Integral dengan Substitusi Trigonometri Part 1

Intergral bentuk $\int {\sqrt {{a^2} - {x^2}} } dx$

Untuk menyelesaikan integral di atas dapat dilakukan dengan pemisalan sebagai berikut

$x = a\sin \theta  \Rightarrow dx = a\cos \theta d\theta $

Dengan menyubtitusikan pemisalan tersebut diperoleh

$\begin{array}{*{20}{l}}{\int {\sqrt {{a^2} - {x^2}} } dx}\\{ = \int {\sqrt {{a^2} - {{\left( {a\sin \theta } \right)}^2}} } .a\cos \theta d\theta }\\{ = \int {\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } } .a\cos \theta d\theta }\\{ = \int {a\sqrt {1 - {{\sin }^2}\theta } } .a\cos \theta d\theta }\\{ = {a^2}\int {\sqrt {{{\cos }^2}\theta } } \cos \theta d\theta }\\{ = {a^2}\int {\sqrt {{{\cos }^2}\theta } } \cos \theta d\theta }\\{ = {a^2}\int {\cos \theta } \cos \theta d\theta }\\{ = {a^2}\int {{{\cos }^2}\theta } d\theta }\\{ = {a^2}\int {\frac{1}{2}\left( {\cos 2\theta  + 1} \right)} d\theta }\\{ = \frac{1}{2}{a^2}\int {\left( {\cos 2\theta  + 1} \right)} d\theta }\\{ = \frac{1}{2}{a^2}\left[ {\frac{1}{2}\sin 2\theta  + \theta } \right] + C}\\{ = \frac{1}{4}{a^2}\sin 2\theta  + \frac{1}{2}{a^2}\theta  + C}\\{ = \frac{1}{4}{a^2}.2\sin \theta \cos \theta  + \frac{1}{2}{a^2}\theta  + C}\\{ = \frac{1}{2}{a^2}\sin \theta \sqrt {1 - {{\sin }^2}\theta }  + \frac{1}{2}{a^2}\theta  + C}\end{array}$

Dari sini kembalikan variabel ke $x$

$\begin{array}{l}
 x = a\sin \theta  \Rightarrow \sin \theta  = \frac{x}{a} \\
 \sin \theta  = \frac{x}{a} \Rightarrow \theta  = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right) \\
 \end{array}$

Diperoleh

$\begin{array}{l}
 \frac{1}{2}{a^2}\sin \theta \sqrt {1 - {{\sin }^2}\theta }  + \frac{1}{2}{a^2}\theta  + C \\
  = \frac{1}{2}{a^2}.\frac{x}{a}\sqrt {1 - {{\left( {\frac{x}{a}} \right)}^2}}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \\
  = \frac{1}{2}ax\sqrt {1 - {{\left( {\frac{x}{a}} \right)}^2}}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \\
  = \frac{1}{2}x\sqrt {{a^2}\left( {1 - {{\left( {\frac{x}{a}} \right)}^2}} \right)}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \\
  = \frac{1}{2}x\sqrt {{a^2}\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \\
 \int {\sqrt {{a^2} - {x^2}} } dx = \frac{1}{2}x\sqrt {{a^2} - {x^2}}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \\
 \end{array}$

KESIMPULAN

\[\int {\sqrt {{a^2} - {x^2}} } dx = \frac{1}{2}x\sqrt {{a^2} - {x^2}}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C\]

CONTOH SOAL

Tentukan hasil dari

$\int {\sqrt {4 - {x^2}} } dx$

Jawab: 

$\begin{array}{l}
 \int {\sqrt {{a^2} - {x^2}} } dx = \frac{1}{2}x\sqrt {{a^2} - {x^2}}  + \frac{1}{2}{a^2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C \\
 \int {\sqrt {4 - {x^2}} } dx = \int {\sqrt {{2^2} - {x^2}} } dx \\
  = \frac{1}{2}x\sqrt {{2^2} - {x^2}}  + \frac{1}{2}{2^2}{\sin ^{ - 1}}\left( {\frac{x}{2}} \right) + C \\
  = \frac{1}{2}x\sqrt {4 - {x^2}}  + \frac{1}{2}.4{\sin ^{ - 1}}\left( {\frac{x}{2}} \right) + C \\
 \int {\sqrt {4 - {x^2}} } dx = \frac{1}{2}x\sqrt {4 - {x^2}}  + 2{\sin ^{ - 1}}\left( {\frac{x}{2}} \right) + C \\
 \end{array}$

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