Find:

Solution (a) $\int \dfrac{dx}{\sqrt{4-x^{2}}}$ resembles $\int \dfrac{1}{\sqrt{1-x^{2}}}\,dx=$ $\sin ^{-1}x+C.$ We begin by rewriting the integrand as $$\dfrac{1}{\sqrt{4-x^{2}}}=\dfrac{1}{\sqrt{4\left( 1-\dfrac{x^{2}}{4}\right) }}=\dfrac{1}{2\sqrt{1-\left( \dfrac{x}{2}\right) ^{2}}}$$

Now we let $u=\dfrac{x}{2}$. Then $du=\dfrac{dx}{2}$, so $dx=2\,du.$ $$\begin{eqnarray*} \int \dfrac{dx}{\sqrt{4-x^{2}}}&=&\int \dfrac{dx}{2\sqrt{1-\left( \dfrac{x}{2}\right) ^{2}}} \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(dx=2du\)}}}}{\color{#0066A7}{\hbox{\(u=\tfrac{x}{2}\)}}}}}{\color{#0066A7}{\uparrow}}} = \int \dfrac{2 du}{2\sqrt{1-u^{2}}}=\int \dfrac{du}{\sqrt{1-u^{2}}}=\sin ^{-1}u+C \\ &=&\sin ^{-1}\left( \dfrac{x}{2}\right) +C \end{eqnarray*}$$

(b) $\int \dfrac{dx}{9+4x^{2}}$ resembles $\int \dfrac{1}{1+x^{2}}dx=\tan ^{-1}x+C$. We rewrite the integrand as $$\dfrac{1}{9+4x^{2}}=\dfrac{1}{9\left( 1+\dfrac{4x^{2}}{9}\right) }=\dfrac{1}{9\left[ 1+\left( \dfrac{2x}{3}\right) ^{2}\right] }$$

Now let $u=\dfrac{2x}{3}.$ Then $du=\dfrac{2}{3}dx$, so $dx=\dfrac{3}{2} du$. $$\begin{eqnarray*} \int \dfrac{dx}{9+4x^{2}}&=&\int \dfrac{dx}{9\left[ 1+\left( \dfrac{2x}{3} \right) ^{2}\right] }=\int \dfrac{\dfrac{3}{2}du}{9( 1+u^{2}) }=\dfrac{1}{6}\int \dfrac{du}{1+u^{2}} \\ &=& \dfrac{1}{6}\tan ^{-1}u+C=\dfrac{1}{6}\tan ^{-1}\left( \dfrac{2x}{3}\right) +C \end{eqnarray*}$$