In this section we will learn to evaluate integrals involving one of the following expressions: \[ \sqrt{x^2+a^2},\sqrt{x^2-a^2},\sqrt{a^2-x^2} \] We use the following trigonometric substitutions: \[ \begin{array}{|lll|} \hline \sqrt{a^2-x^2}=a\cos \theta & \text{ for } x=a\sin \theta & \text{ where } dx=a\cos \theta \;d\theta\\ \hline \sqrt{x^2-a^2}=a\tan \theta & \text{ for } x=a\sec \theta & \text{ where } dx=a\sec \theta \tan \theta \;d\theta\\ \hline \sqrt{x^2+a^2}=a\sec \theta & \text{ for } x=a\tan \theta & \text{ where } dx=a\sec^2 \theta \;d\theta\\ \hline \end{array} \]

Example. Show that \(\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left( \frac{x}{a}\right)+C,\;a >0\).
Solution. Let \(x=a\sin \theta\). Then \(dx=a\cos \theta \;d\theta\) and \[\frac{1}{\sqrt{a^2-x^2}} =\frac{1}{\sqrt{a^2-a^2\sin^2 \theta}} =\frac{1}{\sqrt{a^2(1-\sin^2 \theta})} =\frac{1}{\sqrt{a^2\cos^2 \theta}} =\frac{1}{a\cos \theta}.\] \begin{align*} \int \frac{dx}{\sqrt{a^2-x^2}} &= \int \frac{1}{a\cos \theta} \;a\cos \theta\;d\theta\\ &= \int \;d\theta\\ &= \theta +C\\ &= \sin^{-1}\left( \frac{x}{a}\right) +C. \end{align*}

Example. Evaluate \(\int \sqrt{4-x^2} \,dx\).
Solution. Let \(x=2\sin \theta\). Then \(dx=2\cos \theta \;d\theta\) and \[\sqrt{4-x^2}=\sqrt{4-4\sin^2 \theta}=\sqrt{4(1-\sin^2 \theta)}=\sqrt{4\cos^2 \theta}=2\cos \theta.\] \begin{align*} \int \sqrt{4-x^2} \,dx &= \int 2\cos \theta \;2\cos \theta\;d\theta\\ &= 2\int 2\cos^2 \theta \;d\theta\\ &= 2\int \left(1+\cos(2\theta) \right) d\theta\\ &= 2 \left(\theta+\frac{\sin(2\theta)}{2} \right) +C\\ &= 2\theta+2\sin\theta \cos \theta +C \end{align*} Now we convert the answer to an expression in terms of \(x\). Since \(\sin \theta=\frac{x}{2}\), \(\theta=\sin^{-1}\left( \frac{x}{2}\right)\). To get \(\cos \theta\), draw a right triangle with an angle \(\theta\) where \(\sin \theta=\frac{x}{2}\). Since \(\sin \theta\) is the ratio of the lengths of the opposite side and the hypotenuse, we label the opposite side and the hypotenuse with \(x\) and \(2\) respectively. Then by the Pythagorean Theorem, the length of the adjacent side is \(\sqrt{4-x^2}\). Consequently \(\cos \theta=\frac{\sqrt{4-x^2}}{2}\).

\[\begin{align*} \int \sqrt{4-x^2} \,dx &=2\theta+2\sin\theta \cos \theta +C\\ &=2\sin^{-1}\left( \frac{x}{2}\right)+2\frac{x}{2} \frac{\sqrt{4-x^2}}{2} +C\\ &=2\sin^{-1}\left( \frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{2} +C. \end{align*}\]

Example. Evaluate \(\displaystyle\int\frac{dx}{\sqrt{x^2-4}}\).
Solution. Let \(x=2\sec \theta\). Then \(dx=2\sec \theta \tan \theta\;d\theta\) and \[ \frac{1}{\sqrt{x^2-4}}=\frac{1}{\sqrt{4\sec^2\theta-4}}=\frac{1}{\sqrt{4\tan^2\theta}}=\frac{1}{2\tan\theta}.\] \[\begin{align*} \int\frac{dx}{\sqrt{x^2-4}} &= \int \frac{1}{2\tan\theta} 2\sec \theta \tan \theta\;d\theta\\ &= \int \sec \theta \;d\theta\\ &= \ln| \sec \theta +\tan\theta|+C'\\ &= \ln\left\vert \frac{x}{2}+\frac{\sqrt{x^2-4}}{2} \right\vert+C'\\ &= \ln\left\vert \frac{x+\sqrt{x^2-4}}{2} \right\vert+C'\\ &= \ln\left\vert x+\sqrt{x^2-4} \right\vert-\ln 2+C'\\ &= \ln\left\vert x+\sqrt{x^2-4} \right\vert +C \,\, \left(\text{where } C=C'-\ln 2 \right). \end{align*}\]

Example. Evaluate \(\int \sqrt{x^2+4} \,dx\).
Solution. Let \(x=2\tan \theta\). Then \(dx=2\sec^2 \theta \;d\theta\) and \[\sqrt{x^2+4}=\sqrt{4\tan^2 \theta+4}=\sqrt{4\sec^2 \theta}=2\sec \theta.\] \[\begin{align*} \int \sqrt{x^2+4} \,dx &= \int 2\sec \theta \;2\sec^2 \theta \;d\theta\\ &= 4\int \sec^3 \theta \;d\theta\\ &= 4\cdot \frac{1}{2}\left(\sec \theta \tan \theta+ \ln|\sec \theta+\tan \theta| \right)+C'\\ &= 2\left(\frac{\sqrt{x^2+4}}{2} \frac{x}{2}+ \ln\left\vert\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right\vert \right)+C'\\ &= \frac{x\sqrt{x^2+4}}{2}+ 2\ln\left\vert\frac{x+\sqrt{x^2+4}}{2} \right\vert +C'\\ &=\frac{x\sqrt{x^2+4}}{2}+ 2\ln\left\vert x+\sqrt{x^2+4} \right\vert -2\ln 2+C'\\ &=\frac{x\sqrt{x^2+4}}{2}+ 2\ln\left\vert x+\sqrt{x^2+4} \right\vert+C \,\, \left(\text{where } C=C'-2\ln 2 \right). \end{align*}\]

Note that if the integrand involves \(\sqrt{ax^2+bx+c}\), then we complete the square under the square root and apply the previous techniques of trigonometric substitutions. Recall: \[ax^2+bx+c=a\left[ \left(x+\frac{b}{2a} \right)^2+\frac{4ac-b^2}{4a^2} \right]\]

Example. Evaluate \(\displaystyle\int\frac{dx}{\sqrt{x^2-6x+13}}\).
Solution. First we complete the square: \(x^2-6x+13=(x-3)^2+2^2\). Let \(x-3=2\tan \theta\). Then \(dx=2\sec^2 \theta \;d\theta\) and \[\sqrt{x^2-6x+13}=\sqrt{(x-3)^2+4}=\sqrt{4\tan^2 \theta+4}=\sqrt{4\sec^2 \theta}=2\sec \theta.\] \[\begin{align*} \int\frac{dx}{\sqrt{x^2-6x+13}} &=\int\frac{2\sec^2 \theta \;d\theta}{2\sec \theta}\\ &=\int \sec \theta \;d\theta\\ &= \ln| \sec \theta +\tan\theta|+C'\\ &= \ln\left\vert \frac{x-3}{2}+\frac{\sqrt{(x-3)^2+4}}{2} \right\vert+C'\\ &= \ln\left\vert \frac{x-3+\sqrt{x^2-6x+13}}{2} \right\vert+C'\\ &= \ln\left\vert x-3+\sqrt{x^2-6x+13} \right\vert-\ln 2+C'\\ &= \ln\left\vert x-3+\sqrt{x^2-6x+13} \right\vert +C \,\, \left(\text{where } C=C'-\ln 2 \right). \end{align*}\]

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