Finding the Integral \(\int (4x^{2}+9) ^{1/2}dx\)
Find \(\int ( 4x^{2}+9) ^{1/2}dx\).
We use the substitution \(2x=3\tan \theta\), \(-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi}{2}.\) Then \(dx=\dfrac{3}{2}\sec ^{2}\theta d\theta\) and \[ \begin{eqnarray*} \int ( 4x^{2}+9) ^{1/2}dx &=&\dfrac{3}{2}\int \sqrt{9\tan ^{2}\theta +9}\sec ^{2}\theta\, d\theta =\dfrac{9}{2}\int \sqrt{\tan ^{2}\theta +1}\sec ^{2}\theta\, d\theta \\[5pt] &=&\dfrac{9}{2}\int \sec ^{3}\theta \,d\theta \\[5pt] &=&\dfrac{9}{2}\left[ \dfrac{1}{2}\sec \theta \tan \theta +\dfrac{1}{2}\ln \left\vert \sec \theta +\tan \theta \right\vert \right] +C \end{eqnarray*} \]
To express the solution in terms of \(x\), refer to the right triangles drawn in Figure 6.
Using the Pythagorean Theorem, the hypotenuse of each triangle is \(\sqrt{ (2x) ^{2}+9}=\sqrt{4x^{2}+9}.\) So, \[ \sec \theta =\frac{\sqrt{4x^{2}+9}}{3}\quad \hbox{and}\quad \tan \theta =\frac{2x}{3} \quad -\frac{\pi }{2} \lt \theta \lt \frac{\pi }{2} \]
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Then \[ \begin{eqnarray*} \int ( 4x^{2}+9) ^{1/2}dx &=&\frac{9}{4}\left[ \sec \theta \tan \theta +\ln \left\vert \sec \theta +\tan \theta \right\vert \right] +C \nonumber \\[2pt] &=&\frac{9}{4}\left[ \frac{\sqrt{4x^{2}+9}}{3}\cdot \frac{2x}{3}+\ln \left\vert \frac{\sqrt{4x^{2}+9}}{3}+\frac{2x}{3}\right\vert \right] +C \nonumber \\[3pt] &=&\frac{9}{4}\left[ \frac{2x\sqrt{4x^{2}+9}}{9}+\ln \frac{2x+\sqrt{4x^{2}+9}}{3}\right] +C \end{eqnarray*} \]