$$\bbox[black, 5px, border: 2px solid green] {\int u \text{ dv} = u \cdot v \:-\: \int v \cdot \text{ du}} $$
Contoh 01
Tentukan \(\int x^2 \cdot \cos 3x \text{ dx}\)
Misal \(u = x^2\) dan \(\text{ dv} = \cos 3x \text{ dx}\)
Buat tabel dengan \(u\) pada kolom pertama diturunkan sampai hasilnya 0 dan \(\text{ dv}\) pada kolom kedua diintegralkan.
Perhatikan tanda panah untuk melakukan perkalian \(u\) dengan \(\text{ dv}\) dan tambahkan juga tanda (+), (−), (+), … secara selang-seling.
\(\int x^2 \cdot \cos 3x \text{ dx} = \dfrac{1}{3} x^2 \sin 3x \:-\: 2x (-\dfrac{1}{9} \cos 3x) + 2 (-\dfrac{1}{27} \sin 3x) + C\)
\(\int x^2 \cdot \cos 3x \text{ dx} = \dfrac{1}{3} x^2 \sin 3x + \dfrac{2}{9}x \cos 3x \:-\: \dfrac{2}{27} \sin 3x + C\)
Contoh 02
Tentukan \(\int 6x^5 \cdot \sin x^3 \text{ dx}\)
Misal:
\(w = x^3\)
\(\text{dw} = 3x^2 \text{ dx} \rightarrow \text{ dx} = \dfrac{\text{ dw}}{3x^2}\)
\(\int 6x^2\cdot x^3 \cdot \sin x^3 \text{ dx} = \int \cancelto{2}{6 x^2} \cdot w \cdot \sin w \:\dfrac{\text{ dw}}{\cancel{3x^2}}\)
\(\int 6x^2\cdot x^3 \cdot \sin x^3 \text{ dx} = 2\int w \cdot \sin w \text{ dw}\)
Gunakan rumus integral parsial untuk menyelesaikan \(\int w \cdot \sin w \text{ dw}\)
Misal \(u = w\) dan \(\text{ dv} = \sin w \text{ dw}\)
\(\int w \cdot \sin w \text{ dw} = u \cdot v \:-\: \int v \text{ du}\)
\(\int w \cdot \sin w \text{ dw} = w \cdot (-\cos w) \:-\: \int (-\cos w) \text{ dw}\)
\(\int w \cdot \sin w \text{ dw} = -w \cos w + \sin w + C\)
\(\int 6x^2\cdot x^3 \cdot \sin x^3 \text{ dx} = 2[ -w \cos w + \sin w] + C\)
\(\int 6x^2\cdot x^3 \cdot \sin x^3 \text{ dx} =-2w \cos w + 2\sin w + C\)
Kembalikan lagi \(w = x^3\)
Jadi, \(\int 6 x^5 \cdot \sin x^3 \text{ dx} =-2x^3 \cos x^3 + 2\sin x^3 + C\)
Contoh 03
Tentukan \(\int e^{2x} \cdot \cos x \text{ dx}\)
Misal:
\(u = e^{2x} \rightarrow \dfrac{\text{du}}{\text{dx}} = 2\cdot e^{2x}\)
\(\text{ dv} = \cos x \text{ dx} \rightarrow v = \int \cos x \text{ dx} = \sin x\)
Gunakan rumus integral parsial:
\(\int u \text{ dv} = u \cdot v \:-\: \int v \cdot \text{ du}\)
\(\int e^{2x} \cdot \cos x \text{ dx} = e^{2x} \cdot \sin x\:-\: \int \sin x \cdot 2\cdot e^{2x} \text{ dx}\)
\(\int e^{2x} \cdot \cos x \text{ dx} = e^{2x} \sin x\:-\: 2 \color{blue}\int e^{2x} \cdot \sin x\text{ dx}\)
Untuk menyelesaikan \(\color{blue}\int e^{2x} \cdot \sin x\text{ dx}\) gunakan rumus integral parsial.
Misal:
\(u = e^{2x} \rightarrow \dfrac{\text{du}}{\text{dx}} = 2\cdot e^{2x}\)
\(\text{ dv} = \sin x \text{ dx} \rightarrow v = \int \sin x \text{ dx} = -\cos x\)
\(\int e^{2x} \cdot \cos x \text{ dx} = e^{2x} \sin x\:-\: 2[e^{2x} \cdot (-\cos x) \:-\: \int (-\cos x) \cdot 2 \cdot e^{2x} \text{ dx}]\)
\(\int e^{2x} \cdot \cos x \text{ dx} = e^{2x} \sin x + 2e^{2x} \cos x \:-\: 4\int e^{2x} \cdot \cos x\text{ dx}\)
\(1 \cdot \int e^{2x} \cdot \cos x \text{ dx} + 4 \cdot \int e^{2x} \cdot \cos x\text{ dx} = e^{2x} \sin x + 2e^{2x} \cos x + C\)
\( 5 \cdot \int e^{2x} \cdot \cos x\text{ dx} =e^{2x} \sin x + 2e^{2x} \cos x + C\)
Bagi kedua ruas dengan 5,
\( \int e^{2x} \cdot \cos x\text{ dx} = \dfrac{1}{5} e^{2x} \sin x + \dfrac{2}{5} e^{2x} \cos x + C\)
