Integral dari Invers Fungsi Trigonometri


Untuk integral dari invers fungsi trigonometri, saya akan memanfaatkan Integral Parsial, Integral Sustitusi dan sifat dasar dari Integral Sustitusi Trigonometri. Integral yang dibahas dalam tulisan ini adalah integral dari arc sin x, arc cos x, arc tan x, arc cosec x, arc sec x dan arc cotan x. Berikut integral dari fungsi tersebut.

\int arc sin x dx = …

ambil : u = arc sin x \Rightarrow du = \frac{1}{\sqrt{1-x^2}} dx [bukti]

dv = dx \Rightarrow v = x

\int arc sin x dx = x arc sin x – \int x \frac{1}{\sqrt{1-x^2}} dx

misal : a = 1 – x2 \Rightarrow da = -2x dx

= x arc sin x – \int x \frac{1}{\sqrt{a}} \frac{da}{-2x}

= x arc sin x + \frac{1}{2} \int a-1/2 da

= x arc sin x + a1/2 + C

= x arc sin x + \sqrt{1-x^2} + C

\int arc sin x dx = x arc sin x + \sqrt{1-x^2} + C

.

\int arc cos x dx = …

ambil : u = arc cos x \Rightarrow du = \frac{-1}{\sqrt{1-x^2}} dx [bukti]

dv = dx \Rightarrow v = x

\int arc cos x dx = x arc cos x – \int x \frac{-1}{\sqrt{1-x^2}} dx

misal : a = 1 – x2 \Rightarrow da = -2x dx

= x arc cos x – \int x \frac{-1}{\sqrt{a}} \frac{da}{-2x}

= x arc cos x – \frac{1}{2} \int a-1/2 da

= x arc cos x – a1/2 + C

= x arc cos x – \sqrt{1-x^2} + C

\int arc cos x dx = x arc cos x – \sqrt{1-x^2} + C

.

\int arc tan x dx = …

ambil : u = arc tan x \Rightarrow du = \frac{1}{1+x^2} dx [bukti]

dv = dx \Rightarrow v = x

\int arc tan x dx = x arc tan x – \int x \frac{1}{1+x^2} dx

misal : a = 1 + x2 \Rightarrow da = 2x dx

= x arc tan x – \int x \frac{1}{a} \frac{da}{2x}

= x arc tan x – \dfrac{1}{2} \int  \dfrac{1}{a} da

= x arc tan x – \dfrac{1}{2} ln|a| + C

= x arc tan x – \dfrac{1}{2} ln|1 + x2| + C

\int arc tan x dx = x arc tan x – \dfrac{1}{2} ln|1 + x2| + C

.

\int arc csc x dx = …

ambil : u = arc csc x \Rightarrow du = \frac{-1}{x\sqrt{x^2-1}} dx [bukti]

dv = dx \Rightarrow v = x

\int arc csc x dx = x arc csc x – \int x \frac{-1}{x\sqrt{x^2-1}} dx

= x arc csc x + \int \frac{1}{\sqrt{x^2-1}} dx

misal : x = sec t \Rightarrow dx = sec t tan t dt

= x arc csc x + \int \frac{1}{\sqrt{sec^2 t-1}} sec t tan t dt

= x arc csc x + \int \frac{1}{tan \quad t} sec t tan t dt

= x arc csc x + \int sec t dt

= x arc sec x + ln|sec t + tan t| + C [bukti]

\int arc csc x dx = x arc sec x + ln|sec t + tan t| + C

.

\int arc sec x dx = …

ambil : u = arc sec x \Rightarrow du = \frac{1}{\sqrt{x^2-1}} dx [bukti]

dv = dx \Rightarrow v = x

\int arc sec x dx = x arc sec x – \int x \frac{1}{x\sqrt{x^2-1}} dx

= x arc sec x – \int \frac{1}{\sqrt{x^2-1}} dx

misal : x = sec t \Rightarrow dx = sec t tan t dt

= x arc sec x – \int \frac{1}{\sqrt{sec^2 t-1}} sec t tan t dt

= x arc sec x – \int \frac{1}{tan \quad t} sec t tan t dt

= x arc sec x – \int sec t dt

= x arc sec x – ln|sec t + tan t| + C [bukti]

.

\int arc sec x dx = x arc sec x – ln|sec t + tan t| + C

.

\int arc cot x dx = …

ambil : u = arc cot x \Rightarrow du = \frac{-1}{1+x^2} dx [bukti]

dv = dx \Rightarrow v = x

\int arc cot x dx = x arc cot x – \int x \frac{-1}{1+x^2} dx

misal : a = 1 + x2 \Rightarrow da = 2x dx

= x arc cot x + \int x \frac{1}{a} \frac{da}{2x}

= x arc cot x + \frac{1}{2} \int \frac{1}{a} da

= x arc cot x + ln|a| + C

= x arc cot x + ln|1 + x2| + C

\int arc cot x dx = x arc cot x + ln|1 + x2| + C

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