Matematika

Pertanyaan

Integral sin 6x cos3x dx

1 Jawaban

  • 2 sin a cos b = sin (a + b) + sin (a - b)

    •••
    ∫sin 6x cos 3x dx
    = 1/2 ∫(sin (6x + 3x) + sin (6x - 3x)) dx
    = 1/2 ∫(sin 9x + sin 3x) dx
    = 1/2 (-1/9 cos 9x - 1/3 cos 3x) + C
    = -1/18 cos 9x - 1/6 cos 3x + C

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