Find $\int ( 4x^{2}+9) ^{1/2}dx$.

Solution $\int (4x^{2}+9)^{1/2}dx =\int \sqrt{(2x)^{2}+3^{2}} dx$

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 $$\begin{eqnarray*} \int ( 4x^{2}+9) ^{1/2}dx &=&\dfrac{3}{2}\int \sqrt{9\tan ^{2}\theta +9}\sec ^{2}\theta\, d\theta =\dfrac{9}{2}\int \sqrt{\tan ^{2}\theta +1}\sec ^{2}\theta\, d\theta \\[5pt] &=&\dfrac{9}{2}\int \sec ^{3}\theta \,d\theta \\[5pt] &=&\dfrac{9}{2}\left[ \dfrac{1}{2}\sec \theta \tan \theta +\dfrac{1}{2}\ln \left\vert \sec \theta +\tan \theta \right\vert \right] +C \end{eqnarray*}$$

Figure 6 $\tan \theta =\dfrac{2x}{3}$, $-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi }{2}$

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, $$\sec \theta =\frac{\sqrt{4x^{2}+9}}{3}\quad \hbox{and}\quad \tan \theta =\frac{2x}{3} \quad -\frac{\pi }{2} \lt \theta \lt \frac{\pi }{2}$$

491

Then $$\begin{eqnarray*} \int ( 4x^{2}+9) ^{1/2}dx &=&\frac{9}{4}\left[ \sec \theta \tan \theta +\ln \left\vert \sec \theta +\tan \theta \right\vert \right] +C \nonumber \\[2pt] &=&\frac{9}{4}\left[ \frac{\sqrt{4x^{2}+9}}{3}\cdot \frac{2x}{3}+\ln \left\vert \frac{\sqrt{4x^{2}+9}}{3}+\frac{2x}{3}\right\vert \right] +C \nonumber \\[3pt] &=&\frac{9}{4}\left[ \frac{2x\sqrt{4x^{2}+9}}{9}+\ln \frac{2x+\sqrt{4x^{2}+9}}{3}\right] +C \end{eqnarray*}$$