Trignomometric Substitution
Expression
\(\sqrt{a^2-x^2}\)
Substitution Needed
\(\begin{aligned}x&=a\sin\theta\\dx&=a\cos\theta d\theta\end{aligned}\)
Identity Needed
\(1-\sin^2\theta=\cos^2\theta \)
Right Angle Triangle
\(\sin\theta=\frac{opp}{hyp}=\frac xa\)
\(\sqrt{a^2+x^2}\)
\(\begin{aligned}x&=a\tan\theta\\dx&=a\sec^2\theta d\theta\end{aligned}\)
\(1+\tan^2\theta=\sec^2\theta \)
\(\tan\theta=\frac{opp}{adj}=\frac xa\)
\(\sqrt{x^2-a^2}\)
\(\begin{aligned}x&=a\sec\theta\\dx&=a\sec\theta\tan\theta d\theta\end{aligned}\)
\(\sec^2\theta-1=\tan^2\theta \)
\(\sec\theta=\frac{hyp}{adj}=\frac xa\)
Example. Compute \(\int\frac{\sqrt{x^2-9}}xdx.\)
Solution. \(\mathrm{Let~}x=3\sec\theta.\text{ Then, }dx=3\sec\theta\tan\theta d\theta \)
The trigonometric identity needed is \(\sec^2\theta-1=\tan^2\theta.\)
The right angle triangle needed is:
\(\begin{aligned}
\int\frac{\sqrt{x^2-9}}xdx& =\int\frac{\sqrt{(3\sec\theta)^2-9}}{3\sec\theta}3\sec\theta\tan\theta d\theta \\
&=\int\sqrt{9\sec^2\theta-9}\tan\theta d\theta \\
&=\int\sqrt{9\tan^2\theta}\tan\theta d\theta \\
&=\int3\tan\theta\tan\theta d\theta=3\int\tan^2\theta d\theta \\
&=3\int(\sec^2\theta-1)d\theta=3(\tan\theta-\theta)+C \\
&=3{\left(\frac{\sqrt{x^2-9}}3-arcsec(x/3)\right)}+C
\end{aligned}\)
\(\text{Note that }\theta=\mathrm{arcsec}(x/3)\text{ (from the substitution equation)}.\)