Question

Tolong yang pintar integral mohon dijawab ya.. besok dikumpul..

Answer 1

No 28
∫ 3 cos 2x dx = 3 ∫ cos 2x dx
= 3 × 1/2 sin 2x
= 3/2 sin 2x
Masukkan batas integralnya:
3/2 ×[ sin2(π/4) - 3/2 × sin 2(π/12)\
= 3/2[sin π/2 - sin π/6 ]
= 3/2[1 - 1/2]
= 3/2(1/2)
= 3/4 (B)

No 29
∫2 sin²x dx = 2 ∫ sin²x dx
=2 × ∫(1-cos 2x)/2 dx
= 2 × 1/2 × ∫1 - cos 2x dx
= ∫1 - cos 2x dx
= x - 1/2 sin 2x dx
Masukkan batas integralnya
=(π/2 - 1/2 sin 2(π/2) ) - (0 - 1/2 sin 2(0))
= (π/2 - 1/2 × 0) - (0 - 0)
= π/2 - 0 - 0
= π/2 = 1/2 π (C)