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## Trigonometric Substitutions

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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