BAB 2 : TEKNIK INTEGRASI

2.3 Teknik-teknik Integrasi yang Lain

A) Rangkuman Materi

1 Substitusi trigonometri

Ekspresi dalam integran	Substitusi	Pembatasan θ	Kesamaan trigonometri
\(\sqrt{a^2 - x^2}\)	\(x = a \sin θ\)	\(-\pi/2 \le θ \le \pi/2\)	\(a^2 - a^2 \sin^2 θ = a^2 \cos^2 θ\)
\(\sqrt{a^2 + x^2}\)	\(x = a \tan θ\)	\(-\pi/2 < θ < \pi/2\)	\(a^2 + a^2 \tan^2 θ = a^2 \sec^2 θ\)
\(\sqrt{x^2 - a^2}\)	\(x = a \sec θ\)	\begin{cases} 0 ≤ θ ≤ π/2 (x ≥ a) \\ π ≤ θ < 3π/2 (x ≤ −a) \end{cases}	\(a^2 \sec^2 θ - a^2 = a^2 \tan^2 θ\)

2 Integral yang mencakup \(ax^2+bx+c\)

Misalkan a,b ≠ 0, integrasi yang mengandung \(ax^2+bx+c\) dapat dilakukan dengan membuat kuadrat sempurna.

3 Integral yang Memuat Pangkat Rasional

Integrasi yang mengandung pangkat rasional dapat diselesaikan dengan substitusi \(u=x^{\frac{1}{n}}\), dengan n adalah KPK dari penyebut dalam pangkat (\x\). lebih lanjut \(u=(x+a)^\frac{1}{n}\)

4 Integral yang Memuat Fungsi-fungsi rasional dalam \(sin x\) dan \(cos x\)

\(tan(\pi/2=u\)
\(x=2tan_{-1}u\)
\(sin(\pi /2)=\frac{u}{\sqrt{1+u^2}}\)
\(cos(\pi /2)=\frac{1}{\sqrt{1+u^2}}\)
\(sin x = \frac{2u}{1+u^2}\)
\(cos x = \frac{1-u^2}{1+u^2}\)
\(tan x = \frac{2u}{1-u^2}\)
\(dx = \frac{2}{(1+u^2)^2} du\)

5 Tambahan: Substitusi Hiperbolik

Ekspresi dalam integran	Substitusi	Pembatasan θ	Kesamaan trigonometri
\(\sqrt{a^2 - x^2}\)	\(x = a \sin θ\)	\(-\pi/2 \le θ \le \pi/2\)	\(a^2 - a^2 \sin^2 θ = a^2 \cos^2 θ\)
\(\sqrt{a^2 + x^2}\)	\(x = a \tan θ\)	\(-\pi/2 < θ < \pi/2\)	\(a^2 + a^2 \tan^2 θ = a^2 \sec^2 θ\)
\(\sqrt{x^2 - a^2}\)	\(x = a \sec θ\)	\begin{cases} 0 ≤ θ ≤ π/2 (x ≥ a) \\ π ≤ θ < 3π/2 (x ≤ −a) \end{cases}	\(a^2 \sec^2 θ - a^2 = a^2 \tan^2 θ\)

B) Contoh Soal

1. Soal Kuis Dapatkan

\(\int \frac{1}{x^2 \sqrt{x^2-9}}dx\)

Pembahasan: Perhatikan bahwa \(\sqrt{x^2-9} \) merupakan bentuk \(\sqrt{x^2 - a^2}\), dengan \(a=3\). Substitusi \(x=3 \mathrm{sec} θ,\) sehingga \(dx = 3 sec θ tan θ dθ.\)

\(\int \frac{1}{x^2 \sqrt{x^2-9}}dx = \int \frac{1}{(3 sec θ)^2 \sqrt{(3 sec θ)^2-9}}(3 \mathrm{sec} θ tan θ) dθ\)

=\(\int \frac{1}{9 \mathrm{sec}^2 θ \sqrt{9 \mathrm{sec}^2 θ - 9}}(3 \mathrm{sec} θ tan θ) dθ\)

=\(\int \frac{1}{3 \mathrm{sec}θ \sqrt{9 (\mathrm{sec}^2 θ - 1)}} tan θ dθ\)

=\(\int \frac{1}{3 \mathrm{sec}θ \sqrt{9 tan^2 θ}} tan θ dθ\)

=\(\int \frac{1}{3 \mathrm{sec} θtanθ}tan θ dθ\)

=\(\int \frac{1}{3 \mathrm{sec} θ} dθ\)

=\(\int \frac{cos θ}{9}dθ\)

=\(\frac{1}{9} \int cos θ dθ\)

=\(\frac{1}{9} sin θ + C\)

2. (Soal ETS 2020)
Hitung integral

\(\int \frac{1}{5+5sinx}\)

Pembahasan: Misalkan \(u = 5 + 5 \sin x\), sehingga \(du = 5 \cos x dx\) ↔ \(\frac{1}{5} du = \cos x dx\)

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

=\(\frac{1+u^2}{5(1+u^2+2u)} \cdot \frac{2}{(1+u^2)^2} du\)

=\(\frac{2}{5} \int \frac{1+u^2}{(1+u^2+2u)(1+u^2)^2} du\)

=\(\frac{2}{5} \int \frac{1+u^2}{(u+1)^2(u^2+1)} du\)

Misalkan \(v = u + 1\), sehingga \(du = dv\)

=\(\frac{2}{5} \int \frac{1}{u+1}^2 du\)

=\(\frac{2}{5} \int \frac{1}{v^2} dv\)

=\(\frac{2}{5} \cdot \frac{-1}{v} + C\)

=\(\frac{-2}{5(u+1)} + C\)

=\(\frac{-2}{5(tan(π/2)+1)} + C\)

3. Selesaikan

\(\int \frac{x}{x+3}^{\frac{1}{5}}dx\)

Pembahasan: Misalkan \(u = (x+3)^{\frac{1}{5}}\), sehingga \(du = \frac{1}{5}(x+3)^{-\frac{4}{5}}dx\) ↔ \(dx = 5(x+3)^{\frac{4}{5}}du\) ↔ 5u^4du=dx

\(\int \frac{x}{x+3}^{\frac{1}{5}}dx = \int \frac{(u^5-3)}{u} 5u^4 du\)

=\(5 \int (u^8 - 3u^4) du\)

=\(5 \cdot \frac{1}{9} u^9 - 5 \cdot \frac{3}{5} u^5 + C\)

=\(5 \cdot (\frac{1}{9} u^9 - \frac{3}{4} u^4)+ C\)

=\(\frac{5}{9} (x+3)^{\frac{9}{5}} - \frac{15}{4} (x+3)^{\frac{4}{5}} + C\)

4. Selesaikan

\(\int \frac{1}{\sqrt{5+4x-x^2}}dx\)

Pembahasan: Perhatikan bahwa \(5+4x-x^2\) dapat ditulis sebagai \(-1(x^2-4x-5)\)
Sehingga, \(-1(x^2-4x-5) = -1((x-2)^2-9) = -1((x-2)^2-3^2)\) Maka, kita dapat menggunakan substitusi trigonometri: Misalkan \(x-2 = 3\sin\theta\), sehingga \(dx = 3\cos\theta d\theta\)

\(\int \frac{1}{\sqrt{5+4x-x^2}}dx = \int \frac{1}{\sqrt{-1((x-2)^2-3^2)}}dx\)

=\(\int \frac{1}{\sqrt{-1(9\sin^2\theta-9)}}(3\cos\theta d\theta)\)

=\(\int \frac{3\cos\theta}{\sqrt{-9(\sin^2\theta-1)}}d\theta\)

=\(\int \frac{3\cos\theta}{\sqrt{-9(-\cos^2\theta)}}d\theta\)

=\(\int \frac{3\cos\theta}{\sqrt{9\cos^2\theta}}d\theta\)

=\(\int \frac{3\cos\theta}{3|\cos\theta|}d\theta\)

=\(\int d\theta\)

=\(\theta + C\)

=\(\arcsin\left(\frac{x-2}{3}\right) + C\)

C) Latihan Soal

1. Soal ETS 2024
Hitung integral

\(\int \frac{x^2}{\sqrt{4-x^2}} dx\)

Hint 1: Substitusi \(x = 2\sin\theta\)
Hint 2: \(sin^2 x = \frac{1}{2}(1-cos2x)\)

Pembahasan

\(\int \frac{x^2}{\sqrt{4-x^2}} dx = \int \frac{(2\sin\theta)^2}{\sqrt{4-(2\sin\theta)^2}} (2\cos\theta d\theta)\)

=\(\int \frac{4\sin^2\theta}{\sqrt{4-4\sin^2\theta}} (2\cos\theta d\theta)\)

=\(\int \frac{4\sin^2\theta}{\sqrt{4cos^2\theta}} (2\cos\theta d\theta)\)

=\(\int \frac{4\sin^2\theta}{2|\cos\theta|} (2\cos\theta d\theta)\)

=\(\int 4\sin^2\theta d\theta\)

=\(\int 4\cdot \frac{1}{2}(1-\cos(2\theta)) d\theta\)

=\(2\int (1-\cos(2\theta)) d\theta\)

=\(2\theta - \sin(2\theta) + C\)

2. Soal Kuis
Selesaikan

\(\int \frac{x}{x^2-4x+9}dx\)

Hint 1: \(x^2 -4x+9=(x-2)^2+5\)
Hint 2: Substitusi \(u=x-2\)

Pembahasan

\(\int \frac{x}{x^2-4x+9}dx = \int \frac{x}{(x-2)^2+5}dx\)

Misalkan \(u=x-2\), sehingga \(x=u+2\) dan \(du=dx\)

=\(\int \frac{x}{(x-2)^2+5}dx= \int \frac{x}{(x-2)^2+5}dx\)

=\(\int \frac{u+2}{u^2+5}du\)

Misalkan \(u=\sqrt{5}tan\theta\), sehingga \(du=\sqrt{5}sec^2\theta d\theta\)

=\(\int \frac{u+2}{u^2+5}du\)

=\(\int \frac{\sqrt{5}tan\theta+2}{5sec^2\theta} \cdot \sqrt{5}sec^2\theta d\theta\)

=\(\int \frac{\sqrt{5}tan\theta+2}{5\mathrm{sec}^2\theta} \cdot d\theta\)

=\(\int (\frac{\sqrt{5}}{5}(\sqrt{5}tan\theta+2)) d\theta\)

=\(\int tan \theta d\theta + \frac{2\sqrt{5}{5} d\theta}\)

=\(\frac{2\sqrt{5}}{5}\theta + \int tan \theta d\theta\)

=\(\frac{2\sqrt{5}}{5}\theta + \int \frac{sin\theta}{cos\theta} d\theta\) Misalkan a = cos θ, sehingga \(da=-sin\theta d\theta\)

=\(\frac{2\sqrt{5}}{5}\theta - \int \frac{1}{a} da\)

=\(\frac{2\sqrt{5}}{5}\theta - ln|a| + C\)

=\(\frac{2\sqrt{5}}{5}\theta - ln|cos\theta| + C\)

=\(\frac{2\sqrt{5}}{5}tan^{-1}(\frac{x-2}{\sqrt{5}}) - ln|cos(tan^{-1}(\frac{x-2}{\sqrt{5}}))| + C\)

3. Soal ETS 2024 Hitung integral

\(\int \frac{1}{t^{\frac{1}{2}} - t^\frac{1}{3}} dt\)

Hint 1: KPK (2,3)=6
Hint 2: Substitusi t = u^6

Pembahasan:

Misalkan \(t=u^6,\) sehingga \(dt = 6u^5 du\)

\(\int \frac{1}{t^{\frac{1}{2}} - t^\frac{1}{3}} dt = \int \frac{1}{u^\frac{6}{2}-u^{\frac{6}{3}}6u^5du}\)

\(\int \frac{1}{u^3-u^2}6u^5 du\)

\(6\int \frac{u^3}{u-1}du \)

Perhatikan bahwa \(u^3 = (u^2+u+1)(u-1)+1,\) sehingga

\(\int \frac{1}{t^{\frac{1}{2}} - t^\frac{1}{3}} dt\) = 6 \int \frac{u^3}{u-1}du

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

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

\(6(\frac{1}{3}u^3 + \frac{1}{2}u^2 +u + \ln (u-1)) + C\)

\(2t^{\frac{1}{2}}+3t^{\frac{1}{3}}+6t^{\frac{1}{6}}+6\ln(t^{\frac{1}{6}-1})+ C\)

4. Soal Kuis Selesaikan

\(\int \frac{dx}{4sinx + 3 cos x }\)

Hint 1: Substitusi \(sin x = \frac{2u}{1+u^2}\) dan \(cosx = \frac{1-u^2}{1+u^2}\)

Pembahasan:

\(\int \frac{dx}{4\sin x + 3\cos x} = \int \frac{1}{\frac{4 \cdot 2u}{1+u^2} + 3 \cdot \frac{1-u^2}{1+u^2}} \cdot \frac{2}{(1+u^2)^2} du\)

\(= \int \frac{2}{8u + 3(1-u^2) + 3(1-u^2)} \cdot \frac{1}{(1+u^2)^2} du\)

\(= \int \frac{2}{8u + 3 - 3u^2} \cdot \frac{1}{(1+u^2)} du\)

\(= \int \frac{2}{-3u^2 + 8u + 3} du\)

Pecah menjadi pecahan parsial, lalu integrasikan sesuai bentuknya.

\(= \int \frac{2}{-3(u-1)(u+1)} du\)

\(= \frac{2}{-3} \int \left(\frac{1}{u-1} - \frac{1}{u+1}\right) du\)

\(= \frac{2}{-3} \left(\ln|u-1| - \ln|u+1|\right) + C\)

\(= \frac{2}{-3} \ln\left|\frac{u-1}{u+1}\right| + C\)

\(= \frac{2}{-3} \ln\left|\frac{\frac{2\sin x}{1+\tan^2 x}-1}{\frac{1-\tan^2 x}{1+\tan^2 x}+1}\right| + C\)

\(= \frac{2}{-3} \ln\left|\frac{2\sin x - (1+\tan^2 x)}{(1-\tan^2 x) + (1+\tan^2 x)}\right| + C\)

\(= \frac{2}{-3} \ln\left|\frac{2\sin x - 1 - \tan^2 x}{2}\right| + C\)

\(= \frac{2}{-3} \ln\left|2\sin x - 1 - \tan^2 x\right| + C\)