SUBSTITUSI TRIGONOMETRI
\(\sqrt{b^2 - (ax)^2} \rightarrow ax = b \sin \theta\)
\(\sqrt{(ax)^2 - b^2} \rightarrow ax = b \sec \theta\)
\(\sqrt{(ax)^2 + b^2} \rightarrow ax = b \tan \theta\)
Contoh 01
\begin{equation*} \begin{split} & \int \frac {1}{\sqrt{9 - 4x^2}} \: dx \\\\ & \int \frac {1}{\sqrt{3^2 - (2x)^2}} \: dx \end{split} \end{equation*}
\begin{equation*} \begin{split} 2x & = 3 \sin \theta \rightarrow \sin \theta = \frac {2x}{3} \\\\ 2x & = 3 \sin \theta \quad {\color {blue} \text{(diturunkan)}}\\\\ 2 \: dx & = 3 \cos \theta \: d\theta \\\\ dx & = \frac 32 \cos \theta \: d\theta \end{split} \end{equation*}
\begin{equation*} \begin{split} & \int \frac {1}{\sqrt{9 - 4x^2}} \: dx \\\\ & \int \frac {1}{\sqrt{3^2 - (2x)^2}} \: dx \\\\ & \int \frac {1}{\sqrt{3^2 - (3 \sin \theta)^2}} \:.\: \frac 32 \cos \theta \: d\theta \\\\ & \frac 32 \int \frac {1}{\sqrt{9 - 9 \sin^2 \theta}} \:.\: \cos \theta \: d\theta \\\\ & \frac 32 \int \frac {1}{\sqrt{9 (1 - \sin^2 \theta)}} \:.\: \cos \theta \: d\theta \\\\ & \frac 32 \int \frac {1}{3 \sqrt{\cos^2 \theta}} \:.\: \cos \theta \: d\theta \\\\ & \frac 12 \int \frac {1}{\cos \theta} \:.\: \cos \theta \: d\theta \\\\ &\frac 12 \int \: d\theta \\\\ & \frac 12 \theta + C \\\\ & \bbox[5px, border: 2px solid magenta] {\frac 12 \: \arcsin \left(\frac {2x}{3}\right) + c} \end{split} \end{equation*}
SOAL LATIHAN