1. Integral 4 cos²x
2. Integral 6 sin² 2x
Jawaban Terkonfirmasi
1•
2cos² x = cos 2x + 1
∫4cos² x dx
= 2 ∫2cos² x dx
= 2 ∫(1 + cos 2x) dx
= 2(x + ∫cos 2x dx)
= 2(x + (1/2) ∫2cos 2x dx)
= 2(x + (1/2) ∫dsin 2x)
= 2x + sin 2x + C
2•
2sin² 2x = 1 – cos 4x
∫6 sin² 2x dx
= 3 ∫(1 – cos 4x) dx
= 3(x – 1/4 sin 4x) + C
= 3x – (3/4) sin 4x + C