Finding an Integral Containing \(\sqrt{4-x^{2}}\)

Find \(\int \dfrac{dx}{x^{2}\sqrt{4-x^{2}}}\).

Solution The integrand contains the square root \(\sqrt{4-x^{2}}\) that is of the form \(\sqrt{a^{2}-x^{2}}\), where \(a=2\). We use the substitution \(x=2\sin \theta\), \(-\dfrac{\pi }{2} \lt\theta \lt\dfrac{\pi }{2}\). Then \(dx=2\cos \theta \,d\theta\). Since \[ \begin{eqnarray*} {\sqrt{4-x^{2}}} \underset{\underset{\color{#0066A7}{\hbox{\(x=2\sin \theta\)}}}{\color{#0066A7}{\uparrow }}} {=} {\sqrt{4-4\sin ^{2}\theta }}=2{\sqrt{1-\sin ^{2}\theta }}=2{\sqrt{\cos ^{2}\theta }} \underset{\underset{\color{#0066A7}{\hbox{\(\text {cos}~\theta \gt 0~\text {since} -\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}\)}}}{\color{#0066A7}{\uparrow}}}{=}2\cos \theta \end{eqnarray*} \]

we have \[ \int\! \frac{dx}{x^{2}\sqrt{4-x^{2}}}=\int\! \frac{2\cos \theta \,d\theta }{ (4\sin ^{2}\theta )(2\cos \theta )}=\int\! \dfrac{d\theta }{4\sin ^{2}\theta }= \frac{1}{4}\int \csc ^{2}\theta \,d\theta =-\frac{1}{4}\cot \theta +C \]

The original integral is a function of \(x\), but the solution above is a function of \(\theta\). To express \(\cot \theta\) in terms of \(x\), refer to the right triangles drawn in Figure 4.

Using the Pythagorean Theorem, the third side of each triangle is \(\sqrt{ 2^{2}-x^{2}}=\sqrt{4-x^{2}}\). So, \[ \cot \theta =\dfrac{\sqrt{4-x^{2}}}{x} \qquad -\dfrac{\pi }{2} \lt\theta \lt \dfrac{\pi }{2} \]

Then \[ \int \dfrac{dx}{x^{2}\sqrt{4-x^{2}}}=-\frac{1}{4}\cot \theta +C=-\frac{1}{4}\frac{\sqrt{4-x^{2}}}{x}+C=-\frac{\sqrt{4-x^{2}}}{4x}+C \]