Tag: Aturan Tanzalin

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

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

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

Misalkan
$$\begin{array}{rl}u=t& \Rightarrow \text{d}u=\text{d}t\\ \text{d}v& ={\displaystyle \frac{1}{(3t+4{)}^3}}\text{d}t\end{array}$$Dengan mengintegralkan $\text{d}v$ menggunakan metode substitusi, dalam hal ini, $u=3t+4$, diperoleh
$$\begin{array}{rl}v& ={\displaystyle \int {\displaystyle \frac{1}{(3t+4{)}^3}}\text{d}t}\\ & ={\displaystyle \frac{1}{3}}\int u^{-3}\text{d}u\\ & ={\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{-2}}\cdot u^{-2}\\ & =-{\displaystyle \frac{1}{6(3t+4{)}^2}}\end{array}$$Catatan: Konstanta $C$ tidak perlu ditulis.
Dengan menggunakan rumus integrasi parsial untuk integral tentu, kita peroleh
$$\begin{array}{rl}{\displaystyle {\int }_{-1}^0{\displaystyle \frac{t}{(3t+4{)}^3}}\text{d}t}& ={[t\cdot (-{\displaystyle \frac{1}{6(3t+4{)}^2}})]}_{-1}^0-{\int }_{-1}^0-{\displaystyle \frac{1}{6(3t+4{)}^2}}\text{d}t\\ & =0-(-1\cdot {\displaystyle \frac{-1}{6(1{)}^2}})+{\displaystyle \frac{1}{6}}\cdot {\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{-1}}\cdot {\left[{\displaystyle \frac{1}{3t+4}}\right]}_{-1}^0\\ & =-{\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{18}}[{\displaystyle \frac{1}{4}}-1]\\ & =-{\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{{\cancel{18}}^6}}\cdot (-{\displaystyle \frac{\cancel{3}}{4}})\\ & =-{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{24}}=-{\displaystyle \frac{1}{8}}\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle {\int }_{-1}^0{\displaystyle \frac{t}{(3t+4{)}^3}}\text{d}t=-{\displaystyle \frac{1}{8}}}}}$$(Jawaban D)

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

Misalkan
$$\begin{array}{rl}u=x& \Rightarrow \text{d}u=\text{d}x\\ \text{d}v=\sin 2x\text{d}x& \Rightarrow v=-{\displaystyle \frac{1}{2}}\cos 2x\end{array}$$Dengan menggunakan rumus integrasi parsial, kita peroleh
$$\begin{array}{rl}{\displaystyle \int \underset{u}{\underbrace{x}}\underset{\text{d}v}{\underbrace{\sin 2x\text{d}x}}}& =\underset{u}{\underbrace{x}}\cdot \left(\underset{v}{\underbrace{-{\displaystyle \frac{1}{2}}\cos 2x}}\right)-\int \underset{v}{\underbrace{-{\displaystyle \frac{1}{2}}\cos 2x}}\underset{\text{d}u}{\underbrace{\text{d}x}}\\ & =-{\displaystyle \frac{1}{2}}x\cos 2x+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{2}}\sin 2x+C\\ & =-{\displaystyle \frac{1}{2}}x\cos 2x+{\displaystyle \frac{1}{4}}\sin 2x+C\end{array}$$Jadi, hasil dari $$\boxed{{\displaystyle {\displaystyle \int x\sin 2x\text{d}x=-{\displaystyle \frac{1}{2}}x\cos 2x+{\displaystyle \frac{1}{4}}\sin 2x+C}}}$$(Jawaban A)

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

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

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

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

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

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

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

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

Misalkan
$$\begin{array}{rl}u=x& \Rightarrow \text{d}u=\text{d}x\\ \text{d}v=\cos 3x\text{d}t& \Rightarrow v={\displaystyle \frac{1}{3}}\sin 3x\end{array}$$Dengan menggunakan rumus integrasi parsial untuk integral tentu, kita peroleh
$$\begin{array}{rl}{\displaystyle }& {\int }_{\pi /9}^{\pi /6}x\cos 3x\text{d}x\\ & ={[x\cdot {\displaystyle \frac{1}{3}}\sin 3x]}_{\pi /9}^{\pi /6}-{\displaystyle {\int }_{\pi /6}^{\pi /9}{\displaystyle \frac{1}{3}}\sin 3x\text{d}x}\\ & ={\displaystyle \frac{\pi }{6}}\cdot {\displaystyle \frac{1}{3}}\sin {\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{\pi }{9}}\cdot {\displaystyle \frac{1}{3}}\sin {\displaystyle \frac{\pi }{3}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{3}}{[-\cos 3x]}_{\pi /9}^{\pi /6}\\ & ={\displaystyle \frac{\pi }{18}}(1)-{\displaystyle \frac{\pi }{27}}\cdot {\displaystyle \frac{1}{2}}\sqrt{3}+{\displaystyle \frac{1}{9}}(\cos {\displaystyle \frac{\pi }{2}}-\cos {\displaystyle \frac{\pi }{3}})\\ & ={\displaystyle \frac{\pi }{18}}-{\displaystyle \frac{\pi }{54}}\sqrt{3}+{\displaystyle \frac{1}{9}}(0-{\displaystyle \frac{1}{2}})\\ & ={\displaystyle \frac{\pi }{18}}-{\displaystyle \frac{\pi }{54}}\sqrt{3}-{\displaystyle \frac{1}{18}}\\ & ={\displaystyle \frac{3\pi }{54}}-{\displaystyle \frac{\pi }{54}}\sqrt{3}-{\displaystyle \frac{3}{54}}\\ & ={\displaystyle \frac{(3-\sqrt{3})\pi -3}{54}}\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle {\int }_{\pi /9}^{\pi /6}x\cos 3x\text{d}x={\displaystyle \frac{(3-\sqrt{3})\pi -3}{54}}}}}$$

Diberikan integral berikut.
$${\displaystyle \int \sin x\sin 3x\text{d}x}$$Akan ditunjukkan bahwa hasil integrasinya adalah
$$-{\displaystyle \frac{3}{8}}\sin x\cos 3x+{\displaystyle \frac{1}{8}}\cos x\sin 3x+C$$menggunakan rumus integral parsial.
Misalkan
$$\begin{array}{rl}u=\sin x& \Rightarrow \text{d}u=\cos x\text{d}x\\ \text{d}v=\sin 3x\text{d}x& \Rightarrow v=-{\displaystyle \frac{1}{3}}\cos 3x\end{array}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{array}{rl}{\displaystyle \int \underset{u}{\underbrace{\sin x}}\underset{\text{d}v}{\underbrace{\sin 3x\text{d}x}}}& =\underset{u}{\underbrace{\sin x}}\cdot \left(\underset{v}{\underbrace{-{\displaystyle \frac{1}{3}}\cos 3x}}\right)-\int \underset{v}{\underbrace{-{\displaystyle \frac{1}{3}}\cos 3x}}\cdot \underset{\text{d}u}{\underbrace{\cos x\text{d}x}}\\ & =-{\displaystyle \frac{1}{3}}\sin x\cos 3x+{\displaystyle \frac{1}{3}}\int \cos 3x\cos x\text{d}x\end{array}$$Gunakan rumus integral parsial sekali lagi pada bentuk
$${\displaystyle \int \cos 3x\cos x\text{d}x}$$Misalkan
$$\begin{array}{rl}u=\cos x& \Rightarrow \text{d}u=-\sin x\text{d}x\\ \text{d}v=\cos 3x& \Rightarrow v={\displaystyle \frac{1}{3}}\sin 3x\end{array}$$sehingga kita peroleh
$$\begin{array}{rl}{\displaystyle \int \sin x\sin 3x\text{d}x}& =-{\displaystyle \frac{1}{3}}\sin x\cos 3x+{\displaystyle \frac{1}{3}}[\underset{u}{\underbrace{\cos x}}\cdot \left(\underset{v}{\underbrace{{\displaystyle \frac{1}{3}}\sin 3x}}\right)-\int \underset{v}{\underbrace{{\displaystyle \frac{1}{3}}\sin 3x}}\cdot \underset{\text{d}u}{\underbrace{(-\sin x)\text{d}x}}]\\ \int \sin x\sin 3x\text{d}x& =-{\displaystyle \frac{1}{3}}\sin x\cos 3x+{\displaystyle \frac{1}{9}}\cos x\sin 3x+{\displaystyle \frac{1}{9}}\int \sin 3x\sin x\text{d}x\\ {\displaystyle \frac{8}{9}}\int \sin x\sin 3x\text{d}x& =-{\displaystyle \frac{1}{3}}\sin x\cos 3x+{\displaystyle \frac{1}{9}}\cos x\sin 3x+K\\ \int \sin x\sin 3x\text{d}x& =-{\displaystyle \frac{3}{8}}\sin x\cos 3x+{\displaystyle \frac{1}{8}}\cos x\sin 3x+C\end{array}$$Jadi, berdasarkan integrasi parsial, kita telah menurunkan rumus integral trigonometri berikut.
$$\boxed{{\displaystyle {\displaystyle \int \sin x\sin 3x\text{d}x=-{\displaystyle \frac{3}{8}}\sin x\cos 3x+{\displaystyle \frac{1}{8}}\cos x\sin 3x+C}}}$$

Diberikan integral berikut.
$${\displaystyle \int \cos 5x\sin 7x\text{d}x}$$Akan ditunjukkan bahwa hasil integrasinya adalah
$$-{\displaystyle \frac{7}{24}}\cos 5x\cos 7x-{\displaystyle \frac{5}{24}}\sin 5x\sin 7x+C$$menggunakan rumus integral parsial.
Misalkan
$$\begin{array}{rl}u=\cos 5x& \Rightarrow \text{d}u=-5\sin 5x\text{d}x\\ \text{d}v=\sin 7x\text{d}x& \Rightarrow v=-{\displaystyle \frac{1}{7}}\cos 7x\end{array}$$Dengan menggunakan rumus integral parsial, diperoleh
$$\begin{array}{rl}{\displaystyle \int \underset{u}{\underbrace{\cos 5x}}\underset{\text{d}v}{\underbrace{\sin 7x\text{d}x}}}& =\underset{u}{\underbrace{\cos 5x}}\cdot \left(\underset{v}{\underbrace{-{\displaystyle \frac{1}{7}}\cos 7x}}\right)-\int \underset{v}{\underbrace{-{\displaystyle \frac{1}{7}}\cos 7x}}\cdot \underset{\text{d}u}{\underbrace{(-5\sin 5x)\text{d}x}}\\ & =-{\displaystyle \frac{1}{7}}\cos 5x\cos 7x-{\displaystyle \frac{5}{7}}\int \cos 7x\sin 5x\text{d}x\end{array}$$Gunakan rumus integral parsial sekali lagi pada bentuk
$${\displaystyle \int \cos 7x\sin 5x\text{d}x}$$Misalkan
$$\begin{array}{rl}u=\sin 5x& \Rightarrow \text{d}u=5\cos 5x\text{d}x\\ \text{d}v=\cos 7x& \Rightarrow v={\displaystyle \frac{1}{7}}\sin 7x\end{array}$$sehingga kita peroleh
$$\begin{array}{rl}{\displaystyle \int \cos 5x\sin 7x\text{d}x}& =-{\displaystyle \frac{1}{7}}\cos 5x\cos 7x-{\displaystyle \frac{5}{7}}[\underset{u}{\underbrace{\sin 5x}}\cdot \underset{v}{\underbrace{{\displaystyle \frac{1}{7}}\sin 7x}}-\int \underset{v}{\underbrace{{\displaystyle \frac{1}{7}}\sin 7x}}\cdot \underset{\text{d}u}{\underbrace{5\cos 5x)\text{d}x}}]\\ \int \cos 5x\sin 7x\text{d}x& =-{\displaystyle \frac{1}{7}}\cos 5x\cos 7x-{\displaystyle \frac{5}{49}}\sin 5x\sin 7x+{\displaystyle \frac{25}{49}}\int \cos 5x\sin 7x\text{d}x\\ {\displaystyle \frac{24}{49}}\int \cos 5x\sin 7x\text{d}x& =-{\displaystyle \frac{1}{7}}\cos 5x\cos 7x-{\displaystyle \frac{5}{49}}\sin 5x\sin 7x+K\\ \int \cos 5x\sin 7x\text{d}x& =-{\displaystyle \frac{7}{24}}\cos 5x\cos 7x-{\displaystyle \frac{5}{24}}\sin 5x\sin 7x+C\end{array}$$Jadi, berdasarkan integrasi parsial, kita telah menurunkan rumus integral trigonometri berikut.
$$\boxed{{\displaystyle {\displaystyle \int \cos 5x\sin 7x\text{d}x=-{\displaystyle \frac{7}{24}}\cos 5x\cos 7x-{\displaystyle \frac{5}{24}}\sin 5x\sin 7x+C}}}$$