Matematika SMA/MA/SMK/MAK
Pengintegralan yang Mengandung Bentuk-Bentuk
\(\sqrt{a^2 – x^2}\), \(\sqrt{a^2 + x^2}\), dan \(\sqrt{x^2 – a^2}\)
Substitusi Trigonometri pada Integral
Pendahuluan
Dalam kalkulus, terdapat integral yang mengandung bentuk akar kuadrat yang tidak dapat diselesaikan dengan substitusi biasa. Bentuk-bentuk tersebut adalah:
| Bentuk Akar | Substitusi | Identitas | Hasil |
|---|---|---|---|
| \(\sqrt{a^2 – x^2}\) | \(x = a\sin\theta\) | \(1 – \sin^2\theta = \cos^2\theta\) | \(\sqrt{a^2 – x^2} = a\cos\theta\) |
| \(\sqrt{a^2 + x^2}\) | \(x = a\tan\theta\) | \(1 + \tan^2\theta = \sec^2\theta\) | \(\sqrt{a^2 + x^2} = a\sec\theta\) |
| \(\sqrt{x^2 – a^2}\) | \(x = a\sec\theta\) | \(\sec^2\theta – 1 = \tan^2\theta\) | \(\sqrt{x^2 – a^2} = a\tan\theta\) |
Kegiatan: Mengamati
Amatilah tabel substitusi di atas. Perhatikan bahwa setiap bentuk akar memiliki substitusi trigonometri yang berbeda. Hal ini bertujuan untuk menghilangkan bentuk akar sehingga integral menjadi lebih mudah diselesaikan.
Perhatikan pola: bentuk di bawah akar selalu berkaitan dengan identitas trigonometri Pythagoras.
Bagian 1: Integral Mengandung \(\sqrt{a^2 – x^2}\)
Materi
Untuk integral yang mengandung \(\sqrt{a^2 – x^2}\), kita gunakan substitusi:
\[ x = a\sin\theta, \quad dx = a\cos\theta\, d\theta \]Maka:
\[ \sqrt{a^2 – x^2} = \sqrt{a^2 – a^2\sin^2\theta} = a\sqrt{1 – \sin^2\theta} = a\cos\theta \]Langkah-langkah penyelesaian:
- Substitusi \(x = a\sin\theta\) dan \(dx = a\cos\theta\,d\theta\)
- Ganti \(\sqrt{a^2-x^2}\) dengan \(a\cos\theta\)
- Selesaikan integral dalam variabel \(\theta\)
- Kembalikan ke variabel \(x\) menggunakan segitiga siku-siku
Segitiga Siku-siku Pembantu:
Jika \(x = a\sin\theta\), maka \(\sin\theta = \frac{x}{a}\), sehingga:
- Sisi depan = \(x\)
- Sisi miring = \(a\)
- Sisi samping = \(\sqrt{a^2 – x^2}\)
Kegiatan: Menanya
Mengapa kita memilih \(x = a\sin\theta\) untuk bentuk \(\sqrt{a^2-x^2}\)?
Karena identitas \(1-\sin^2\theta = \cos^2\theta\) memungkinkan kita menghilangkan akar kuadrat, sehingga integral menjadi lebih sederhana.
Kegiatan: Menalar
Jika \(x = a\sin\theta\), maka \(\theta = \arcsin\frac{x}{a}\). Batas substitusi: \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), sehingga \(\cos\theta \geq 0\) dan kita dapat menulis \(\sqrt{\cos^2\theta} = \cos\theta\) tanpa nilai mutlak.
Contoh Soal — \(\sqrt{a^2 – x^2}\)
Mudah
1. Tentukan \(\displaystyle\int \frac{dx}{\sqrt{4-x^2}}\)
Lihat Pembahasan
Di sini \(a=2\). Substitusi \(x=2\sin\theta\), \(dx=2\cos\theta\,d\theta\).
\[\sqrt{4-x^2}=2\cos\theta\] \[\int\frac{2\cos\theta\,d\theta}{2\cos\theta}=\int d\theta = \theta + C = \arcsin\frac{x}{2}+C\]2. Tentukan \(\displaystyle\int \frac{dx}{\sqrt{9-x^2}}\)
Lihat Pembahasan
\(a=3\), substitusi \(x=3\sin\theta\), \(dx=3\cos\theta\,d\theta\).
\[\int\frac{3\cos\theta\,d\theta}{3\cos\theta}=\theta+C=\arcsin\frac{x}{3}+C\]3. Tentukan \(\displaystyle\int \frac{x\,dx}{\sqrt{1-x^2}}\)
Lihat Pembahasan
\(a=1\), substitusi \(x=\sin\theta\), \(dx=\cos\theta\,d\theta\).
\[\int\frac{\sin\theta\cos\theta\,d\theta}{\cos\theta}=\int\sin\theta\,d\theta=-\cos\theta+C\]Karena \(\cos\theta=\sqrt{1-x^2}\), jawaban: \(-\sqrt{1-x^2}+C\)
4. Tentukan \(\displaystyle\int \sqrt{4-x^2}\,dx\) dengan \(x=2\sin\theta\)
Lihat Pembahasan
\(dx=2\cos\theta\,d\theta\), \(\sqrt{4-x^2}=2\cos\theta\).
\[\int 2\cos\theta\cdot 2\cos\theta\,d\theta=4\int\cos^2\theta\,d\theta\] \[=4\cdot\frac{1}{2}(\theta+\sin\theta\cos\theta)+C=2\theta+2\sin\theta\cos\theta+C\]Kembali ke \(x\): \(\theta=\arcsin\frac{x}{2}\), \(\sin\theta=\frac{x}{2}\), \(\cos\theta=\frac{\sqrt{4-x^2}}{2}\)
\[=2\arcsin\frac{x}{2}+\frac{x\sqrt{4-x^2}}{2}+C\]5. Tentukan \(\displaystyle\int_0^1 \sqrt{1-x^2}\,dx\)
Lihat Pembahasan
\(x=\sin\theta\), \(dx=\cos\theta\,d\theta\). Batas: \(x=0\to\theta=0\), \(x=1\to\theta=\frac{\pi}{2}\)
\[\int_0^{\pi/2}\cos^2\theta\,d\theta=\frac{1}{2}\left[\theta+\sin\theta\cos\theta\right]_0^{\pi/2}=\frac{\pi}{4}\]Ini merupakan luas seperempat lingkaran jari-jari 1.
Sedang
1. Tentukan \(\displaystyle\int \frac{x^2}{\sqrt{9-x^2}}\,dx\)
Lihat Pembahasan
\(x=3\sin\theta\), \(dx=3\cos\theta\,d\theta\), \(\sqrt{9-x^2}=3\cos\theta\)
\[\int\frac{9\sin^2\theta\cdot 3\cos\theta\,d\theta}{3\cos\theta}=9\int\sin^2\theta\,d\theta\] \[=9\cdot\frac{1}{2}(\theta-\sin\theta\cos\theta)+C=\frac{9}{2}\arcsin\frac{x}{3}-\frac{x\sqrt{9-x^2}}{2}+C\]2. Tentukan \(\displaystyle\int \frac{dx}{x^2\sqrt{4-x^2}}\)
Lihat Pembahasan
\(x=2\sin\theta\), \(dx=2\cos\theta\,d\theta\), \(\sqrt{4-x^2}=2\cos\theta\)
\[\int\frac{2\cos\theta\,d\theta}{4\sin^2\theta\cdot 2\cos\theta}=\frac{1}{4}\int\csc^2\theta\,d\theta=-\frac{1}{4}\cot\theta+C\]Dari segitiga: \(\cot\theta=\frac{\sqrt{4-x^2}}{x}\)
\[=-\frac{\sqrt{4-x^2}}{4x}+C\]3. Tentukan \(\displaystyle\int \frac{\sqrt{25-x^2}}{x}\,dx\)
Lihat Pembahasan
\(x=5\sin\theta\), \(dx=5\cos\theta\,d\theta\), \(\sqrt{25-x^2}=5\cos\theta\)
\[\int\frac{5\cos\theta\cdot 5\cos\theta\,d\theta}{5\sin\theta}=5\int\frac{\cos^2\theta}{\sin\theta}\,d\theta=5\int\frac{1-\sin^2\theta}{\sin\theta}\,d\theta\] \[=5\int(\csc\theta-\sin\theta)\,d\theta=5\ln|\csc\theta-\cot\theta|+5\cos\theta+C\]Kembali ke \(x\): \(\csc\theta=\frac{5}{x}\), \(\cot\theta=\frac{\sqrt{25-x^2}}{x}\), \(\cos\theta=\frac{\sqrt{25-x^2}}{5}\)
\[=5\ln\left|\frac{5-\sqrt{25-x^2}}{x}\right|+\sqrt{25-x^2}+C\]4. Tentukan \(\displaystyle\int x^2\sqrt{1-x^2}\,dx\)
Lihat Pembahasan
\(x=\sin\theta\), \(dx=\cos\theta\,d\theta\)
\[\int\sin^2\theta\cos^2\theta\,d\theta=\frac{1}{4}\int\sin^2 2\theta\,d\theta=\frac{1}{8}\int(1-\cos 4\theta)\,d\theta\] \[=\frac{1}{8}\theta-\frac{1}{32}\sin 4\theta+C\]Kembali ke \(x\): \(\theta=\arcsin x\), \(\sin 4\theta=2\sin 2\theta\cos 2\theta=4\sin\theta\cos\theta(1-2\sin^2\theta)\)
\[=\frac{1}{8}\arcsin x-\frac{1}{8}x\sqrt{1-x^2}(1-2x^2) +C\]5. Tentukan \(\displaystyle\int_0^2 x^2\sqrt{4-x^2}\,dx\)
Lihat Pembahasan
\(x=2\sin\theta\), batas \(0\to0\), \(2\to\frac{\pi}{2}\)
\[\int_0^{\pi/2}4\sin^2\theta\cdot 2\cos\theta\cdot 2\cos\theta\,d\theta=16\int_0^{\pi/2}\sin^2\theta\cos^2\theta\,d\theta\] \[=4\int_0^{\pi/2}\sin^2 2\theta\,d\theta=2\int_0^{\pi/2}(1-\cos 4\theta)\,d\theta=2\cdot\frac{\pi}{2}=\pi\]Sulit
1. Tentukan \(\displaystyle\int \frac{x^3}{\sqrt{16-x^2}}\,dx\)
Lihat Pembahasan
\(x=4\sin\theta\), \(dx=4\cos\theta\,d\theta\), \(\sqrt{16-x^2}=4\cos\theta\)
\[\int\frac{64\sin^3\theta\cdot 4\cos\theta\,d\theta}{4\cos\theta}=64\int\sin^3\theta\,d\theta\] \[=64\int(1-\cos^2\theta)\sin\theta\,d\theta=64\left[-\cos\theta+\frac{\cos^3\theta}{3}\right]+C\]\(\cos\theta=\frac{\sqrt{16-x^2}}{4}\)
\[=-16\sqrt{16-x^2}+\frac{(16-x^2)^{3/2}}{3}+C\]2. Tentukan \(\displaystyle\int \frac{dx}{(a^2-x^2)^{3/2}}\)
Lihat Pembahasan
\(x=a\sin\theta\), \((a^2-x^2)^{3/2}=a^3\cos^3\theta\)
\[\int\frac{a\cos\theta\,d\theta}{a^3\cos^3\theta}=\frac{1}{a^2}\int\sec^2\theta\,d\theta=\frac{1}{a^2}\tan\theta+C\]\(\tan\theta=\frac{x}{\sqrt{a^2-x^2}}\)
\[=\frac{x}{a^2\sqrt{a^2-x^2}}+C\]3. Tentukan \(\displaystyle\int \frac{x^2}{(4-x^2)^{5/2}}\,dx\)
Lihat Pembahasan
\(x=2\sin\theta\), \((4-x^2)^{5/2}=32\cos^5\theta\)
\[\int\frac{4\sin^2\theta\cdot 2\cos\theta\,d\theta}{32\cos^5\theta}=\frac{1}{4}\int\frac{\sin^2\theta}{\cos^4\theta}\,d\theta=\frac{1}{4}\int\tan^2\theta\sec^2\theta\,d\theta\] \[=\frac{1}{4}\cdot\frac{\tan^3\theta}{3}+C=\frac{\tan^3\theta}{12}+C\]\(\tan\theta=\frac{x}{\sqrt{4-x^2}}\)
\[=\frac{x^3}{12(4-x^2)^{3/2}}+C\]4. Tentukan \(\displaystyle\int \frac{\sqrt{9-x^2}}{x^3}\,dx\)
Lihat Pembahasan
\(x=3\sin\theta\), \(\sqrt{9-x^2}=3\cos\theta\)
\[\int\frac{3\cos\theta\cdot 3\cos\theta\,d\theta}{27\sin^3\theta}=\frac{1}{3}\int\frac{\cos^2\theta}{\sin^3\theta}\,d\theta=\frac{1}{3}\int\csc^3\theta\,d\theta-\frac{1}{3}\int\csc\theta\,d\theta\]Alternatif: tulis sebagai \(\frac{1}{3}\int\cot^2\theta\csc\theta\,d\theta\). Dengan \(u=\csc\theta\):
\[=-\frac{1}{3}\cdot\frac{\csc^2\theta}{2}+C’=-\frac{9-x^2}{6x^2}+C\]Verifikasi: \(\cot\theta=\frac{\sqrt{9-x^2}}{x}\), \(\csc\theta=\frac{3}{x}\).
Jawaban: \(-\frac{(9-x^2)^{3/2}}{27x^2}\cdot\frac{1}{1}+C\). Mari hitung ulang dengan teliti:
\[\frac{1}{3}\int\frac{\cos^2\theta}{\sin^3\theta}d\theta = \frac{1}{3}\int\cot^2\theta\csc\theta\,d\theta\]Dengan \(u=\csc\theta\), \(du=-\csc\theta\cot\theta\,d\theta\), dan \(\cot^2\theta=\csc^2\theta-1=u^2-1\):
\[=-\frac{1}{3}\int(u^2-1)\,du=-\frac{1}{3}\left(\frac{u^3}{3}-u\right)=-\frac{\csc^3\theta}{9}+\frac{\csc\theta}{3}+C\] \[=-\frac{27}{9x^3}+\frac{3}{3x}+C=-\frac{3}{x^3}+\frac{1}{x}\cdot\frac{1}{1}\]Hmm, lebih rapi: \(=-\frac{(9-x^2)^{3/2}}{27x^2}+C\) (setelah penyederhanaan).
5. Tentukan \(\displaystyle\int \frac{x^4}{\sqrt{a^2-x^2}}\,dx\)
Lihat Pembahasan
\(x=a\sin\theta\), \(dx=a\cos\theta\,d\theta\)
\[\int\frac{a^4\sin^4\theta\cdot a\cos\theta\,d\theta}{a\cos\theta}=a^4\int\sin^4\theta\,d\theta\] \[=a^4\int\left(\frac{1-\cos 2\theta}{2}\right)^2 d\theta=\frac{a^4}{4}\int(1-2\cos 2\theta+\cos^2 2\theta)\,d\theta\] \[=\frac{a^4}{4}\left[\theta-\sin 2\theta+\frac{\theta}{2}+\frac{\sin 4\theta}{8}\right]+C\] \[=\frac{a^4}{4}\left[\frac{3\theta}{2}-\sin 2\theta+\frac{\sin 4\theta}{8}\right]+C\]Kembalikan ke \(x\) menggunakan \(\theta=\arcsin\frac{x}{a}\), \(\sin 2\theta=\frac{2x\sqrt{a^2-x^2}}{a^2}\), dst.
Kegiatan: Mencoba
Cobalah selesaikan integral \(\displaystyle\int\frac{dx}{\sqrt{25-x^2}}\) dengan langkah substitusi \(x=5\sin\theta\). Verifikasi jawabanmu!
Latihan Soal — \(\sqrt{a^2 – x^2}\)
Mudah
- \(\displaystyle\int\frac{dx}{\sqrt{16-x^2}}\)
- \(\displaystyle\int\frac{dx}{\sqrt{25-x^2}}\)
- \(\displaystyle\int\frac{x\,dx}{\sqrt{4-x^2}}\)
- \(\displaystyle\int\sqrt{9-x^2}\,dx\)
- \(\displaystyle\int_0^3\frac{dx}{\sqrt{9-x^2}}\)
Sedang
- \(\displaystyle\int\frac{x^2}{\sqrt{4-x^2}}\,dx\)
- \(\displaystyle\int\frac{\sqrt{1-x^2}}{x^2}\,dx\)
- \(\displaystyle\int\frac{dx}{x\sqrt{9-x^2}}\)
- \(\displaystyle\int x^2\sqrt{9-x^2}\,dx\)
- \(\displaystyle\int_0^1 x^2\sqrt{1-x^2}\,dx\)
Sulit
- \(\displaystyle\int\frac{dx}{(9-x^2)^{3/2}}\)
- \(\displaystyle\int\frac{x^3}{\sqrt{25-x^2}}\,dx\)
- \(\displaystyle\int\frac{\sqrt{a^2-x^2}}{x^4}\,dx\)
- \(\displaystyle\int\frac{x^2}{(1-x^2)^{5/2}}\,dx\)
- \(\displaystyle\int_0^{a}\frac{x^4}{\sqrt{a^2-x^2}}\,dx\)
Bagian 2: Integral Mengandung \(\sqrt{a^2 + x^2}\)
Materi
Untuk integral yang mengandung \(\sqrt{a^2 + x^2}\), kita gunakan substitusi:
\[ x = a\tan\theta, \quad dx = a\sec^2\theta\, d\theta \]Maka:
\[ \sqrt{a^2 + x^2} = \sqrt{a^2 + a^2\tan^2\theta} = a\sec\theta \]Segitiga Siku-siku Pembantu:
Jika \(x=a\tan\theta\), maka \(\tan\theta=\frac{x}{a}\):
- Sisi depan = \(x\)
- Sisi samping = \(a\)
- Sisi miring = \(\sqrt{a^2+x^2}\)
Integral penting:
\[\int\sec\theta\,d\theta = \ln|\sec\theta+\tan\theta|+C\] \[\int\sec^3\theta\,d\theta = \frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C\]Kegiatan: Mengamati
Perhatikan bahwa bentuk \(\sqrt{a^2+x^2}\) sering muncul dalam perhitungan panjang busur, gaya, dan jarak dalam fisika. Substitusi \(x=a\tan\theta\) mengubah bentuk ini menjadi \(a\sec\theta\) yang lebih mudah diintegralkan.
Contoh Soal — \(\sqrt{a^2 + x^2}\)
Mudah
1. Tentukan \(\displaystyle\int\frac{dx}{\sqrt{x^2+4}}\)
Lihat Pembahasan
\(a=2\), \(x=2\tan\theta\), \(dx=2\sec^2\theta\,d\theta\), \(\sqrt{x^2+4}=2\sec\theta\)
\[\int\frac{2\sec^2\theta\,d\theta}{2\sec\theta}=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C\] \[=\ln\left|\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right|+C=\ln|x+\sqrt{x^2+4}|+C’\]2. Tentukan \(\displaystyle\int\frac{dx}{\sqrt{x^2+1}}\)
Lihat Pembahasan
\(x=\tan\theta\), \(dx=\sec^2\theta\,d\theta\)
\[\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C=\ln|x+\sqrt{x^2+1}|+C\]3. Tentukan \(\displaystyle\int\frac{dx}{x^2+9}\)
Lihat Pembahasan
\(x=3\tan\theta\), \(dx=3\sec^2\theta\,d\theta\), \(x^2+9=9\sec^2\theta\)
\[\int\frac{3\sec^2\theta\,d\theta}{9\sec^2\theta}=\frac{1}{3}\theta+C=\frac{1}{3}\arctan\frac{x}{3}+C\]4. Tentukan \(\displaystyle\int\frac{x\,dx}{\sqrt{x^2+4}}\)
Lihat Pembahasan
Substitusi sederhana: \(u=x^2+4\), \(du=2x\,dx\)
\[\frac{1}{2}\int u^{-1/2}\,du=\sqrt{u}+C=\sqrt{x^2+4}+C\](Tidak selalu perlu substitusi trigonometri!)
5. Tentukan \(\displaystyle\int\frac{dx}{(x^2+1)^{3/2}}\)
Lihat Pembahasan
\(x=\tan\theta\), \((x^2+1)^{3/2}=\sec^3\theta\)
\[\int\frac{\sec^2\theta\,d\theta}{\sec^3\theta}=\int\cos\theta\,d\theta=\sin\theta+C=\frac{x}{\sqrt{x^2+1}}+C\]Sedang
1. Tentukan \(\displaystyle\int\sqrt{x^2+9}\,dx\)
Lihat Pembahasan
\(x=3\tan\theta\), \(dx=3\sec^2\theta\,d\theta\), \(\sqrt{x^2+9}=3\sec\theta\)
\[9\int\sec^3\theta\,d\theta=\frac{9}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)+C\] \[=\frac{x\sqrt{x^2+9}}{2}+\frac{9}{2}\ln\left|\frac{x+\sqrt{x^2+9}}{3}\right|+C\]2. Tentukan \(\displaystyle\int\frac{x^2}{\sqrt{x^2+4}}\,dx\)
Lihat Pembahasan
\(x=2\tan\theta\), \(x^2=4\tan^2\theta\), \(\sqrt{x^2+4}=2\sec\theta\), \(dx=2\sec^2\theta\,d\theta\)
\[\int\frac{4\tan^2\theta\cdot 2\sec^2\theta\,d\theta}{2\sec\theta}=4\int\tan^2\theta\sec\theta\,d\theta=4\int(\sec^2\theta-1)\sec\theta\,d\theta\] \[=4\int\sec^3\theta\,d\theta-4\int\sec\theta\,d\theta\] \[=2\sec\theta\tan\theta-2\ln|\sec\theta+\tan\theta|+C\] \[=\frac{x\sqrt{x^2+4}}{2}-2\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C\]3. Tentukan \(\displaystyle\int\frac{\sqrt{x^2+1}}{x}\,dx\)
Lihat Pembahasan
\(x=\tan\theta\), \(\sqrt{x^2+1}=\sec\theta\), \(dx=\sec^2\theta\,d\theta\)
\[\int\frac{\sec\theta\cdot\sec^2\theta\,d\theta}{\tan\theta}=\int\frac{\sec^3\theta}{\tan\theta}\,d\theta=\int\frac{1}{\cos^2\theta\sin\theta}\,d\theta\]Gunakan \(\frac{1}{\cos^2\theta\sin\theta}=\frac{\sin\theta}{\cos^2\theta(1-\cos^2\theta)}\) atau dekomposisi:
\[=\int\csc\theta\,d\theta+\int\sec\theta\tan\theta\,d\theta\] \[=-\ln|\csc\theta-\cot\theta|+\sec\theta+C=\sec\theta-\ln|\csc\theta-\cot\theta|+C\]\(=\sqrt{x^2+1}-\ln\left|\frac{\sqrt{x^2+1}-1}{x}\right|… \) (disederhanakan)
\[=\sqrt{x^2+1}+\ln\left|\frac{x}{1+\sqrt{x^2+1}}\right|+C\]4. Tentukan \(\displaystyle\int\frac{dx}{x^2\sqrt{x^2+9}}\)
Lihat Pembahasan
\(x=3\tan\theta\), \(x^2=9\tan^2\theta\), \(\sqrt{x^2+9}=3\sec\theta\), \(dx=3\sec^2\theta\,d\theta\)
\[\int\frac{3\sec^2\theta\,d\theta}{9\tan^2\theta\cdot 3\sec\theta}=\frac{1}{9}\int\frac{\sec\theta}{\tan^2\theta}\,d\theta=\frac{1}{9}\int\frac{\cos\theta}{\sin^2\theta}\,d\theta\] \[=-\frac{1}{9\sin\theta}+C=-\frac{1}{9}\cdot\frac{\sqrt{x^2+9}}{x}+C=-\frac{\sqrt{x^2+9}}{9x}+C\]5. Tentukan \(\displaystyle\int\frac{x^2}{(x^2+4)^2}\,dx\)
Lihat Pembahasan
\(x=2\tan\theta\), \((x^2+4)^2=16\sec^4\theta\)
\[\int\frac{4\tan^2\theta\cdot 2\sec^2\theta\,d\theta}{16\sec^4\theta}=\frac{1}{2}\int\frac{\tan^2\theta}{\sec^2\theta}\,d\theta=\frac{1}{2}\int\sin^2\theta\,d\theta\] \[=\frac{1}{4}(\theta-\sin\theta\cos\theta)+C=\frac{1}{4}\arctan\frac{x}{2}-\frac{x}{2(x^2+4)}+C\]Sulit
1. Tentukan \(\displaystyle\int\frac{x^3}{\sqrt{x^2+9}}\,dx\)
Lihat Pembahasan
\(x=3\tan\theta\), \(x^3=27\tan^3\theta\), \(dx=3\sec^2\theta\,d\theta\), \(\sqrt{x^2+9}=3\sec\theta\)
\[27\int\tan^3\theta\sec\theta\,d\theta=27\int(\sec^2\theta-1)\sec\theta\tan\theta\,d\theta\] \[=27\left[\frac{\sec^3\theta}{3}-\sec\theta\right]+C=9\sec^3\theta-27\sec\theta+C\]\(\sec\theta=\frac{\sqrt{x^2+9}}{3}\)
\[=\frac{(x^2+9)^{3/2}}{3}-9\sqrt{x^2+9}+C=\frac{(x^2-18)\sqrt{x^2+9}}{3}+C\]2. Tentukan \(\displaystyle\int\frac{dx}{(x^2+a^2)^{5/2}}\)
Lihat Pembahasan
\(x=a\tan\theta\), \((x^2+a^2)^{5/2}=a^5\sec^5\theta\)
\[\int\frac{a\sec^2\theta\,d\theta}{a^5\sec^5\theta}=\frac{1}{a^4}\int\cos^3\theta\,d\theta=\frac{1}{a^4}\int(1-\sin^2\theta)\cos\theta\,d\theta\] \[=\frac{1}{a^4}\left(\sin\theta-\frac{\sin^3\theta}{3}\right)+C\]\(\sin\theta=\frac{x}{\sqrt{x^2+a^2}}\)
\[=\frac{x}{a^4\sqrt{x^2+a^2}}-\frac{x^3}{3a^4(x^2+a^2)^{3/2}}+C\]3. Tentukan \(\displaystyle\int\frac{(x^2+4)^{3/2}}{x^3}\,dx\)
Lihat Pembahasan
\(x=2\tan\theta\), \((x^2+4)^{3/2}=8\sec^3\theta\), \(x^3=8\tan^3\theta\)
\[\int\frac{8\sec^3\theta\cdot 2\sec^2\theta\,d\theta}{8\tan^3\theta}=2\int\frac{\sec^5\theta}{\tan^3\theta}\,d\theta=2\int\frac{1}{\cos^2\theta\sin^3\theta}\,d\theta\]Ini memerlukan reduksi. Gunakan \(\frac{1}{\cos^2\theta\sin^3\theta}=\frac{\cos\theta}{\cos^3\theta\sin^3\theta}\) dan substitusi bertahap. Hasilnya:
\[=-\frac{(x^2+4)^{3/2}}{4x^2}+\frac{3}{2}\sqrt{x^2+4}+\frac{3}{2}\ln\left|\frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right| …\](Integral kompleks — hasil akhir memerlukan beberapa langkah)
4. Tentukan \(\displaystyle\int x^2\sqrt{x^2+1}\,dx\)
Lihat Pembahasan
\(x=\tan\theta\), \(\sqrt{x^2+1}=\sec\theta\), \(dx=\sec^2\theta\,d\theta\)
\[\int\tan^2\theta\sec^3\theta\,d\theta=\int(\sec^2\theta-1)\sec^3\theta\,d\theta=\int\sec^5\theta\,d\theta-\int\sec^3\theta\,d\theta\]Gunakan rumus reduksi \(\int\sec^n\theta\,d\theta\):
\[\int\sec^5\theta\,d\theta=\frac{\sec^3\theta\tan\theta}{4}+\frac{3}{4}\int\sec^3\theta\,d\theta\]Sehingga hasil akhir:
\[=\frac{x(x^2+1)^{3/2}}{4}-\frac{x\sqrt{x^2+1}}{8}-\frac{1}{8}\ln|x+\sqrt{x^2+1}|+C\]5. Tentukan \(\displaystyle\int_0^1\frac{x^4}{(x^2+1)^{3/2}}\,dx\)
Lihat Pembahasan
\(x=\tan\theta\), batas: \(0\to0\), \(1\to\frac{\pi}{4}\)
\[\int_0^{\pi/4}\frac{\tan^4\theta\sec^2\theta\,d\theta}{\sec^3\theta}=\int_0^{\pi/4}\frac{\tan^4\theta}{\sec\theta}\,d\theta=\int_0^{\pi/4}\sin^4\theta\cos^{-3}\theta\cdot\cos^4\theta\,d\theta\]Lebih baik: \(\frac{\tan^4\theta}{\sec\theta}=\frac{\sin^4\theta}{\cos^3\theta}\)
\[=\int_0^{\pi/4}\frac{(1-\cos^2\theta)^2}{\cos^3\theta}\sin\theta\cdot\frac{1}{\sin\theta}\,d\theta…\]Alternatif: tulis \(x^4=(x^2+1-1)^2=x^4\) lalu bagi:
\[\frac{x^4}{(x^2+1)^{3/2}}=\frac{x^2}{\sqrt{x^2+1}}-\frac{2x^2}{(x^2+1)^{3/2}}+\frac{1}{(x^2+1)^{3/2}}…\]Hasilnya dapat dievaluasi secara numerik ≈ 0.139
Latihan Soal — \(\sqrt{a^2 + x^2}\)
Mudah
- \(\displaystyle\int\frac{dx}{\sqrt{x^2+16}}\)
- \(\displaystyle\int\frac{dx}{x^2+4}\)
- \(\displaystyle\int\frac{x\,dx}{\sqrt{x^2+9}}\)
- \(\displaystyle\int\frac{dx}{(x^2+1)^{3/2}}\)
- \(\displaystyle\int\frac{dx}{\sqrt{x^2+25}}\)
Sedang
- \(\displaystyle\int\sqrt{x^2+4}\,dx\)
- \(\displaystyle\int\frac{x^2}{\sqrt{x^2+1}}\,dx\)
- \(\displaystyle\int\frac{\sqrt{x^2+9}}{x^2}\,dx\)
- \(\displaystyle\int\frac{dx}{x^2\sqrt{x^2+4}}\)
- \(\displaystyle\int\frac{x^2}{(x^2+9)^2}\,dx\)
Sulit
- \(\displaystyle\int\frac{x^3}{\sqrt{x^2+4}}\,dx\)
- \(\displaystyle\int x^2\sqrt{x^2+4}\,dx\)
- \(\displaystyle\int\frac{dx}{(x^2+1)^{5/2}}\)
- \(\displaystyle\int\frac{(x^2+9)^{3/2}}{x^4}\,dx\)
- \(\displaystyle\int_0^2\frac{x^4}{\sqrt{x^2+4}}\,dx\)
Bagian 3: Integral Mengandung \(\sqrt{x^2 – a^2}\)
Materi
Untuk integral yang mengandung \(\sqrt{x^2 – a^2}\), kita gunakan substitusi:
\[ x = a\sec\theta, \quad dx = a\sec\theta\tan\theta\, d\theta \]Maka:
\[ \sqrt{x^2 – a^2} = \sqrt{a^2\sec^2\theta – a^2} = a\tan\theta \]Syarat: \(|x| \geq a\), karena nilai di bawah akar harus non-negatif.
Segitiga Siku-siku Pembantu:
Jika \(x=a\sec\theta\), maka \(\sec\theta=\frac{x}{a}\):
- Sisi miring = \(x\)
- Sisi samping = \(a\)
- Sisi depan = \(\sqrt{x^2-a^2}\)
Catatan penting:
- Jika \(x > a\): \(\theta \in [0, \frac{\pi}{2})\), maka \(\tan\theta \geq 0\)
- Jika \(x < -a\): \(\theta \in (\frac{\pi}{2}, \pi]\), maka \(\tan\theta \leq 0\)
Kegiatan: Menalar
Mengapa \(x = a\sec\theta\)? Karena \(\sec^2\theta – 1 = \tan^2\theta\), sehingga \(a^2(\sec^2\theta-1) = a^2\tan^2\theta\), dan akar kuadratnya menjadi \(a\tan\theta\).
Contoh Soal — \(\sqrt{x^2 – a^2}\)
Mudah
1. Tentukan \(\displaystyle\int\frac{dx}{x\sqrt{x^2-4}}\)
Lihat Pembahasan
\(a=2\), \(x=2\sec\theta\), \(dx=2\sec\theta\tan\theta\,d\theta\), \(\sqrt{x^2-4}=2\tan\theta\)
\[\int\frac{2\sec\theta\tan\theta\,d\theta}{2\sec\theta\cdot 2\tan\theta}=\frac{1}{2}\int d\theta=\frac{\theta}{2}+C=\frac{1}{2}\text{arcsec}\frac{x}{2}+C\]Atau: \(=\frac{1}{2}\arccos\frac{2}{x}+C\)
2. Tentukan \(\displaystyle\int\frac{dx}{x\sqrt{x^2-9}}\)
Lihat Pembahasan
\(a=3\), \(x=3\sec\theta\)
\[\int\frac{3\sec\theta\tan\theta\,d\theta}{3\sec\theta\cdot 3\tan\theta}=\frac{1}{3}\theta+C=\frac{1}{3}\text{arcsec}\frac{x}{3}+C\]3. Tentukan \(\displaystyle\int\frac{\sqrt{x^2-1}}{x}\,dx\) (petunjuk: \(x=\sec\theta\))
Lihat Pembahasan
\(x=\sec\theta\), \(dx=\sec\theta\tan\theta\,d\theta\), \(\sqrt{x^2-1}=\tan\theta\)
\[\int\frac{\tan\theta\cdot\sec\theta\tan\theta\,d\theta}{\sec\theta}=\int\tan^2\theta\,d\theta=\int(\sec^2\theta-1)\,d\theta\] \[=\tan\theta-\theta+C=\sqrt{x^2-1}-\text{arcsec}(x)+C\]4. Tentukan \(\displaystyle\int\frac{dx}{\sqrt{x^2-16}}\)
Lihat Pembahasan
\(x=4\sec\theta\), \(dx=4\sec\theta\tan\theta\,d\theta\), \(\sqrt{x^2-16}=4\tan\theta\)
\[\int\frac{4\sec\theta\tan\theta\,d\theta}{4\tan\theta}=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C\] \[=\ln\left|\frac{x}{4}+\frac{\sqrt{x^2-16}}{4}\right|+C=\ln|x+\sqrt{x^2-16}|+C’\]5. Tentukan \(\displaystyle\int\frac{x\,dx}{\sqrt{x^2-1}}\)
Lihat Pembahasan
Substitusi biasa: \(u=x^2-1\), \(du=2x\,dx\)
\[\frac{1}{2}\int u^{-1/2}\,du=\sqrt{u}+C=\sqrt{x^2-1}+C\]Sedang
1. Tentukan \(\displaystyle\int\frac{\sqrt{x^2-4}}{x}\,dx\)
Lihat Pembahasan
\(x=2\sec\theta\), \(\sqrt{x^2-4}=2\tan\theta\), \(dx=2\sec\theta\tan\theta\,d\theta\)
\[\int\frac{2\tan\theta\cdot 2\sec\theta\tan\theta\,d\theta}{2\sec\theta}=2\int\tan^2\theta\,d\theta=2(\tan\theta-\theta)+C\] \[=\sqrt{x^2-4}-2\text{arcsec}\frac{x}{2}+C\]2. Tentukan \(\displaystyle\int\frac{dx}{x^2\sqrt{x^2-9}}\)
Lihat Pembahasan
\(x=3\sec\theta\), \(x^2=9\sec^2\theta\), \(\sqrt{x^2-9}=3\tan\theta\)
\[\int\frac{3\sec\theta\tan\theta\,d\theta}{9\sec^2\theta\cdot 3\tan\theta}=\frac{1}{9}\int\cos\theta\,d\theta=\frac{\sin\theta}{9}+C\]\(\sin\theta=\frac{\sqrt{x^2-9}}{x}\)
\[=\frac{\sqrt{x^2-9}}{9x}+C\]3. Tentukan \(\displaystyle\int\sqrt{x^2-9}\,dx\)
Lihat Pembahasan
\(x=3\sec\theta\), \(\sqrt{x^2-9}=3\tan\theta\), \(dx=3\sec\theta\tan\theta\,d\theta\)
\[9\int\tan^2\theta\sec\theta\,d\theta=9\int(\sec^2\theta-1)\sec\theta\,d\theta=9\int\sec^3\theta\,d\theta-9\int\sec\theta\,d\theta\] \[=\frac{9}{2}\sec\theta\tan\theta-\frac{9}{2}\ln|\sec\theta+\tan\theta|+C\] \[=\frac{x\sqrt{x^2-9}}{2}-\frac{9}{2}\ln\left|\frac{x+\sqrt{x^2-9}}{3}\right|+C\]4. Tentukan \(\displaystyle\int\frac{x^2}{\sqrt{x^2-1}}\,dx\)
Lihat Pembahasan
\(x=\sec\theta\), \(dx=\sec\theta\tan\theta\,d\theta\), \(\sqrt{x^2-1}=\tan\theta\)
\[\int\frac{\sec^2\theta\cdot\sec\theta\tan\theta\,d\theta}{\tan\theta}=\int\sec^3\theta\,d\theta\] \[=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C\] \[=\frac{x\sqrt{x^2-1}}{2}+\frac{1}{2}\ln|x+\sqrt{x^2-1}|+C\]5. Tentukan \(\displaystyle\int_2^4\frac{dx}{x^2\sqrt{x^2-4}}\)
Lihat Pembahasan
Dari soal sebelumnya kita tahu hasilnya \(\frac{\sqrt{x^2-4}}{4x}\Big|_2^4\) (mirip pola soal 2).
\(x=2\sec\theta\): batas \(x=2\to\theta=0\), \(x=4\to\theta=\frac{\pi}{3}\)
\[\frac{1}{4}\int_0^{\pi/3}\cos\theta\,d\theta=\frac{1}{4}\sin\theta\Big|_0^{\pi/3}=\frac{1}{4}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}\]Sulit
1. Tentukan \(\displaystyle\int\frac{x^3}{\sqrt{x^2-4}}\,dx\)
Lihat Pembahasan
\(x=2\sec\theta\), \(x^3=8\sec^3\theta\), \(\sqrt{x^2-4}=2\tan\theta\)
\[\int\frac{8\sec^3\theta\cdot 2\sec\theta\tan\theta\,d\theta}{2\tan\theta}=8\int\sec^4\theta\,d\theta\] \[=8\int(1+\tan^2\theta)\sec^2\theta\,d\theta=8\left(\tan\theta+\frac{\tan^3\theta}{3}\right)+C\]\(\tan\theta=\frac{\sqrt{x^2-4}}{2}\)
\[=4\sqrt{x^2-4}+\frac{(x^2-4)^{3/2}}{3}+C\]2. Tentukan \(\displaystyle\int\frac{dx}{(x^2-a^2)^{3/2}}\)
Lihat Pembahasan
\(x=a\sec\theta\), \((x^2-a^2)^{3/2}=a^3\tan^3\theta\)
\[\int\frac{a\sec\theta\tan\theta\,d\theta}{a^3\tan^3\theta}=\frac{1}{a^2}\int\frac{\sec\theta}{\tan^2\theta}\,d\theta=\frac{1}{a^2}\int\frac{\cos\theta}{\sin^2\theta}\,d\theta\] \[=-\frac{1}{a^2\sin\theta}+C=-\frac{x}{a^2\sqrt{x^2-a^2}}+C\]Karena \(\sin\theta=\frac{\sqrt{x^2-a^2}}{x}\), maka \(\frac{1}{\sin\theta}=\frac{x}{\sqrt{x^2-a^2}}\).
\[=-\frac{x}{a^2\sqrt{x^2-a^2}}+C\]3. Tentukan \(\displaystyle\int\frac{\sqrt{x^2-4}}{x^3}\,dx\)
Lihat Pembahasan
\(x=2\sec\theta\), \(\sqrt{x^2-4}=2\tan\theta\), \(x^3=8\sec^3\theta\)
\[\int\frac{2\tan\theta\cdot 2\sec\theta\tan\theta\,d\theta}{8\sec^3\theta}=\frac{1}{2}\int\frac{\tan^2\theta}{\sec^2\theta}\,d\theta=\frac{1}{2}\int\sin^2\theta\,d\theta\] \[=\frac{1}{4}(\theta-\sin\theta\cos\theta)+C\]\(\sin\theta=\frac{\sqrt{x^2-4}}{x}\), \(\cos\theta=\frac{2}{x}\), \(\theta=\text{arcsec}\frac{x}{2}\)
\[=\frac{1}{4}\text{arcsec}\frac{x}{2}-\frac{\sqrt{x^2-4}}{2x^2}+C\]4. Tentukan \(\displaystyle\int\frac{x^2}{(x^2-1)^{3/2}}\,dx\)
Lihat Pembahasan
\(x=\sec\theta\), \((x^2-1)^{3/2}=\tan^3\theta\), \(dx=\sec\theta\tan\theta\,d\theta\)
\[\int\frac{\sec^2\theta\cdot\sec\theta\tan\theta\,d\theta}{\tan^3\theta}=\int\frac{\sec^3\theta}{\tan^2\theta}\,d\theta=\int\frac{1}{\cos\theta\sin^2\theta}\,d\theta\]Tulis \(\frac{1}{\cos\theta\sin^2\theta}=\frac{\cos\theta}{\cos^2\theta\sin^2\theta}=\frac{\cos\theta}{(1-\sin^2\theta)\sin^2\theta}\)
Dengan \(u=\sin\theta\): \(\int\frac{du}{(1-u^2)u^2}=\int\left(\frac{1}{u^2}+\frac{1}{1-u^2}\right)du\)
\[=-\frac{1}{u}+\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C=-\csc\theta+\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+C\]\(\sin\theta=\frac{\sqrt{x^2-1}}{x}\), \(\csc\theta=\frac{x}{\sqrt{x^2-1}}\)
\[=-\frac{x}{\sqrt{x^2-1}}+\frac{1}{2}\ln\left|\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\right|+C\]5. Tentukan \(\displaystyle\int x^2\sqrt{x^2-9}\,dx\)
Lihat Pembahasan
\(x=3\sec\theta\), \(\sqrt{x^2-9}=3\tan\theta\), \(dx=3\sec\theta\tan\theta\,d\theta\)
\[9\sec^2\theta\cdot 3\tan\theta\cdot 3\sec\theta\tan\theta\,d\theta=81\int\sec^3\theta\tan^2\theta\,d\theta\] \[=81\int\sec^3\theta(\sec^2\theta-1)\,d\theta=81\int\sec^5\theta\,d\theta-81\int\sec^3\theta\,d\theta\]Gunakan rumus reduksi:
\[\int\sec^5\theta\,d\theta=\frac{\sec^3\theta\tan\theta}{4}+\frac{3}{8}\sec\theta\tan\theta+\frac{3}{8}\ln|\sec\theta+\tan\theta|+C\]Setelah substitusi kembali dan penyederhanaan:
\[=\frac{x(x^2-9)^{3/2}}{4}+\frac{9x\sqrt{x^2-9}}{8}-\frac{81}{8}\ln\left|\frac{x+\sqrt{x^2-9}}{3}\right|… \](Jawaban lengkap memerlukan penyederhanaan aljabar)
Latihan Soal — \(\sqrt{x^2 – a^2}\)
Mudah
- \(\displaystyle\int\frac{dx}{x\sqrt{x^2-1}}\)
- \(\displaystyle\int\frac{dx}{\sqrt{x^2-9}}\)
- \(\displaystyle\int\frac{x\,dx}{\sqrt{x^2-4}}\)
- \(\displaystyle\int\frac{dx}{x\sqrt{x^2-25}}\)
- \(\displaystyle\int\frac{\sqrt{x^2-1}}{x}\,dx\)
Sedang
- \(\displaystyle\int\sqrt{x^2-16}\,dx\)
- \(\displaystyle\int\frac{x^2}{\sqrt{x^2-4}}\,dx\)
- \(\displaystyle\int\frac{\sqrt{x^2-9}}{x^2}\,dx\)
- \(\displaystyle\int\frac{dx}{x^2\sqrt{x^2-1}}\)
- \(\displaystyle\int_3^6\frac{\sqrt{x^2-9}}{x}\,dx\)
Sulit
- \(\displaystyle\int\frac{x^3}{\sqrt{x^2-9}}\,dx\)
- \(\displaystyle\int\frac{dx}{(x^2-4)^{3/2}}\)
- \(\displaystyle\int\frac{\sqrt{x^2-1}}{x^3}\,dx\)
- \(\displaystyle\int x^2\sqrt{x^2-4}\,dx\)
- \(\displaystyle\int\frac{x^4}{(x^2-1)^{3/2}}\,dx\)
Kegiatan: Mengkomunikasikan
Tugas: Buatlah ringkasan dalam bentuk tabel atau peta konsep yang menjelaskan:
- Kapan menggunakan substitusi \(x=a\sin\theta\), \(x=a\tan\theta\), dan \(x=a\sec\theta\)
- Segitiga siku-siku pembantu untuk masing-masing substitusi
- Hasil akhir dalam variabel \(x\) untuk setiap kasus
Presentasikan hasil kerjamu di depan kelas atau diskusikan dengan teman sekelompokmu.
Rangkuman
| Bentuk | Substitusi | \(dx\) | Akar menjadi |
|---|---|---|---|
| \(\sqrt{a^2-x^2}\) | \(x=a\sin\theta\) | \(a\cos\theta\,d\theta\) | \(a\cos\theta\) |
| \(\sqrt{a^2+x^2}\) | \(x=a\tan\theta\) | \(a\sec^2\theta\,d\theta\) | \(a\sec\theta\) |
| \(\sqrt{x^2-a^2}\) | \(x=a\sec\theta\) | \(a\sec\theta\tan\theta\,d\theta\) | \(a\tan\theta\) |
Tips mengingat:
- \(a^2 – x^2\) → “minus di tengah” → sin (karena \(1-\sin^2=\cos^2\))
- \(a^2 + x^2\) → “plus” → tan (karena \(1+\tan^2=\sec^2\))
- \(x^2 – a^2\) → “x di depan minus” → sec (karena \(\sec^2-1=\tan^2\))