BAB 2 : TEKNIK INTEGRASI

2.1 Integrasi Parsial dan Integrasi Fungsi Trigonometri

A) Rangkuman Materi

1 Integrasi Parsial

\(\int u \, dv = uv - \int v \, du\)

2 Reduksi untuk Integral Fungsi Trigonometri

untuk \(n \geq 2\) dan \(n \in Z\), berlaku

\(\int sin^n x dx = -\frac{1}{n} sin^{n-1}x cos x+ \frac{n-1}{n} \int sin^{n-2} x dx\)

\(\int cos^n x dx = \frac{1}{n} cos^{n-1}x sin x+ \frac{n-1}{n} \int cos^{n-2} x dx\)

3 Pengintegralan Perpangkatan Sinus dan Cosinus

\(\int \sin^m x \cos^n x \, dx\)	LANGKAH
m ganjil	Substitusi \(m = 2k + 1\). Substitusi \(\sin^2 x = 1 - \cos^2 x\), sehingga \(\sin^{2k} x \sin x \cos^n x\) menjadi \((1 - \cos^2 x)^k \sin x \cos^n x\). Substitusi \(u = \cos x\).
n ganjil	Substitusi \(n = 2k + 1\). Substitusi \(\cos^2 x = 1 - \sin^2 x\), sehingga \(\sin^m x \cos^{2k} x \cos x\) menjadi \(\sin^m x (1 - \sin^2 x)^k \cos x\). Substitusi \(u = \sin x\).
m dan n genap	Substitusi \(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\) dan \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\). Ingat \((1 - \cos 2x)(1 + \cos 2x) = 1 - \cos^2 2x\).

\(\sin mx \cos nx\)	\(\displaystyle \frac{\sin((m+n)x) + \sin((m-n)x)}{2}\)
\(\sin mx \sin nx\)	\(\displaystyle \frac{\cos((m-n)x) - \cos((m+n)x)}{2}\)
\(\cos mx \cos nx\)	\(\displaystyle \frac{\cos((m+n)x) + \cos((m-n)x)}{2}\)
Ingat: \(\displaystyle \int \sin ax = -\frac{1}{a} \cos ax\) dan \(\displaystyle \int \cos ax = \frac{1}{a} \sin ax\)

4 Integrasi Perpangkatan Secan dan Tangen

\(\int \tan^m x \sec^n x \, dx\)	LANGKAH
n genap	Substitusi \(m = 2l,\, l \geq 0\). Substitusi \(\sec^2 x = \tan^2 x + 1\), sehingga \(\tan^m x (\sec^2 x)^{l-1} \sec^2 x\) menjadi \(\tan^m x (\tan^2 x + 1)^{l-1} \sec^2 x\). Substitusi \(u = \tan x\).
m ganjil	Substitusi \(m = 2l + 1,\, l \geq 0\). Substitusi \(\tan^2 x = \sec^2 x - 1\), sehingga \((\tan^2 x)^k \sec^{n-1} x \sec x \tan x\) menjadi \((\sec^2 x - 1)^k \sec^{n-1} x \sec x \tan x\). Substitusi \(u = \sec x\).
m genap dan n ganjil	Substitusi \(m = 2l,\, l \geq 0\). Substitusi \(\tan^2 x = \sec^2 x - 1\), sehingga \((\tan^2 x)^k \sec^n x\) menjadi \((\sec^2 x - 1)^k \sec^n x\). Substitusi \(u = \sec x\).

5 Rumus-rumus penting Trigonometri

\(\tan x = \frac{\sin x}{\cos x}\)	\(\cos 2x = 2\cos^2 x - 1\)
\(\cot x = \frac{\cos x}{\sin x}\)	\(\cos 2x = 1 - 2\sin^2 x\)
\(\sec x = \frac{1}{\cos x}\)	\(\tan 2x = \frac{2\tan x}{1 - \tan^2 x}\)
\(\csc x = \frac{1}{\sin x}\)	\(\sin 3x = 3\sin x - 4\sin^3 x\)
\(\sin^2 x + \cos^2 x = 1\)	\(\cos 3x = 4\cos^3 x - 3\cos x\)
\(1 + \tan^2 x = \sec^2 x\)	\(2\sin x \cos y = \sin(x+y) + \sin(x-y)\)
\(1 + \cot^2 x = \csc^2 x\)	\(2\cos x \sin y = \sin(x+y) - \sin(x-y)\)
\(\sin 2x = 2\sin x \cos x\)	\(2\cos x \cos y = \cos(x+y) + \cos(x-y)\)
\(\cos 2x = \cos^2 x - \sin^2 x\)	\(2\sin x \sin y = \cos(x-y) - \cos(x+y)\)

B) Contoh Soal

1. Hitung integral \(\int x \ln (x)\) dari
Pembahasan: Misalkan

\(u = \ln (x)\) dan \(dv = x \, dx\)

Sehingga,

\(du = \frac{1}{x} \, dx\) dan \(v = \frac{x^2}{2}\)

Dengan menggunakan integral parsial, diperoleh

\(\int u dv = uv - \int v du\)

\(= \int x \ln (x) \, dx = \ln(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx\)

\(= \frac{x^2}{2} \ln (x) - \frac{1}{2} \int x \, dx\)

\(= \frac{x^2}{2} \ln (x) - \frac{1}{2} \cdot \frac{x^2}{2} + C\)

\(= \frac{x^2}{2} \ln (x) - \frac{x^2}{4} + C\)

2. (Soal ETS 2020)
Dapatkan integral \(\int \sin^2 x \cos^3 x dx\)
Pembahasan:

\(\int (\sin^2 x \cos^2 x) \cos x dx\)

=\(\int \sin^2 x(1-\sin^2 x) \cos x dx\)

Misalkan u = \(\sin x\), sehingga \(du = \cos x dx\)

=\(\int sin^2 x cos^3 x dx= \int (sin^2 x(1-sin^2 x)) cos x dx\)

=\(\int (u^2(1-u^2)) du\)

=\(\int (u^2 - u^4) du\)

=\(\frac{u^3}{3} - \frac{u^5}{5} + C\)

=\(\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C\)

3. Selesaikan \(\int \mathrm{sec}^4 x \, dx\)
Pembahasan: Gunakan integral perpangkatan secant dan tangent untuk \(n\) genap,

\(\int \mathrm{sec}^4 x \, dx=\int (\mathrm{sec}^2 x)(\mathrm{sec}^2 x) \, dx\)

=\(\int (tan^2 x + 1)(\mathrm{sec}^2 x) \, dx\)

Misalkan \(u = \tan x\), sehingga \(du = \mathrm{sec}^2 x \, dx\)

=\(\int \mathrm{sec}^2 x \, dx = \int (u^2 +1) \, du\)

=\(\frac{u^3}{3} + u + C\)

=\(\frac{\tan^3 x}{3} + \tan x + C\)

C) Latihan Soal

1. Soal Kuis
Dapatkan \(\int (e^{2x}sin(3x)) dx\)
Hint 1: Gunakan integrasi parsial dengan memisahkan \(u = e^{2x}\) dan \(dv = \sin(3x)dx\)
Hint 2: Integrasi parsial sebanyak dua kali

Pembahasan

Misalkan

\(u = e^{2x}\) dan \(dv = \sin(3x)dx\)

Sehingga,

\(du = 2e^{2x}dx\) dan \(v = -\frac{1}{3}\cos(3x)\)

Dengan menggunakan integral parsial, diperoleh

\(\int u \, dv = uv - \int v \, du\)

\(\int e^{2x} \sin(3x)dx = -\frac{1}{3}e^{2x}\cos(3x) - \int -\frac{1}{3}\cos(3x)(2e^{2x})dx\)

\(= -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}\int e^{2x}\cos(3x)dx\)

Perhatikan bahwa integral \(\int e^{2x}\cos(3x)dx\) harus dilakukan integrasi parsial lagi. Misalkan

\(u = e^{2x}\) dan \(dv = \cos(3x)dx\)

Sehingga,

\(du = 2e^{2x}dx\) dan \(v = \frac{1}{3}\sin(3x)\)

Dengan menggunakan integral parsial lagi, diperoleh

\(\int e^{2x} \cos(3x)dx = \frac{1}{3}e^{2x}\sin(3x) + \int \frac{2}{3}\cos(3x)(e^{2x})dx\)

\(= \frac{1}{3}e^{2x}\sin(3x) + \frac{2}{3}\int e^{2x}\cos(3x)dx\)

Sehingga, kita dapat menyusun kembali integral awal:

\(\int e^{2x} \sin(3x)dx = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin(3x) + \frac{2}{3}\int e^{2x}\cos(3x)dx\right)\)

\(= -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{9}e^{2x}\sin(3x) + \frac{4}{9}\int e^{2x}\cos(3x)dx\)

Dengan menyusun kembali, kita mendapatkan:

\(\frac{13}{9}\int e^{2x}\sin(3x)dx = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{9}e^{2x}\sin(3x)\)

Sehingga, integralnya adalah:

\(\int e^{2x}\sin(3x)dx = -\frac{3}{13}e^{2x}\cos(3x) + \frac{2}{13}e^{2x}\sin(3x) + C\)

2. Soal ETS 2024
Hitung integral \(\int \ln (t^2 + 1 dt)\)
Hint 1: Gunakan integrasi parsial dengan \( u = \ln(t^2 + 1) \)
Hint 2: \(\int \frac{1}{x^2+1} dx = \arctan x\)

Pembahasan

Misalkan

\(u = \ln(t^2 + 1)\) dan \(dv = dt\)

Sehingga,

\(du = \frac{2t}{t^2 + 1} dt\) dan \(v = t\)

Dengan menggunakan integral parsial, diperoleh

\(\int u \, dv = uv - \int v \, du\)

\(\int \ln(t^2 + 1) dt = t \ln(t^2 + 1) - \int t \cdot \frac{2t}{t^2 + 1} dt\)

\(= t \ln(t^2 + 1) - 2\int \frac{t^2}{t^2 + 1} dt\)

Perhatikan bahwa integral \(\int \frac{t^2}{t^2 + 1} dt\) dapat disederhanakan:

\(\int \frac{t^2}{t^2 + 1} dt = \int \left(1 - \frac{1}{t^2 + 1}\right) dt\)

\(= \int dt - \int \frac{1}{t^2 + 1} dt\)

\(= t - \arctan(t) + C\)

Sehingga, kita dapat menyusun kembali integral awal:

\(\int \ln(t^2 + 1) dt = t \ln(t^2 + 1) - 2\left(t - \arctan(t) + C\right)\)

\(= t \ln(t^2 + 1) - 2t + 2\arctan(t) + C\)

Dengan demikian, integralnya adalah:

\(\int \ln(t^2 + 1) dt = t \ln(t^2 + 1) - 2t + 2\arctan(t) + C\)

3. Soal ETS 2020 Selesaikan \(\int (sin x cos 3x)\)
Hint 1: Gunakan |(sin x cos y = \(\frac{1}{2}(\sin(x+y) + \sin(x-y))\)|

Pembahasan:

\(\int (sin x cos 3x) dx = \int \frac{sin(4x) +sin (-2x)}{2} dx\)

\(= \frac{1}{2} \int sin(4x) dx + \frac{1}{2} \int sin(-2x) dx\)

\(= -\frac{1}{8} cos(4x) - \frac{1}{4} cos(-2x) + C\)

4. Soal ETS 2020 Selesaikan \(\int (sin^5(4x)cos^2(4x) \)
Hint 1: Gunakan "Pengintegralan Perpangkatan Sinus dan Cosinus" dengan m ganjil

Pembahasan:

\(\int (sin^5(4x)cos^2(4x)) dx = \int ((sin^2(4x)))^2sin(4x)cos^2(4x) dx\)

\(= \int ((1-cos^2(4x)))^2sin(4x)cos^2(4x) dx\)

Misalkan \(u = cos(4x)\), sehingga \(du = -4sin(4x)dx\) ↔ \(\frac{-1}{4}du = sin(4x)dx\)

\(= \int sin^5(4x)cos^2(4x) dx = \int (1-u^2)^2u^2 \cdot \frac{-1}{4}du\)

\((1-u^2)^2 (\frac{-1}{4}du)

\(= \frac{-1}{4} \int (1-2u^2+u^4)u^2 du\)

\(= \frac{-1}{4} \int (u^2 - 2u^4 + u^6) du\)

\(= \frac{-1}{4} (\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}) + C\)

\(= \frac{-1}{12} cos^3(4x) + \frac{2}{10} cos^5(4x) - \frac{1}{28} cos^7(4x) + C\)

5. Selesaikan \(\int (\mathrm{sec}^5 x tan^3 x) dx\)
Hint 1: Gunakan "Pengintegralan Perpangkatan Secan dan Tangen" dengan m ganjil

Pembahasan:

\(\int (\mathrm{sec}^5 x tan^3 x) dx = \int (\mathrm{sec}^4 x tan^2 x) \mathrm{sec} x tan x dx\)

\(= \int (\mathrm{sec}^4 x (\mathrm{sec}^2 x - 1)) \mathrm{sec} x tan x dx\)

Misalkan \(u = \mathrm{sec} x\), sehingga \(du = \mathrm{sec} x \tan x dx\)

\(= \int (\mathrm{sec}^4 x (\mathrm{sec}^2 x - 1)) \mathrm{sec} x tan x dx\)

\(= \int (u^4(u^2 - 1)) du\)

\(= \int (u^6 - u^4) du\)

\(= \frac{u^7}{7} - \frac{u^5}{5} + C\)

\(= \frac{\mathrm{sec}^7 x}{7} - \frac{\mathrm{sec}^5 x}{5} + C\)

6. Dapatkan \(\int (\cot^3 x csc^3 x) dx\)
Hint 1: Substitusi \(\cot^2 x = \csc^2 x - 1\)
Hint 2: Substitusi \(u = \csc x\)

Pembahasan:

\(\int (\cot^3 x \mathrm{csc}^3 x) dx = \int (\cot^2 x \csc^2 x \cot x \mathrm{csc} x) dx\)

\(= \int ((\csc^2 x - 1) \csc^2 x \cot x \mathrm{csc} x) dx\)

Misalkan \(u = \csc x\), sehingga \(du = -\csc x \cot x dx\) ↔ \(-du = \csc x \cot x dx\)

\(= -\int ((u^2 - 1) u^2) du\)

\(= -\int (u^4 - u^2) du\)

\(= -\left(\frac{u^5}{5} - \frac{u^3}{3}\right) + C\)

\(= -\left(\frac{\mathrm{csc}^5 x}{5} - \frac{\mathrm{csc}^3 x}{3}\right) + C\)