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\begin{center}\textbf{Integration by Trigonometric Substitution}\end{center}
\Large Consider the following integral $\int \sqrt{a^2 - x^2}\; dx$ \nolinebreak where $a > 0$ . If the integral were $\int x \sqrt{a^2 - x^2} \;dx$, the substitution $u = a^2 - x^2$ would be effective. However, as it stands, $\int \sqrt{a^2 - x^2} \;dx$ is more difficult. If we make a change of variables where we use the substitution $x = a\sin\theta$, then the identity $\sin^2\theta + \cos^2\theta = 1$ would allow us to get rid of the root sign.\\\\
Observe\\\\
$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2 \theta}$ $= \sqrt{a^2(1 - \sin^2\theta)} =$ $\sqrt{a^2\cos^2\theta} = a|\cos\theta|$\\\\
Notice how the substitution $u = a^2 - x^2$ would not have gotten you anywhere. So in general, we can make a substitution of the form $x = h(u)$ by using the reverse chain rule (remember this from Calculus I?!). To simplify our calculations, we assume \textit{h} has an inverse function (and hence it must not be even and is one-to-one).\\\\
So we have\\\\
$\int \textit{f}(x)\;dx$ = $\int \textit{f}(\textit{g}(u))\textit{g}'(u)\;du$\\\\
It should noted that when we made the substitution $x = a\sin\theta$, we restricted the values of $\theta$ in $\left [ \frac{-\pi}{2}, \frac{\pi}{2}\right ]$
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Here are some general strategies for tackling on similar cases\\\\
Case \# 1 $\sqrt{a^2 - x^2}$\\
Use $x = a\sin\theta$ or $x = a\cos\theta$ and manipulate the identity $\sin^2\theta + \cos^2\theta =1$\\\\
Case \# 2 $\sqrt{a^2 + x^2}$\\
Use $x = a\tan\theta$ and manipulate the identity $\tan^2\theta + 1 = \sec^2\theta$\\\\
Case \#3 $\sqrt{x^2 - a^2}$\\
Use $x = a\sec\theta$ and manipulate the identity $\tan^2\theta + 1 = \sec^2\theta$\\\\
Notice that even though there are two different substitutions in the first case, either one will get you to the correct answer. This then brings us to a conclusion.\\\\
\begin{center}\textbf{There is no such thing as unique substitutions}\end{center}
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