Teknik Pengintegralan

A. Teknik Pengintegralan dengan Substitusi Aljabar

\LARGE\boxed{\int U^{n}.{U}'\: dx=\int U^{n}\: du=\displaystyle \frac{1}{n+1}.U^{n+1}+C}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Selesaikanlah}\\ &\textrm{a}.\quad \int 12x\left ( x^{2}+3 \right )^{5}\: dx\\ &\textrm{b}.\quad \int \left ( x^{2}+1 \right )\left ( x^{3}+3x+2 \right )^{5}\: dx\\ &\textrm{c}.\quad \int \displaystyle \frac{x^{5}}{x^{6}+a}\: dx\\ &\textrm{d}.\quad \int x\sqrt{x-1}\: dx\\ &\textrm{e}.\quad \int_{1}^{\frac{3}{2}}x\sqrt[3]{3-x}\: dx\\ &\textrm{f}.\quad \int_{0}^{\frac{\pi }{2}}\left ( \cos x.\sin ^{2}x \right )\: dx \end{array}.

Jawab:

\begin{array}{|ll|l|l|}\hline \multicolumn{4}{|l|}{1.a.\: \int 12x\left ( x^{2}+3 \right )^{5}\: dx=}\\\hline &\begin{aligned}\textrm{Misalkan}\\ u&=x^{2}+3\\ du&=2x\qquad dx \end{aligned}&\begin{aligned}\textrm{Sehingga}\\ \int 12x\left ( x^{2}+3 \right )^{5}\: dx&=\int \left ( x^{2}+3 \right )^{5}.6.2x\: dx\\ &=6\int \underset{u^{5}}{\underbrace{\left ( x^{2}+3 \right )^{5}}}.\underset{du}{\underbrace{2x\: dx}}\\ &=6\int u^{5}\: du\\ &=6.\displaystyle \frac{u^{5+1}}{5+1}+C\\ &=u^{6}+C\\ &=\left ( x^{2}+3 \right )^{6}+C\end{aligned}&\begin{aligned}&\textrm{Jadi},\\ &\int 12x\left ( x^{2}+3 \right )^{5}\: dx\\ &=\left ( x^{2}+3 \right )^{6}+C \end{aligned}\\\hline \end{array}.

\begin{array}{|ll|l|l|}\hline \multicolumn{4}{|l|}{1.b.\: \int \left ( x^{2}+1 \right )\left ( x^{3}+3x+2 \right )^{5}\: dx=}\\\hline &\begin{aligned}&\textrm{Misalkan}\\ u&=x^{3}+3x+2\\ du&=\left (3x^{2}+3 \right ) dx\\ du&=3\left ( x^{2}+1 \right )dx\\ \displaystyle \frac{1}{3}du&=\left ( x^{2}+1 \right )\: dx \end{aligned}&\begin{aligned}&\textrm{Sehingga}\\ \int &\left ( x^{2}+1 \right )\left ( x^{3}+3x+2 \right )^{5}\: dx\\&=\int \left ( x^{3}+3x+2 \right )^{5}.\left ( x^{2}+1 \right )\: dx\\ &=\int \underset{u^{5}}{\underbrace{\left ( x^{3}+3x+2 \right )^{5}}}.\underset{\frac{1}{3}du}{\underbrace{\left ( x^{2}+1 \right )\: dx}}\\ &=\displaystyle \frac{1}{3}\int u^{5}\: du\\ &=\displaystyle \frac{1}{3}.\displaystyle \frac{u^{5+1}}{5+1}+C\\ &=\displaystyle \frac{1}{18}.u^{6}+C\\ &=\displaystyle \frac{1}{18}\left ( x^{3}+3x+2 \right )^{6}+C\end{aligned}&\begin{aligned}&\textrm{Jadi},\\ &\int \left ( x^{2}+1 \right )\left ( x^{3}+3x+2 \right )^{5}\: dx\\ &=\displaystyle \frac{1}{18}.\left ( x^{3}+3x+2 \right )^{6}+C \end{aligned}\\\hline \end{array}.

\begin{array}{|ll|l|l|}\hline \multicolumn{4}{|l|}{1.c.\: \int \displaystyle \frac{x^{5}}{x^{6}+a}\: dx=}\\\hline &\begin{aligned}\textrm{Misalkan}\\ u&=x^{6}+a\\ du&=6x^{5}\quad dx\\ \displaystyle \frac{1}{6}du&=x^{5}\quad dx \end{aligned}&\begin{aligned}\textrm{Sehingga}\\ \int \displaystyle \frac{x^{5}}{x^{6}+a}\: dx&=\int \displaystyle \frac{1}{x^{6}+a}.x^{5}\: dx\\ &=\int \displaystyle \frac{1}{\underset{u}{\underbrace{x^{6}+a}}}.\underset{\frac{1}{6}du}{\underbrace{x^{5}\: dx}}\\ &=\displaystyle \frac{1}{6}\int \frac{1}{u}\: du\\ &=\displaystyle \frac{1}{6}\ln \left | u \right |+C\\ &=\displaystyle \frac{1}{6}\ln \left | x^{6}+a \right |+C\end{aligned}&\begin{aligned}&\textrm{Jadi},\\ &\int \displaystyle \frac{x^{5}}{x^{6}+a}\: dx\\ &=\displaystyle \frac{1}{6}\ln \left | x^{6}+a \right |+C \end{aligned}\\\hline \end{array}.

\begin{array}{|ll|l|l|}\hline \multicolumn{4}{|l|}{1.d.\: \int x\sqrt{x-1}\: dx=}\\\hline &\begin{aligned}&\textrm{Misalkan}\\ u&=x-1\\ du&=1\quad dx\\ du&=dx \end{aligned}&\begin{aligned}&\textrm{Sehingga}\\ &\int x\sqrt{x-1}\: dx\\&=\int \left ( x-1+1 \right )\sqrt{x-1}\: dx\\ &=\int \left ( \underset{u}{\underbrace{\left (x-1 \right )}}+1 \right )\sqrt{\underset{u}{\underbrace{x-1}}}\: dx \\ &=\int \left ( u+1 \right )\sqrt{u}\: du\\ &=\int \left (u\sqrt{u}+\sqrt{u}\: \right )\: du\\ &=\int \left (u^{\frac{3}{2}}+u^{\frac{1}{2}} \right )\: du\\ &=\displaystyle \frac{1}{\left ( \frac{3}{2}+1 \right )}u^{\left (\frac{3}{2}+1 \right )}+\displaystyle \frac{1}{\left (\frac{1}{2}+1 \right )}u^{\left (\frac{1}{2}+1 \right )}+C\\ &=\displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C\end{aligned}&\begin{aligned}&\textrm{Jadi},\\ &\int x\sqrt{x-1}\: dx\\ &=\displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C \end{aligned}\\\hline \end{array}.

B. Teknik Pengintegralan dengan Substitusi Trigonometri

Hal ini terjadi apabila fungsinya memuat bentuk

f(x)=\begin{cases} \sqrt{a^{2}-u^{2}} & \textrm{maka}\: u=a\sin \theta \: ,\textrm{sehingga}\: \sqrt{a^{2}-u^{2}}=a\cos \theta \\ \sqrt{a^{2}+u^{2}} & \textrm{maka}\: u=a\tan \theta \: ,\textrm{sehingga}\: \sqrt{a^{2}-u^{2}}=a\cos \theta \\ \sqrt{u^{2}-a^{2}} & \textrm{maka}\: u=a\sec \theta \: ,\textrm{sehingga}\: \sqrt{a^{2}-u^{2}}=a\cos \theta \end{cases}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

 \begin{array}{ll}\\ \fbox{2}.&\textrm{Selesaikanlah}\\ &\textrm{a}.\quad \displaystyle \int \displaystyle \frac{dx}{\sqrt{1-x^{2}}}\\ &\textrm{b}.\quad \displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\sqrt{1-x^{2}}}\\ &\textrm{c}.\quad \displaystyle \int \displaystyle \sqrt{9-x^{2}}\: dx\\ &\textrm{d}.\quad \displaystyle \int \displaystyle \frac{dx}{\sqrt{1+x^{2}}}\\ &\textrm{e}.\quad \displaystyle \int_{0}^{1} \displaystyle \frac{dx}{\sqrt{1+x^{2}}}\\ &\textrm{f}.\quad \displaystyle \int \displaystyle \frac{x^{2}}{\sqrt{x^{2}-9}}\: dx \end{array}.

Jawab:

\begin{array}{|l|ll|}\hline 2.a.&1-x^{2}=a^{2}-x^{2}=\begin{cases} a & =1 \\ x & = a.\sin \theta =1.\sin \theta =\sin \theta \end{cases}&\\ &&\\&\textrm{Selanjutnya}&\\ &\begin{aligned}x&=\sin \theta \\ dx&=\cos \theta \: d\theta \end{aligned}&\begin{aligned}\int \displaystyle \frac{dx}{\sqrt{1-x^{2}}}&=\int \displaystyle \frac{\cos \theta \: d\theta }{\sqrt{1-\sin ^{2}\theta }}\\ &=\int \displaystyle \frac{\cos \theta \: d\theta }{\sqrt{\cos ^{2}\theta }}\\ &=\int \displaystyle \frac{\cos \theta \: d\theta }{\cos \theta }\\ &=\int d\theta \\ &=\theta +C \end{aligned}\\\hline \end{array}.

\begin{array}{|l|ll|}\hline 2.b.&1-x^{2}=a^{2}-x^{2}=\begin{cases} a & =1 \\ x & = a.\sin \theta =1.\sin \theta =\sin \theta \end{cases}&\\ &&\\&\textrm{Selanjutnya}&\\ &\begin{aligned}x&=\sin \theta \\ dx&=\cos \theta \: d\theta\\\\ &\textbf{Untuk\: batas\: Integral}\\ &\begin{cases} x=1 & \sin \theta =1,\: \textrm{maka}\: \: \theta =\displaystyle \frac{\pi }{2} \\ x=0 & \sin \theta =0,\: \textrm{maka}\: \: \theta =0 \end{cases} \end{aligned}&\begin{aligned}\textrm{Sehingga,}&\\ \int_{0}^{1} \displaystyle \frac{dx}{\sqrt{1-x^{2}}}&=\int_{0}^{\frac{\pi }{2}} \displaystyle \frac{\cos \theta \: d\theta }{\sqrt{1-\sin ^{2}\theta }}\\ &=\int_{0}^{\frac{\pi }{2}} \displaystyle \frac{\cos \theta \: d\theta }{\sqrt{\cos ^{2}\theta }}\\ &=\int_{0}^{\frac{\pi }{2}} \displaystyle \frac{\cos \theta \: d\theta }{\cos \theta }\\ &=\int_{0}^{\frac{\pi }{2}} d\theta \\ &=\theta |_{0}^{\frac{\pi }{2}}\\ &=\displaystyle \frac{\pi }{2}-0\\ &=\displaystyle \frac{\pi }{2}\\ \textrm{Jadi},\: \: \int_{0}^{\frac{\pi }{2}}\displaystyle \frac{dx}{\sqrt{1-x^{2}}}&=\displaystyle \frac{\pi }{2} \end{aligned}\\\hline \end{array}.

\begin{array}{|l|ll|}\hline 2.c.&9-x^{2}=a^{2}-x^{2}=\begin{cases} a & =3 \\ x & = a.\sin \theta =3.\sin \theta =3\sin \theta \end{cases}&\\ &&\\&\textrm{Selanjutnya}&\\ &\begin{aligned}x&=3\sin \theta \\ dx&=3\cos \theta \: d\theta \end{aligned}&\begin{aligned}\int \displaystyle \sqrt{9-x^{2}}\: dx&=\int \sqrt{9-\left ( 9\sin ^{2}\theta \right )}.\: \underset{dx}{\underbrace{3\cos \theta \: d\theta}} \\ &=\int \sqrt{9\cos ^{2}\theta }.\: 3\cos \theta \: d\theta\\ &=\int 3\cos \theta .\: 3\cos \theta \: d\theta\\ &=9\int \cos ^{2}\theta \: d\theta =9\int \left ( \displaystyle \frac{1 +\cos 2\theta }{2} \right )\: dx\\ &=9\left ( \displaystyle \frac{1}{2}\theta +\frac{1}{2}.\frac{1}{2}\sin 2\theta \right )+C\\ &=\displaystyle \frac{9}{2}\theta +\displaystyle \frac{9}{4}\sin 2\theta +C \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \textbf{\underline{Catatan:}}&\\ \begin{aligned}x=3\sin \theta &\Rightarrow \theta =\arcsin \displaystyle \frac{x}{3}.\\ &\\ &\\ &\textbf{Perhatikan}\\ &\textbf{gambar}\\ &\textbf{di\: bawah}\\ &\textbf{ini}\\ \end{aligned}&\begin{aligned}\textrm{Sehingga}&\: \textrm{untuk\: jawaban\: poin\: 2.c\: dapat\: juga\: dinyatakan\: dengan}\\ \int \sqrt{9-x^{2}}\: dx&=\displaystyle \frac{9}{2}\arcsin \frac{x}{3}+\displaystyle \frac{9}{4}\sin 2\theta +C\\ &=\displaystyle \frac{9}{2}\arcsin \frac{x}{3}+\displaystyle \frac{9}{4}\left ( 2\sin \theta .\cos \theta \right )+C\\ &=\displaystyle \frac{9}{2}\arcsin \frac{x}{3}+\displaystyle \frac{9}{2}\sin \theta \cos \theta +C\\ &=\displaystyle \frac{9}{2}\arcsin \frac{x}{3}+\displaystyle \frac{9}{2}\left ( \frac{x}{3} \right )\left ( \frac{\sqrt{9-x^{2}}}{3} \right )+C\\ &=\displaystyle \frac{9}{2}\arcsin \frac{x}{3}+\displaystyle \frac{x}{2}\sqrt{9-x^{2}}+C \end{aligned}\\ \end{array}.

263

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

Kerjakanlah Soal

  1. Lihat soal pada poin 1 yang terdapat pada CONTOH SOAL Yang belum dibahas.
  2. Lihat soal pada poin 2 yang terdapat pada CONTOH SOAL Yang belum dibahas.

C. Integral Parsial

\LARGE\boxed{\int u\: dv=uv-\int v\: du}.

Contoh Soal 1

Perhatikanlah CONTOH SOAL 1.d di atas, soal tersebut juga dapat kita selesaikan dengan dengan cara integral parsial yaitu;

\begin{array}{|l|l|}\hline \multicolumn{2}{|l|}{\int x\sqrt{x-1}\: dx=....}\\\hline \int x\sqrt{x-1}\: dx=\int \underset{\displaystyle \underset{u}{\mid }}{x}.\underset{\displaystyle \underset{dv}{\mid }}{\underbrace{\sqrt{x-1}\: dx}}&\\ \begin{matrix} u=x &&& dv=\sqrt{x-1}\: dx\\ du=dx &&& \int dv=\int \sqrt{x-1}\: dx\\ &&&v=\int \left ( x-1 \right )^{\frac{1}{2}}\: dx\\ &&&v=\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}\\ \end{matrix}&\begin{aligned}\int x\sqrt{x-1}\: dx&=\int \underset{\displaystyle \underset{u}{\mid }}{x}.\underset{\displaystyle \underset{dv}{\mid }}{\underbrace{\sqrt{x-1}\: dx}}= u.v-\int v.du\\ &=x.\left ( \displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}} \right )-\int \left ( \displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}} \right )dx\\ &=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{2}{3}\times \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+C\\ &=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )^{\frac{5}{2}}+C \end{aligned}\\\hline \end{array}.

Sekilas hasilnya berbeda dengan penyelesaian pada CONTOH SOAL 1.d sebelumnya, tetapi jika kita sederhanakan sama, yaitu;

\begin{array}{|c|c|}\hline \textrm{Hasil akhir 1.d}&\textrm{Contoh Soal Itegral Parsial}\\\hline \displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C&\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )^{\frac{5}{2}}+C\\\hline \multicolumn{2}{|c|}{\textrm{Jika kita sejajarkan dengan ruas yang berbeda, maka}}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C&=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )^{\frac{5}{2}}+C\\ \displaystyle \frac{2}{5}\left ( x-1 \right )\left ( x-1 \right )^{\frac{3}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}&=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )\left ( x-1 \right )^{\frac{3}{2}}\\ \left ( \displaystyle \frac{2}{5}\left ( x-1 \right )+\frac{2}{3} \right )\left ( x-1 \right )^{\frac{3}{2}}&=\left ( \displaystyle \frac{2x}{3}-\frac{4}{15}\left ( x-1 \right ) \right )\left ( x-1 \right )^{\frac{3}{2}}\\ \displaystyle \frac{2x}{5}-\frac{2}{5}+\frac{2}{3}&=\displaystyle \frac{2x}{3}-\frac{4x}{15}+\frac{4}{15}\\ \displaystyle \frac{2x}{5}+\frac{4}{15}&=\displaystyle \frac{2x}{5}+\frac{4}{15}\\ \textrm{ruas kiri}\: &=\: \textrm{ruas kanan} \end{aligned}}\\\hline \end{array}.

Jadi hasil yang kita dapatkan adalah sama antara CONTOH SOAL 1.d dengan Contoh Soal pada Integral Parsial, cuma beda bentuknya saja.

Contoh Soal 2

Tentukanlah  \int x^{2}\: \cos x\: \: dx.

Jawab:

\begin{array}{l|l}\\ \int x^{2}\: \cos x\: \: dx=\int \underset{\displaystyle \underset{u}{\mid }}{x^{2}}.\underset{du}{\underbrace{\cos x\: \: dx}}&\\ \begin{matrix} u=x^{2} &&& dv=\cos x\: \: dx\\ du=2x\: dx &&& \int dv=\int \cos x\: \: dx\\ &&& v=\sin x \end{matrix}&\begin{aligned}\int x^{2}\: \cos x\: \: dx&=\int \underset{\displaystyle \underset{u}{\mid }}{x^{2}}.\underset{du}{\underbrace{\cos x\: \: dx}}=uv-\int v.du\\ &=x^{2}.\sin x-\int 2x.\: \sin x\: dx\\ &=x^{2}.\sin x-\left ( \textbf{integral parsial juga} \right )\\ &=x^{2}.\sin x-\left ( 2x\left ( -\cos x \right )-\int 2\left ( -\cos x \right )\: \: dx \right )\\ &=x^{2}.\sin x+2x\cos x-2\sin x+C \end{aligned} \end{array}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\textrm{Selesaikan soal berikut ini}\\ \begin{array}{llllll}\\ .\quad &a.&\displaystyle \int x\: \cos x\: \: dx&f.&\displaystyle \int_{0}^{\frac{\pi }{4}}x\: \sin 2x\: \: dx\\ &b.&\displaystyle \int x\: \sin x\: \: dx&g.&\displaystyle \int_{0}^{1}x^{4}\sqrt{1+x^{5}}\: dx\\ &c.&\displaystyle \int x\sqrt{x+1}\: dx&h.&\displaystyle \int_{0}^{\frac{\pi }{2}}\cos^{3} \theta \: \: d\theta \\ &d.&\displaystyle \int_{1}^{2}x\sqrt{x-1}\: dx&i.&\displaystyle \int_{\frac{\pi }{2}}^{\pi }\sin ^{3}t\: dt\\ &e.&\displaystyle \int_{0}^{\pi }x\: \sin x\: \: dx&j.&\displaystyle \int_{-1}^{1}x\left ( 3x-2 \right )^{4}\: dx \end{array}.

Sumber Referensi

  1. Kuntarti, Sulistiyono dan Sri Kurnianingsih. 2007. Matematika  SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: esis.
  2. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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