Problem (4) : Integral Trigonometri


soal ini diambil dari kolom komentar pengunjung

Nilai dari \int tan2 x dx = …

PENYELESAIAN :

\int (tan x – sec x)2 dx = \int (tan2 x – 2 tan x sec x + sec2 x) dx

= \int tan2 x dx – 2 \int tan x sec x dx + \int sec2 x dx

= (tan x – x) – 2 sec x + tan x + c

= 2 tan x – 2 sec x + c

NOTE :

sec2 x = 1 + tan2 x

\int tan2 x dx = \int (sec2x – 1) dx

= \int sec2x dx – \int 1 dx

= tan x – x + c

\int sec2x dx = tan x (BUKTI)

\int tan x sec x dx = \int 2\frac{sin \quad x}{cos \quad x} \frac{1}{cos \quad x} dx

= \int 2 \frac{1}{cos^2x} sin x dx (menggunakan Integral Substitusi)

misal cos x = u maka -sin x dx = du

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

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

= \int -2 u-2 du

= (-2/-1) u-1 + C

substitusi u = cos x

= 2 \frac{1}{cos \quad x} + C

= 2 sec x + C

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