 2x + cos 3x
= (2 sin (2.π/2) + cos 3.π/2) - (2 sin (2.π/3) + cos 3.π/3)
= 0 - (√3 - 1)
= 1 - √3

∫sin (2x - π/4) dx [π/4_0]
= -1/2 ∫dcos (2x - π/4)
= -1/2 cos (2x - π/4)
= (-1/2 cos (2.π/4 - π/4) - (-1/2 cos (2.0 - π/4))
= (-1/4 √2) - (-1/4 √2)
= 0