Matematika

Pertanyaan

1. Integral 4 cos²x
2. Integral 6 sin² 2x

1 Jawaban

  • 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

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