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10 Contoh Soal Integral Subtitusi

Ngerjainpr - Setelah sebelumnya teman-teman baca artikel tentang 10 contoh soal integral tak tentu. Sekarang kita sama-sama pelajari tentang salah satu metode integral, yakni integral subtitusi. Nah untuk itu guna menambah wawasan teman-teman sekalian tentang integral, artikel ini akan membahas 10 contoh soal integral subtitusi dengan caranya. Semoga artikel ini bermanfaat ya.

10 contoh soal integral subtitusi

Contoh Soal Integral Subtitusi


Integral subtitusi adalah salah satu metode pengintegralan yang teman-teman harus ketahui, apa saja ciri2 fungsi yang bisa digunakan metode subtitusi ini, dan bagaimana cara penggunaan metode subtitusi ini ? Untuk menjawab itu semua langsung saja ke contoh soal di bawah ini ya. Cekidot..
  1. \(\int(2x-3)(x^{2}-3x)^{3}\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    u = \(x^{2}-3x\)
    du = (2x - 3) dx
    dx = \(\frac{du}{2x-3}\)
    Maka subtitusi pemisalan di atas pada soal, sehinga diperoleh :
    \(\begin{eqnarray}\int(2x-3)(x^{2}-3x)^{3}\hspace{1mm}dx&=&\int(2x-3)u^{3}\frac{du}{(2x-3)}\\&=&\int u^{3}\hspace{1mm}du\\&=&\frac{1}{4}u^{4}+C\\&=&\frac{1}{4}(x^{2}-3x)^{4}+C\end{eqnarray}\)
  2. \(\int 2x(\sqrt[3]{x^{2}-7})\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    u = \(x^{2}-7\)
    du = 2x dx
    dx = \(\frac{du}{2x}\)
    Maka :
    \(\begin{eqnarray}\int 2x(\sqrt[3]{x^{2}-7})\hspace{1mm}dx&=&\int 2x(x^{2}-7)^{\frac{1}{3}}\hspace{1mm}dx\\&=&\int 2x(u^{\frac{1}{3}})\frac{du}{2x}\\&=&\int u^{\frac{1}{3}}\hspace{1mm}du\\&=&\frac{1}{\frac{1}{3}+1}u^{\frac{1}{3}+1}+C\\&=&\frac{3}{4}u^{\frac{4}{3}}+C\\&=&\frac{3}{4}(x^{2}-7)^{\frac{4}{3}}+C\\&=&\frac{3}{4}(x^{2}-7)\sqrt[3]{x^{2}-7}+C\end{eqnarray}\)
  3. \(\int\frac{6x^{2}-4x+2}{(x^{3}-x^{2}+x-2)^{2}}\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    u = \(x^{3}-x^{2}+x-2\)
    du = (\(3x^{2}-2x+1\)) dx
    dx = \(\frac{du}{3x^{2}-2x+1}\)
    Maka :
    \(\begin{eqnarray}\int\frac{6x^{2}-4x+2}{(x^{3}-x^{2}+x-20)^{2}}\hspace{1mm}dx&=&\int(\frac{6x^{2}-4x+2}{u^{2}})\frac{du}{3x^{2}-2x+1}\\&=&\int\frac{2}{u^{2}}\hspace{1mm}du\\&=&\int 2u^{-2}\hspace{1mm}du\\&=&\frac{2}{-1}u^{-1}+C\\&=&\frac{-2}{u}+C\\&=&\frac{-2}{x^{3}-x^{2}+x-2}+C\end{eqnarray}\)
  4. \(\int\frac{x-3}{\sqrt[4]{(x^{2}-6x+10)^{3}}}\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    t = \(x^{2}-6x+10\)
    dt = (2x - 6) dx
    dx = \(\frac{dt}{2x-6}\)
    \(\begin{eqnarray}\int\frac{x-3}{\sqrt[4]{(x^{2}-6x+10)^{3}}}\hspace{1mm}dx&=&\frac{x-3}{t^{\frac{3}{4}}}\frac{dt}{2x-6}\\&=&\frac{1}{2}t^{-\frac{3}{4}}\hspace{1mm}dt\\&=&\frac{\frac{1}{2}}{-\frac{3}{4}+1}t^{-\frac{3}{4}+1}+C\\&=&2t^{\frac{1}{4}}+C\\&=&2(x^{2}-6x+10)^{\frac{1}{4}}+C\\&=&2\sqrt[4]{x^{2}-6x+10}+C\end{eqnarray}\)
  5.  \(\int p^{2}\hspace{1mm}sin(p^{3}-2)\hspace{1mm}dp=\) ...
    Jawaban :
    Misal :
    x = \(p^{3}-2\)
    dx = \(3p^{2}\) dp
    dp = \(\frac{dx}{3p^{2}}\)
    Maka :
    \(\begin{eqnarray}\int p^{2}\hspace{1mm}sin(p^{3}-2)\hspace{1mm}dp&=&\int p^{2}\hspace{1mm}sin\hspace{1mm}x\hspace{1mm}\frac{dx}{3p^{2}}\\&=&\int\frac{1}{3}sin\hspace{1mm}x\hspace{1mm}dx\\&=&-\frac{1}{3}cos\hspace{1mm}x+C\\&=&-\frac{1}{3}cos(p^{3}-2)+C\end{eqnarray}\)
  6.  \(\int sin^{3}\hspace{1mm}x\hspace{1mm}\cdot cos\hspace{1mm}x\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    p = sin x
    dp = cos x dx
    dx = \(\frac{dp}{cos\hspace{1mm}x}\)
    \(\begin{eqnarray}\int sin^{3}\hspace{1mm}x\hspace{1mm}\cdot cos\hspace{1mm}x\hspace{1mm}dx&=&\int p^{3}\hspace{1mm}cos\hspace{1mm}x\frac{dp}{cos\hspace{1mm}x}\\&=&p^{3}\hspace{1mm}dp\\&=&\frac{1}{4}p^{4}+C\\&=&\frac{1}{4}sin^{4}x+C\end{eqnarray}\)
  7. \(\int sec\hspace{1mm}\cdot tan\hspace{1mm}x\hspace{1mm}dx=\) ...
    Jawaban :
    \(sec\hspace{1mm}x\cdot tan\hspace{1mm}x=\frac{1}{cos\hspace{1mm}x}\cdot \frac{sin\hspace{1mm}x}{cos\hspace{1mm}x}=\frac{sin\hspace{1mm}x}{cos^{2}x}\)
    Misal :
    u = \(cos\hspace{1mm}x\)
    du = \(-sin\hspace{1mm}x\) dx
    dx = \(\frac{du}{-sin\hspace{1mm}x}\)
    Maka :
    \(\begin{eqnarray}\int sec\hspace{1mm}\cdot tan\hspace{1mm}x\hspace{1mm}dx&=&\int\frac{sin\hspace{1mm}x}{cos^{2}x}\hspace{1mm}dx\\&=&\int\frac{sin\hspace{1mm}x}{u^{2}}\frac{du}{-sin\hspace{1mm}x}\\&=&\int -u^{-2}\hspace{1mm}du\\&=&\frac{-1}{-1}u^{-1}+C\\&=&\frac{1}{u}+C\\&=&\frac{1}{cos\hspace{1mm}x}+C\end{eqnarray}\)
  8.  \(\int\frac{6x}{\sqrt{2+9x^{2}}}\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    u = \(2+9x^{2}\)
    du = 18x dx
    dx = \(\frac{du}{18x}\)
    Maka :
    \(\begin{eqnarray}\int\frac{6x}{\sqrt{2+9x^{2}}}\hspace{1mm}dx&=&\frac{6x}{u^{\frac{1}{2}}}\frac{du}{18x}\\&=&\frac{1}{3}u^{-\frac{1}{2}}du\\&=&\frac{\frac{1}{3}}{-\frac{1}{2}+1}u^{-\frac{1}{2}+1}+C\\&=&\frac{2}{3}u^{\frac{1}{2}}+C\\&=&\frac{2}{3}\sqrt{2+9x^{2}}+C\end{eqnarray}\)
  9. \(\int 8\hspace{1mm}sin\hspace{1mm}x\cdot cos\hspace{1mm}x\sqrt{cos^{2}x+5}\hspace{1mm}dx=\) ...
    Jawaban :
    Misal :
    u = \(cos^{2}x+5\)
    du = 2 cos x \(\cdot\) (-sin x) dx
    dx = \(\frac{du}{-2\hspace{1mm}sin\hspace{1mm}x\cdot cos\hspace{1mm}x}\)
    Maka :
    \(\begin{eqnarray}\int 8\hspace{1mm}sin\hspace{1mm}x\cdot cos\hspace{1mm}x\sqrt{cos^{2}x+5}\hspace{1mm}dx&=&\int 8\hspace{1mm}sin\hspace{1mm}x\cdot cos\hspace{1mm}x\cdot u^{\frac{1}{2}}\frac{du}{-2\hspace{1mm}sin\hspace{1mm}x\hspace{1mm}cos\hspace{1mm}x}\\&=&-4\hspace{1mm}u^{\frac{1}{2}}\hspace{1mm}du\\&=&\frac{-4}{\frac{1}{2}+1}u^{\frac{1}{2}+1}+C\\&=&-\frac{8}{3}u^{\frac{3}{2}}+C\\&=&-\frac{8}{3}(cos^{2}x+5)\sqrt{cos^{2}x+5}+C\end{eqnarray}\)
  10. \(\int \frac{1}{sin\hspace{1mm}x}\hspace{1mm}dx=\) ...
    Jawaban :
    \(\begin{eqnarray}\int\frac{1}{sin\hspace{1mm}x}&=&cosec\hspace{1mm}x\times\frac{cotan\hspace{1mm}x+cosec\hspace{1mm}x}{cotan\hspace{1mm}x+cosec\hspace{1mm}x}\\&=&\frac{cosec\hspace{1mm}x\cdot cotan\hspace{1mm}x+cosec^{2}\hspace{1mm}x}{cot\hspace{1mm}x+cosec\hspace{1mm}x}\end{eqnarray}\)
    Misal :
    u = cotan x + cosec x
    du = \(-cosec^{2}\hspace{1mm}x-cotan\hspace{1mm}x\cdot cosec\hspace{1mm}x\) dx
    dx = \(\frac{du}{-(cosec^{2}\hspace{1mm}x+cotan\hspace{1mm}x\cdot cosec\hspace{1mm}x)}\)
    Maka :
    \(\begin{eqnarray}&=&\int\frac{cosec\hspace{1mm}x\cdot cotan\hspace{1mm}x+cosec^{2}\hspace{1mm}x}{cot\hspace{1mm}x+cosec\hspace{1mm}x}\hspace{1mm}dx\\&=&\int\frac{cosec\hspace{1mm}x\cdot cotan\hspace{1mm}x+cosec^{2}\hspace{1mm}x}{u}\frac{du}{-(cosec^{2}\hspace{1mm}x+cotan\hspace{1mm}x\cdot cosec\hspace{1mm}x)}\\&=&\int\frac{-1}{u}\hspace{1mm}du\\&=&-u^{-1}\hspace{1mm}du\\&=&-ln\hspace{1mm}|u|+C\\&=&-ln\hspace{1mm}|cotan\hspace{1mm}x+cosec\hspace{1mm}x|+C\end{eqnarray}\)
Itulah 10 contoh soal integral subtitusi beserta pembahasannya. Kalian bisa cek artikel di situs ini jika ingin soal integral lainnya. Share artikel ini jika bermanfaat dan semoga bisa bermanfaat untuk teman-teman semua. 

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