Kalkulus : Integral Subtitusi

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Masih ingatkah anda dengan pelajaran Integral?? Yuuuuuuk belajar lagi…!!!! Kali ini khusus kita bahas tentang integral subtitusi, contoh soa...

Masih ingatkah anda dengan pelajaran Integral??
Yuuuuuuk belajar lagi…!!!!
Kali ini khusus kita bahas tentang integral subtitusi, contoh soal dan pembahasannya ok…!!!
Jangan sampai ketinggalan ya…

Jika suatu fungsi yang dapat didiferensialkan dan suatu bilangan rasional tak nol, maka

\int {\color{Red} (u(x))^r.\;u'(x)}\;dx={\color{Red} \frac {1}{r+1}(u(x))^{r+1}+c}   di mana c   adalah konstanta dan r\neq -1



Nah…udah lihat rumus integral yang di atas sono tuh…???
Pusing,tidak..??? hehehe…lebih baik langsung di contohin aja  ya….
contoh soal dan pembahasan integral subtitusi :
 1.   \int (5x-3)^4dx=....
Jawab :
*  kita misalkan u=5x-3  dan fungsi u dapat diturunkan menjadi
\begin{align*}{\color{Red} u}&=&{\color{Red} 5x-3}\\\frac{du}{dx}&=&5\\dx&=&{\color{Blue} \frac 15\;du} \end{align*}
*  Baru kita subtitusikan ke soal :
\begin{align*}\int({\color{Red} 5x-3})^4{\color{Blue} dx}&=&\int {\color{Red} u}^4.{\color{Blue} \frac 15\;du}\\&=&{\color{Blue} \frac 15}.\frac{1}{4+1}.{\color{Red} u}^{4+1}+C\\&=&\frac{1}{25}\;{\color{Red} u}^5+C\\&=&\frac{1}{25}\;({\color{Red} 5x-3})^5+C\end{align*}
Jangan sampai lupa untuk mengembalikan permisalan kita  u={\color{Red} 5x-3}    ya…..
2.  \int (2x-1)(3x^2-3x+5)^8\;dx=...
Jawab :
*  kita misalkan  u=3x^2-3x+5   dan fungsi u dapat diturunkan menjadi :
\begin{align*}{\color{Red} u}&=&{\color{Red} 3x^2-3x+5}\\\frac {du}{dx}&=&6x-3\\dx&=&{\color{Blue} \frac{1}{6x-3}\;du}\end{align*}
 *  Baru kita subtitusikan ke soal :
\begin{align*}\int (2x-1)(3x^2-3x+5)^8\;dx&=&\int (2x-1).{\color{Red} u}^8\;{\color{Blue} \frac{1}{6x-3}\;du}\\&=&\int \frac{2x-1}{{\color{Blue} 3(2x-1)}}\;{\color{Red} u}^8\;{\color{Blue} du}\\&=&\int \frac{1}{3}\;{\color{Red} u}^8\;{\color{Blue} du}\\&=&\frac 13.\frac{1}{8+1}.{\color{Red} u}^{8+1}+C\\&=&\frac{1}{27}.{\color{Red} u}^9 +C\\&=&\frac{1}{27}({\color{Red} 3x^2-3x+5})^9+C\end{align*}
3.   \int x^2\sqrt{2x^3+1}\;dx=...
Jawab :
*  kita misalkan u=2x^3+1  dan fungsi u dapat diturunkan menjadi
\begin{align*}{\color{Red} u}&=&{\color{Red} 2x^3+1}\\\frac{du}{dx}&=&6x^2\\dx&=&{\color{Blue} \frac{1}{6x^2}\;du} \end{align*}
*  Baru kita subtitusikan ke soal :
\begin{align*}\int x^2\sqrt{2x^3+1}\;dx&=&\int x^2.\sqrt{{\color{Red} u}}\;.{\color{Blue} \frac{1}{6x^2}\;du}\\&=&\int \frac{x^2}{{\color{Blue} 6x^2}}.{\color{Red} u}^{\frac 12}\;{\color{Blue} du}\\&=&\int \frac{1}{6}.{\color{Red} u}^{\frac 12}\;{\color{Blue} du}\\&=&\frac 16.\frac{1}{\frac 12+1}\;{\color{Red} u}^{\frac 12+1}+C\\&=&\frac 16.\frac 23\;{\color{Red} u}^{\frac 32}+C\\&=&\frac 19\;{\color{Red} u}\sqrt {\color{Red} u}+C\\&=&\frac 19({\color{Red} 2x^3+1})\sqrt{{\color{Red} 2x^3+1}}+C\end{align*}
4.   \int sin\;x.cos^2x\;dx  = …
Jawab :
* kita misalkan u=cos\;x  maka
 \begin{align*}{\color{Red} u}&=&{\color{Red} cos\;x}\\\frac{du}{dx}&=&-sin\;x\\du&=&-sin\;x\;dx \end{align*}
*sehingga :
\begin{align*}\int sin\;x.{\color{Red} cos}^2{\color{Red} x}\;dx&=&\int -{\color{Red} u}^2\;du\\&=&-\frac 13.{\color{Red} u}^3+C\\&=&-\frac 13.{\color{Red} cos}^3{\color{Red} x}+C\end{align*}
5.   \int cos\;5x\;sin^4\;5x\;dx=  …
Jawab :
* kita misalkan u=sin\;5x   maka :
\begin{align*}{\color{Red} u}&=&{\color{Red} sin\;5x}\\\frac{du}{dx}&=&5.cos\;5x\\\frac{du}{5}&=&cos\;5x\;dx\end{align*}
*sehingga :
\begin{align*}\int cos\;5x\;{\color{Red} sin}^4\;{\color{Red} 5x}\;dx&=&\int \frac 15.{\color{Red} u}^4\;du\\&=&\frac 15.\frac 15.{\color{Red} u}^5+C\\&=&\frac {1}{25}{\color{Red} sin}^5\;{\color{Red} 5x}+C\end{align*}
Wheheheehe…latihan soal dan pembahasan integral subtitusi aljabar dan trigonometrinya dicukupkan dulu yaaaaaa….kapan-kapan ditambah lagi soalnya….
Selamat belajar ………..!!!!



Sumber : www.meetmath.com

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Kalkulus : Integral Subtitusi
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