Integral Trigonometri

November 12, 2011

Integral Trigonometri merupakan hasil kebalikan dari turunan trigonometri. Sebelum kita mencoba mengingat rumus-rumus integral triogonometri maka sebaiknya kita ingat dulu turunan trigonometri. Turunan trigonometri bisa kita tuliskan sebagai berikut
y = sin x maka y' = cos x
y = cos x maka y' = - sin x
y = tan x maka y' = sec² x
y = cot x maka y' = -csc² x
y = sec x maka y' = sec x tan x
y = csc x maka y' = -csc x cot x

Dengan demikian jika rumus-rumus ini kita balik akan menjadi
$\int cos x dx = sin x + c$
$\int sin x dx = -cos x + c$
$\int sec^{2} x dx = tan x + c$
$\int csc^{2} x dx = -cot x + c$
$\int sec x tan x dx = sec x + c$
$\int csc x cot x dx = -csc x + c$

Rumus-rumus tersebut bisa dibuat lebih umum sebagai berikut
$\int cos (ax+b) dx = \frac{1}{a}sin(ax+b) + c$
$\int sin (ax+b) dx = -\frac{1}{a}cos(ax+b) + c$
$\int sec^{2} (ax+b) dx = \frac{1}{a}tan(ax+b) + c$
$\int csc^{2} (ax+b) dx = -\frac{1}{a}cot(ax+b) + c$
$\int sec (ax+b)tan (ax+b) dx = \frac{1}{a}sec(ax+b) + c$
$\int csc (ax+b)cot (ax+b) dx = -\frac{1}{a}csc(ax+b) + c$

Untuk lebih jelasnya kita bisa membuktikan sebagai berikut
misalkan : y = ax + b
maka
$\frac{dy}{dx}=a\rightarrow dx=\frac{dy}{a}$
Jadi :
$\int cos (ax+b)dx = \int cos y \frac{dy}{a}=\frac{1}{a}\int cosy dy = \frac{1}{a}siny+c$
$=\frac{1}{a}sin(ax+b) + c$

$\int sin (ax+b)dx = \int sin y \frac{dy}{a}=\frac{1}{a}\int sin y dy = -\frac{1}{a}cosy+c$
$= -\frac{1}{a}cos(ax+b)+c$

$\int sec^{2} (ax+b)dx = \int sec^{2} y \frac{dy}{a}=\frac{1}{a}\int sec^{2} y dy = \frac{1}{a}tany+c$
$= \frac{1}{a}tan(ax+b)+c$

$\int csc^{2} (ax+b)dx = \int csc^{2} y \frac{dy}{a}=\frac{1}{a}\int csc^{2} y dy = -\frac{1}{a}coty+c$
$= -\frac{1}{a}cot(ax+b)+c$

$\int sec (ax+b)tan (ax+b)dx = \int sec y tan y \frac{dy}{a}=\frac{1}{a}\int sec y tan y dy$
$= \frac{1}{a}sec y+c = \frac{1}{a}sec(ax+b)+c$

$\int csc (ax+b)cot (ax+b)dx = \int csc y cot y \frac{dy}{a}=\frac{1}{a}\int csc y cot y dy$

$= -\frac{1}{a}csc y+c = -\frac{1}{a}csc(ax+b)+c$

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