Pembuktian Integral sec x tan x dx = sec x + C


\int sec x tan x dx = \int \frac{1}{cos \quad x} \quad \frac{sin \quad x}{cos \quad x} dx

= \int \frac{sin \quad x}{cos^2 \quad x} \quad d(\frac{cos \quad x}{-sin \quad x})

= \int -\frac{1}{cos^2 \quad x} d(cos x)

misal : cos x = u kemudian substitusi

= \int -\frac{1}{u^2} du

= -\frac{1}{-1} \quad \frac{1}{u} + C

kembalikan dalam bentuk trigonometri

= \frac{1}{cos \quad x} + C

= sec x + C \blacksquare

One comment on “Pembuktian Integral sec x tan x dx = sec x + C

  1. Ping-balik: Problem (23) : Integral | Math IS Beautiful

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