Materi, Soal, dan Pembahasan – Integral Parsial

Misalkan
$$\begin{aligned} u = x & \Rightarrow \text{d}u = \text{d}x \\ \text{d}v = (x+4)^5~\text{d}x & \Rightarrow v = \dfrac16(x+4)^6 \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{(x}_{u}~\underbrace{(x+4)^5~\text{d}x}_{\text{d}v} & = \underbrace{x}_{u} \cdot \underbrace{\dfrac16(x+4)^6}_{v}- \int \underbrace{\dfrac16(x+4)^6}_{v} \cdot~\underbrace{\text{d}x}_{\text{d}u} \\ & = \dfrac16x(x+4)^6-\dfrac16 \cdot \dfrac17(x+4)^7 + C \\ & = \dfrac16x(x+4)^6-\dfrac{1}{42}(x+4)^7 + C \\ & = \dfrac{1}{42}(x+4)^6(7x-(x+4)) + C \\ & = \dfrac{1}{42}(x+4)^6(6x-4) + C \\ & = \dfrac{1}{21}(3x-2)(x+4)^6 + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int x(x+4)^5~\text{d}x = \dfrac{1}{21}(3x-2)(x+4)^6 + C}$$(Jawaban A)

Misalkan
$$\begin{aligned} u & = t \Rightarrow \text{d}u = \text{d}t \\ \text{d}v & = \sqrt[3]{2t+7}~\text{d}t \end{aligned}$$Dengan mengintegralkan $\text{d}v$ menggunakan metode substitusi, dalam hal ini, $u = 2t + 7$, diperoleh
$$\begin{aligned} v & = \displaystyle \int \sqrt[3]{2t+7}~\text{d}t \\ & = \dfrac12 \int u^{1/3}~\text{d}u \\ & = \dfrac12 \cdot \dfrac34 \cdot u^{4/3} \\ & = \dfrac38(2t+7)^{4/3} \end{aligned}$$Catatan: Konstanta $C$ tidak perlu ditulis.
Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{t}_{u} \underbrace{\sqrt[3]{2t+7}~\text{d}t}_{\text{d}v} & = \underbrace{t}_{u} \cdot \underbrace{\dfrac38(2t+7)^{4/3}}_{v}- \int \underbrace{\dfrac38(2t+7)^{4/3}}_{v}~\underbrace{\text{d}t}_{\text{d}u} \\ & = \dfrac38t(2t+7)^{4/3}-\dfrac38 \cdot \dfrac12 \cdot \dfrac37(2t+7)^{7/3} + C \\ & = \dfrac{42}{112}t(2t+7)^{4/3}-\dfrac{9}{112}(2t+7)^{7/3}+C \\ & = \dfrac{3}{112}(2t+7)^{4/3}(14t-3(2t+7))+C \\ & = \dfrac{3}{112}(2t+7)^{4/3}(8t-21) + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int t\sqrt[3]{2t+7}~\text{d}t =\dfrac{3}{112}(2t+7)^{4/3}(8t-21) + C}$$(Jawaban A)

Misalkan
$$\begin{aligned} u = t & \Rightarrow \text{d}u = \text{d}t \\ \text{d}v = \sqrt{t+1}~\text{d}t & \Rightarrow v = \displaystyle \sqrt{t+1}~\text{d}t = \dfrac23(t+1)^{3/2} \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{t}_{u} \underbrace{\sqrt{t+1}~\text{d}t}_{\text{d}v} & = \underbrace{t}_{u} \cdot \underbrace{\dfrac23(t+1)^{3/2}}_{v}-\int \underbrace{\dfrac23(t+1)^{3/2}}_{v}~\underbrace{\text{d}t}_{\text{d}u} \\ & = \dfrac23t(t+1)^{3/2}-\dfrac23 \cdot \dfrac25(t+1)^{5/2} + C \\ & = \dfrac23t(t+1)^{3/2}-\dfrac{4}{15}(t+1)^{5/2} + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int t\sqrt{t+1}~\text{d}t = \dfrac23t(t+1)^{3/2}-\dfrac{4}{15}(t+1)^{5/2} + C}$$(Jawaban A)

Misalkan
$$\begin{aligned} u = t & \Rightarrow \text{d}u = \text{d}t \\ \text{d}v & = \dfrac{1}{(3t+4)^3}~\text{d}t \end{aligned}$$Dengan mengintegralkan $\text{d}v$ menggunakan metode substitusi, dalam hal ini, $u = 3t + 4$, diperoleh
$$\begin{aligned} v & = \displaystyle \int \dfrac{1}{(3t+4)^3}~\text{d}t \\ & = \dfrac13 \int u^{-3}~\text{d}u \\ & = \dfrac13 \cdot \dfrac{1}{-2} \cdot u^{-2} \\ & = -\dfrac{1}{6(3t+4)^2} \end{aligned}$$Catatan: Konstanta $C$ tidak perlu ditulis.
Dengan menggunakan rumus integrasi parsial untuk integral tentu, kita peroleh
$$\begin{aligned} \displaystyle \int_{-1}^0 \dfrac{t}{(3t+4)^3}~\text{d}t & = \left[t \cdot \left(-\dfrac{1}{6(3t+4)^2}\right)\right]_{-1}^0-\int_{-1}^0 -\dfrac{1}{6(3t+4)^2}~\text{d}t \\ & = 0-\left(-1 \cdot \dfrac{-1}{6(1)^2}\right) + \dfrac16 \cdot \dfrac13 \cdot \dfrac{1}{-1} \cdot \left[\dfrac{1}{3t+4}\right]_{-1}^0 \\ & = -\dfrac16-\dfrac{1}{18}\left[\dfrac14-1\right] \\ & = -\dfrac16-\dfrac{1}{\cancelto{6}{18}} \cdot \left(-\dfrac{\cancel{3}}{4}\right) \\ & = -\dfrac16+\dfrac{1}{24} = -\dfrac18 \end{aligned}$$Jadi, nilai dari $$\boxed{\displaystyle \int_{-1}^0 \dfrac{t}{(3t+4)^3}~\text{d}t = -\dfrac18}$$(Jawaban D)

Kita tuliskan $x \cos x~\text{d}x$ sebagai $u~\text{d}v$.
Misalkan
$$\begin{aligned} u = x & \Rightarrow \text{d}u = \text{d}x \\ \text{d}v = \cos x~\text{d}x & \Rightarrow v = \sin x \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{x}_{u} \underbrace{\cos x~\text{d}x}_{\text{d}v} & = \underbrace{x}_{u} \underbrace{\sin x}_{v}- \int \underbrace{\sin x}_{v}~\underbrace{\text{d}x}_{\text{d}u} \\ & = x \sin x-(-\cos x)+C \\ & = x \sin x + \cos x + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int x \cos x~\text{d}x =x \sin x + \cos x + C}$$(Jawaban B)

Misalkan
$$\begin{aligned} u = x & \Rightarrow \text{d}u = \text{d}x \\ \text{d}v = \sin 2x~\text{d}x & \Rightarrow v = -\dfrac12 \cos 2x \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{x}_{u} \underbrace{\sin 2x~\text{d}x}_{\text{d}v} & = \underbrace{x}_{u} \cdot \left(\underbrace{-\dfrac12 \cos 2x}_{v}\right)- \int \underbrace{-\dfrac12 \cos 2x}_{v}~\underbrace{\text{d}x}_{\text{d}u} \\ & = -\dfrac12x \cos 2x + \dfrac12 \cdot \dfrac12 \sin 2x + C \\ & = -\dfrac12x \cos 2x + \dfrac14 \sin 2x + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int x \sin 2x~\text{d}x = -\dfrac12x \cos 2x + \dfrac14 \sin 2x + C}$$(Jawaban A)

Misalkan
$$\begin{aligned} u = x^2-1 & \Rightarrow \text{d}u = 2x~\text{d}x \\ \text{d}v = \cos x~\text{d}x & \Rightarrow v = \sin x\end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{(x^2-1)}_{u}~\underbrace{\cos x~\text{d}x}_{\text{d}v} & = \underbrace{(x^2-1)}_{u} \cdot \underbrace{\sin x}_{v}- \int \underbrace{\sin x}_{v} \cdot~\underbrace{2x~\text{d}x}_{\text{d}u} \\ & = (x^2-1) \sin x-2 \color{red}{\int x~\sin x~\text{d}x} \end{aligned}$$Untuk mengintegralkan bentuk yang ditandai dengan warna merah di atas, gunakan kembali rumus integral parsial untuk
$$\begin{aligned} u = x & \Rightarrow \text{d}u = \text{d}x \\ \text{d}v = \sin x~\text{d}x & \Rightarrow v = -\cos x\end{aligned}$$Kita akan peroleh
$$\begin{aligned} (x^2-1) \sin x-2 \color{red}{\int x~\sin x~\text{d}x} & = (x^2-1) \sin x-2\left[x(-\cos x)-\int (-\cos x)~\text{d}x\right] \\ & = (x^2-1) \sin x-2\left(-x \cos x+\sin x\right)+C \\ & = (x^2-1) \sin x+2x \cos x-2 \sin x+C \\ & = (x^2-3) \sin x + 2x \cos x + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int (x^2-1) \cos x~\text{d}x = (x^2-3) \sin x+2x \cos x + C}$$(Jawaban C)

Misalkan
$$\begin{aligned} u = t-3 & \Rightarrow \text{d}u = \text{d}t \\ \text{d}v = \cos (t-3)~\text{d}t & \Rightarrow v = \sin (t-3) \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int \underbrace{(t-3)}_{u} \underbrace{\cos (t-3)~\text{d}t}_{\text{d}v} & = \underbrace{t-3}_{u} \cdot \underbrace{\sin (t-3)}_{v}- \int \underbrace{\sin (t-3)}_{v}~\underbrace{\text{d}t}_{\text{d}u} \\ & = (t-3) \sin (t-3)+\cos (t-3) + C \end{aligned}$$Dari bentuk terakhir, diperoleh nilai $a = 1$, $b = 3$, dan $c = 1$ sehingga $\boxed{a+b+c = 1+3+1 = 5}$
(Jawaban E)

Misalkan
$$\begin{aligned} u = \ln (3x) = \ln 3 + \ln x & \Rightarrow \text{d}u = \dfrac{1}{x}~ \text{d}x \\ \text{d}v = \text{d}x & \Rightarrow v = x \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int \underbrace{\ln (3x)}_{u} \underbrace{\text{d}x}_{\text{d}v} & = \underbrace{\ln (3x)}_{u} \cdot \underbrace{x}_{v}- \int \underbrace{x}_{v}\cdot ~\underbrace{\dfrac{1}{x}~\text{d}x}_{\text{d}u} \\ & = x \ln (3x)-\int \text{d}x \\ & = x \ln (3x)-x + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int \ln (3x)~\text{d}x = x \ln (3x)-x + C}$$(Jawaban A)

Misalkan
$$\begin{aligned} u = x & \Rightarrow \text{d}u = \text{d}x \\ \text{d}v = e^x~\text{d}x & \Rightarrow v = e^x \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{x}_{u} \underbrace{e^x~\text{d}x}_{\text{d}v} & = \underbrace{x}_{u} \cdot \underbrace{e^x}_{v}- \int \underbrace{e^x}_{v}~\underbrace{\text{d}x}_{\text{d}u} \\ & = x \cdot e^x-e^x+C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int x \cdot e^x~\text{d}x = x \cdot e^x-e^x+C}$$(Jawaban B)

Misalkan
$$\begin{aligned} u = t & \Rightarrow \text{d}u = \text{d}t \\ \text{d}v = e^{5t+\pi}~\text{d}t & \Rightarrow v = \dfrac15e^{5t+\pi} \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{t}_{u} \underbrace{e^{5t+\pi}~\text{d}t}_{\text{d}v} & = \underbrace{t}_{u} \cdot \underbrace{\dfrac15e^{5t+\pi}}_{v}- \int \underbrace{\dfrac15e^{5t+\pi}}_{v}~\underbrace{\text{d}t}_{\text{d}u} \\ & = \dfrac15te^{5t+\pi}-\dfrac15 \cdot \dfrac15e^{5t+\pi} + C \\ & = \dfrac15te^{5t+\pi}-\dfrac{1}{25}e^{5t+\pi} + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int t \cdot e^{5t+\pi}~\text{d}t = \dfrac15te^{5t+\pi}-\dfrac{1}{25}e^{5t+\pi} + C}$$(Jawaban D)

Misalkan
$$\begin{aligned} u = 2x+5 & \Rightarrow \text{d}u = 2~\text{d}x \\ \text{d}v =\dfrac{1}{(x-2)^3}~\text{d}x & \Rightarrow v = -\dfrac{1}{2(x-2)^2} \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \dfrac{2x+5}{(x-2)^3}~\text{d}x & = \int \underbrace{(2x+5}_{u} \cdot ~\underbrace{\dfrac{1}{(x-2)^3}~\text{d}x}_{\text{d}v} \\ & = \underbrace{2x+5}_{u} \cdot \left(\underbrace{-\dfrac{1}{2(x-2)^2}}_{v}\right)- \int \underbrace{-\dfrac{1}{\cancel{2}(x-2)^2}}_{v} \cdot~\underbrace{\cancel{2}~\text{d}x}_{\text{d}u} \\ & = -\dfrac{2x+5}{2(x-2)^2}+\int \dfrac{1}{(x-2)^2}~\text{d}x \\ & = -\dfrac{2x+5}{2(x-2)^2}-\dfrac{1}{x-2}+C \\ & = -\dfrac{2x+5}{2(x-2)^2}-\dfrac{2(x-2)}{2(x-2)^2} + C \\ & = -\dfrac{4x+1}{2(x-2)^2} + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int \dfrac{2x+5}{(x-2)^3}~\text{d}x = -\dfrac{4x+1}{2(x-2)^2} + C}$$

Perhatikan bahwa bentuk integran di atas dapat ditulis kembali menjadi
$$\displaystyle 7 \int t(2t-1)^{-5}~\text{d}t.$$Misalkan
$$\begin{aligned} u & = t \Rightarrow \text{d}u = \text{d}t \\ \text{d}v & = (2t-1)^{-5}~\text{d}t \end{aligned}$$Dengan mengintegralkan $\text{d}v$ menggunakan metode substitusi, dalam hal ini, $u = 2t-1$, diperoleh
$$\begin{aligned} v & = \displaystyle \int (2t-1)^{-5}~\text{d}t \\ & = \dfrac12 \cdot \dfrac{1}{-4} (2t-1)^{-4} \\ & = -\dfrac18 (2t-1)^{-4} \end{aligned}$$Catatan: Konstanta $C$ tidak perlu ditulis.
Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle 7 \int \underbrace{t}_{u} \underbrace{(2t-1)^{-5}~\text{d}t}_{\text{d}v} & = 7\left[\underbrace{t}_{u} \cdot \left(\underbrace{-\dfrac18(2t-1)^{-4}}_{v}\right)- \int \underbrace{-\dfrac18(2t-1)^{-4}}_{v}~\underbrace{\text{d}t}_{\text{d}u}\right] \\ & = -\dfrac78t(2t-1)^{-4} + 7 \cdot \dfrac18 \cdot \dfrac12 \cdot \dfrac{1}{-3} (2t-1)^{-3} + C \\ & = -\dfrac78t(2t-1)^{-4}-\dfrac{7}{48}(2t-1)^{-3} + C \\ & = -\dfrac{7}{48}(2t-1)^{-3}\left(6t(2t-1)^{-1} + 1\right) + C \\ & = -\dfrac{7}{48(2t-1)^3}\left(\dfrac{6t}{2t-1}+\dfrac{2t-1}{2t-1}\right)+C \\ & = -\dfrac{7(8t-1)}{48(2t-1)^4}+C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int \dfrac{7t}{(2t-1)^5}~\text{d}t = -\dfrac{7(8t-1)}{48(2t-1)^4}+C}$$

Untuk mencari hasil integral tersebut, kita akan menggunakan teknik integral parsial sebanyak $3$ kali.
Misalkan
$$\begin{aligned} u = t^3 & \Rightarrow \text{d}u = 3t^2 ~\text{d}t \\ \text{d}v = \sin t~\text{d}t & \Rightarrow v = -\cos t\end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle \int \underbrace{t^3}_{u} \underbrace{\sin t~\text{d}t}_{\text{d}v} & = \underbrace{t^3}_{u} \cdot \left(\underbrace{-\cos t}_{v}\right)- \int \underbrace{-\cos t}_{v}~\underbrace{3t^2~\text{d}t}_{\text{d}u} \\ & = -t^3 \cos t + 3 \color{red}{\int t^2~\cos t~\text{d}t} \end{aligned}$$Sekarang, misalkan
$$\begin{aligned} u = t^2 & \Rightarrow \text{d}u = 2t ~\text{d}t \\ \text{d}v = \cos t~\text{d}t & \Rightarrow v = \sin t\end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} \displaystyle -t^3 \cos t + 3 \color{red}{\int t^2~\cos t~\text{d}t} & = -t^3 \cos t + 3\left[t^2 \sin t-\int \sin t \cdot 2t~\text{d}t\right] \\ & = -t^3 \cos t + 3t^2 \sin t-6 \color{red}{\int t \sin t~\text{d}t} \end{aligned}$$Terakhir, misalkan
$$\begin{aligned} u = t & \Rightarrow \text{d}u = \text{d}t \\ \text{d}v = \sin t~\text{d}t & \Rightarrow v = -\cos t\end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{aligned} & -t^3 \cos t + 3t^2 \sin t-6 \color{red}{\int t \sin t~\text{d}t} \\ & = -t^3 \cos t + 3t^2 \sin t-6\left[-t \cos t-\int -\cos t~\text{d}t\right] \\ & = -t^3 \cos t + 3t^2 \sin t+6t \cos t-6 \sin t \\ & = (-t^3+6t)~\cos t + (3t^2-6)~\sin t + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int t^3 \sin t~\text{d}t = (-t^3+6t)~\cos t + (3t^2-6)~\sin t + C }$$

Misalkan
$$\begin{aligned} u = x & \Rightarrow \text{d}u = \text{d}x \\ \text{d}v = \cos 3x~\text{d}t & \Rightarrow v = \dfrac13 \sin 3x \end{aligned}$$Dengan menggunakan rumus integrasi parsial untuk integral tentu, kita peroleh
$$\begin{aligned} \displaystyle & \int_{\pi/9}^{\pi/6} x \cos 3x~\text{d}x \\ & = \left[x \cdot \dfrac13 \sin 3x\right]_{\pi/9}^{\pi/6}-\displaystyle \int_{\pi/6}^{\pi/9} \dfrac13 \sin 3x~\text{d}x \\ & = \dfrac{\pi}{6} \cdot \dfrac13 \sin \dfrac{\pi}{2}-\dfrac{\pi}{9} \cdot \dfrac13 \sin \dfrac{\pi}{3}-\dfrac13 \cdot \dfrac13 \left[-\cos 3x\right]_{\pi/9}^{\pi/6} \\ & = \dfrac{\pi}{18}(1)-\dfrac{\pi}{27} \cdot \dfrac12\sqrt3+\dfrac19\left(\cos \dfrac{\pi}{2}-\cos \dfrac{\pi}{3}\right) \\ & = \dfrac{\pi}{18}-\dfrac{\pi}{54}\sqrt3 + \dfrac19\left(0-\dfrac12\right) \\ & = \dfrac{\pi}{18}-\dfrac{\pi}{54}\sqrt3-\dfrac{1}{18} \\ & = \dfrac{3\pi}{54}-\dfrac{\pi}{54}\sqrt3-\dfrac{3}{54} \\ & = \dfrac{(3-\sqrt3)\pi-3}{54} \end{aligned}$$Jadi, nilai dari $$\boxed{\displaystyle \int_{\pi/9}^{\pi/6} x \cos 3x~\text{d}x = \dfrac{(3-\sqrt3)\pi-3}{54}}$$

Diberikan integral berikut.
$$ \displaystyle \int \sin x \sin 3x~\text{d}x$$Akan ditunjukkan bahwa hasil integrasinya adalah
$$-\dfrac38 \sin x \cos 3x + \dfrac18 \cos x \sin 3x + C$$menggunakan rumus integral parsial.
Misalkan
$$\begin{aligned} u = \sin x & \Rightarrow \text{d}u = \cos x~ \text{d}x \\ \text{d}v = \sin 3x~\text{d}x & \Rightarrow v = -\dfrac13 \cos 3x \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int \underbrace{\sin x}_{u} \underbrace{\sin 3x~\text{d}x}_{\text{d}v} & = \underbrace{\sin x}_{u} \cdot \left(\underbrace{-\dfrac13 \cos 3x}_{v}\right)- \int \underbrace{-\dfrac13 \cos 3x}_{v} \cdot ~\underbrace{\cos x~\text{d}x}_{\text{d}u} \\ & = -\dfrac13 \sin x \cos 3x+\dfrac13 \int \cos 3x \cos x~\text{d}x \end{aligned}$$Gunakan rumus integral parsial sekali lagi pada bentuk
$$\displaystyle \int \cos 3x \cos x~\text{d}x$$Misalkan
$$\begin{aligned} u = \cos x & \Rightarrow \text{d}u = -\sin x~\text{d}x \\ \text{d}v = \cos 3x & \Rightarrow v = \dfrac13 \sin 3x \end{aligned}$$sehingga kita peroleh
$$\begin{aligned} \displaystyle \int \sin x \sin 3x~\text{d}x & = -\dfrac13 \sin x \cos 3x+\dfrac13\left[\underbrace{\cos x}_{u} \cdot \left(\underbrace{\dfrac13 \sin 3x}_{v}\right)- \int \underbrace{\dfrac13 \sin 3x}_{v} \cdot ~\underbrace{(-\sin x)~\text{d}x}_{\text{d}u}\right] \\ \int \sin x \sin 3x~\text{d}x & = -\dfrac13 \sin x \cos 3x+\dfrac19 \cos x \sin 3x+\dfrac19 \int \sin 3x \sin x~\text{d}x \\ \dfrac89 \int \sin x \sin 3x~\text{d}x & = -\dfrac13 \sin x \cos 3x+\dfrac19 \cos x \sin 3x+K \\ \int \sin x \sin 3x~\text{d}x & = -\dfrac38 \sin x \cos 3x+\dfrac18 \cos x \sin 3x + C \end{aligned}$$Jadi, berdasarkan integrasi parsial, kita telah menurunkan rumus integral trigonometri berikut.
$$\boxed{\displaystyle \int \sin x \sin 3x~\text{d}x = -\dfrac38 \sin x \cos 3x + \dfrac18 \cos x \sin 3x + C}$$

Diberikan integral berikut.
$$\displaystyle \int \cos 5x \sin 7x~\text{d}x$$Akan ditunjukkan bahwa hasil integrasinya adalah
$$-\dfrac{7}{24} \cos 5x \cos 7x -\dfrac{5}{24} \sin 5x \sin 7x + C$$menggunakan rumus integral parsial.
Misalkan
$$\begin{aligned} u = \cos 5x & \Rightarrow \text{d}u = -5 \sin 5x~ \text{d}x \\ \text{d}v = \sin 7x~\text{d}x & \Rightarrow v = -\dfrac17 \cos 7x \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int \underbrace{\cos 5x}_{u} \underbrace{\sin 7x~\text{d}x}_{\text{d}v} & = \underbrace{\cos 5x}_{u} \cdot \left(\underbrace{-\dfrac17 \cos 7x}_{v}\right)- \int \underbrace{-\dfrac17 \cos 7x}_{v} \cdot \underbrace{(-5 \sin 5x)~\text{d}x}_{\text{d}u} \\ & = -\dfrac17 \cos 5x \cos 7x-\dfrac57 \int \cos 7x \sin 5x~\text{d}x \end{aligned}$$Gunakan rumus integral parsial sekali lagi pada bentuk
$$\displaystyle \int \cos 7x \sin 5x~\text{d}x$$Misalkan
$$\begin{aligned} u = \sin 5x & \Rightarrow \text{d}u = 5 \cos 5x~\text{d}x \\ \text{d}v = \cos 7x & \Rightarrow v = \dfrac17 \sin 7x \end{aligned}$$sehingga kita peroleh
$$\begin{aligned} \displaystyle \int \cos 5x \sin 7x~\text{d}x & = -\dfrac17 \cos 5x \cos 7x-\dfrac57\left[\underbrace{\sin 5x}_{u} \cdot \underbrace{\dfrac17 \sin 7x}_{v}- \int \underbrace{\dfrac17 \sin 7x}_{v} \cdot ~\underbrace{5 \cos 5x)~\text{d}x}_{\text{d}u}\right] \\ \int \cos 5x \sin 7x~\text{d}x & = -\dfrac17 \cos 5x \cos 7x-\dfrac{5}{49} \sin 5x \sin 7x + \dfrac{25}{49} \int \cos 5x \sin 7x~\text{d}x \\ \dfrac{24}{49} \int \cos 5x \sin 7x~\text{d}x & = -\dfrac17 \cos 5x \cos 7x-\dfrac{5}{49} \sin 5x \sin 7x + K \\ \int \cos 5x \sin 7x~\text{d}x & = -\dfrac{7}{24} \cos 5x \cos 7x-\dfrac{5}{24} \sin 5x \sin 7x + C \end{aligned}$$Jadi, berdasarkan integrasi parsial, kita telah menurunkan rumus integral trigonometri berikut.
$$\boxed{\displaystyle \int \cos 5x \sin 7x~\text{d}x = -\dfrac{7}{24} \cos 5x \cos 7x -\dfrac{5}{24} \sin 5x \sin 7x + C}$$

Misalkan
$$\begin{aligned} u = e^x & \Rightarrow \text{d}u = e^x~\text{d}x \\ \text{d}v = \sin x~\text{d}x & \Rightarrow v = -\cos x \end{aligned}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\displaystyle \int e^x \sin x~\text{d}x = -e^x \cos x + \color{red}{\int e^x \cos x~\text{d}x}~~~(\cdots 1)$$Selanjutnya, gunakan rumus integrasi parsial sekali lagi pada bentuk integralnya (ditandai dengan warna merah di atas).
$$\begin{aligned} u = e^x & \Rightarrow \text{d}u = e^x~\text{d}x \\ \text{d}v = \cos x~\text{d}x & \Rightarrow v = \sin x \end{aligned}$$Kita akan peroleh
$$\displaystyle \int e^x \cos x~\text{d}x = e^x \sin x- \color{blue}{\int e^x \sin x~\text{d}x}$$Jika disubstitusikan pada persamaan $(1)$ di atas, kita peroleh
$$\begin{aligned} \displaystyle \int e^x \sin x~\text{d}x & = -e^x \cos x + e^x \sin x-\int e^x \sin x~\text{d}x \\ 2 \int e^x \sin x~\text{d}x & = -e^x \cos x + e^x \sin x+C \\ \int e^x \sin x~\text{d}x & = \dfrac{ -e^x \cos x + e^x \sin x}{2}+K \\ \int e^x \sin x~\text{d}x & = \dfrac{e^x(\sin x-\cos x)}{2}+K \end{aligned}$$Catatan: Perhatikan bahwa notasi konstanta berubah dari $C$ menjadi $K = \dfrac{C}{2}$. Penggunaan notasi konstanta bisa disesuaikan dengan memilih huruf kapital yang lain.
Fakta bahwa integral yang hendak kita cari muncul kembali di ruas kanan membuat kita dapat mencari hasil integralnya.

Misalkan
$$\begin{aligned} u = \ln (ax^b) = \ln a + b \ln x & \Rightarrow \text{d}u = \dfrac{b}{x}~ \text{d}x \\ \text{d}v = \text{d}x & \Rightarrow v = x \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int \underbrace{\ln (ax^b)}_{u} \underbrace{\text{d}x}_{\text{d}v} & = \underbrace{\ln (ax^b)}_{u} \cdot \underbrace{x}_{v}- \int \underbrace{x}_{v}\cdot ~\underbrace{\dfrac{b}{x}~\text{d}x}_{\text{d}u} \\ & = x \ln (ax^b)-\int b~\text{d}x \\ & = x \ln (ax^b)-bx + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int \ln (ax^b)~\text{d}x = x \ln (ax^b)-bx + C}$$

Misalkan
$$\begin{aligned} u = \arctan x & \Rightarrow \text{d}u = \dfrac{1}{1+x^2}~ \text{d}x \\ \text{d}v = \text{d}x & \Rightarrow v = x \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int \underbrace{\arctan x}_{u} \underbrace{\text{d}x}_{\text{d}v} & = \underbrace{\arctan x}_{u} \cdot \underbrace{x}_{v}- \int \underbrace{x}_{v} \cdot ~\underbrace{\dfrac{1}{1+x^2}~\text{d}x}_{\text{d}u} \\ & = x \arctan x-\int \dfrac{x}{1+x^2}~\text{d}x \end{aligned}$$Gunakan metode substitusi. Misalkan $u = 1+x^2$, maka $\text{d}u = 2x~\text{d}x$ sehingga diperoleh
$$\begin{aligned} \displaystyle x \arctan x-\int \dfrac{x}{1+x^2}~\text{d}x & = x \arctan x-\dfrac12 \int \dfrac{1}{u}~\text{d}u \\ & = x \arctan x-\dfrac12 \ln u + C \\ & = x \arctan x-\dfrac12 \ln (1+x^2) + C \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int \arctan x~\text{d}x = x \arctan x-\dfrac12 \ln (1+x^2) + C}$$

Misalkan
$$\begin{aligned} u = \ln (t) & \Rightarrow \text{d}u = \dfrac{1}{t}~ \text{d}t \\ \text{d}v = \sqrt{t}~\text{d}t & \Rightarrow v = \dfrac23t^{3/2} \end{aligned}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{aligned} \displaystyle \int_1^e \underbrace{\ln t}_{u} \underbrace{\sqrt{t}~\text{d}t}_{\text{d}v} & = \left[\underbrace{\ln t}_{u} \cdot \underbrace{\dfrac23t^{3/2}}_{v}\right]_1^e- \int_1^e \underbrace{\dfrac23t^{3/2}}_{v} \cdot ~\underbrace{\dfrac{1}{t}~\text{d}t}_{\text{d}u} \\ & = \left[\dfrac23t^{3/2} \cdot \ln t\right]_1^e-\dfrac23 \int_1^e t^{1/2}~\text{d}t \\ & = \left[\dfrac23t^{3/2} \cdot \ln t\right]_1^e-\dfrac49 \left[t^{3/2}\right]_1^e \\ & = \left(\dfrac23e^{3/2} \cdot \ln e-\dfrac23 \cdot 1^{3/2} \cdot \ln 1\right)-\dfrac49\left(e^{3/2}-1^{3/2}\right) \\ & = \left(\dfrac23e^{3/2}-0\right)-\dfrac49\left(e^{3/2}-1\right) \\ & = \dfrac29e^{3/2}+\dfrac49 \\ & = \dfrac29\left(e^{3/2} + 2\right) \end{aligned}$$Jadi, hasil dari $$\boxed{\displaystyle \int \sqrt{t} \ln t~\text{d}t = \dfrac29\left(e^{3/2} + 2\right)}$$

Dengan menggunakan aturan dasar integral tak tentu, kita seharusnya tahu bahwa
$$\displaystyle \int f(x)~\text{d}x = F(x) + C$$ untuk suatu konstanta $C$. Ini menunjukkan setiap proses pengintegrasian integral tak tentu, konstanta $C$ harus dimunculkan.
Pada langkah terakhir pembuktian di atas, konstanta $C$ tidak dimunculkan. Misalkan hasil integralnya adalah $F(x) + C_i$, maka diperoleh
$$\begin{aligned} \cancel{F(x)} + C_1 & = 1 + \cancel{F(x)} + C_2 \\ C_1 & = 1 + C_2 \\ 0 & = 1 + (C_2-C_1) \\ 0 & = 1 + C \end{aligned}$$Pernyataan ini akan benar apabila $C_2-C_1 = C = -1$.
Catatan:
Pembuktian yang menghasilkan pernyataan yang keliru seperti kasus ini termasuk dalam ranah kelancungan matematis (mathematical fallacy).