Matematika

Pertanyaan

integral dari 4 cos^2 3x dx

1 Jawaban

  • cos 6x = 2cos² 3x - 1
    2cos² 3x = cos 6x + 1

    Int 4 cos² 3x dx
    = 2 Int (1 + cos 6x) dx
    = 2(Int dx + (1/6) Int dsin 6x)
    = 2(x + (1/6)sin 6x) + C
    = 2x + (1/3)sin 6x + C

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