Roots like $\sqrt{9 - x^{2}}$ come from circles, and $\sqrt{x^{2} + 4}$ shows up in arc lengths. A plain substitution fails when there is no spare factor of $x$. Trigonometric substitution uses a Pythagorean identity to remove the root, then the trig integral skills finish the job. You return to $x$ with the right-triangle reasoning from compositions with inverse trig.
Why it works
If $x = a\sin\theta$ with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, then
$$\begin{aligned} a^{2} - x^{2} &= a^{2} - a^{2}\sin^{2}\theta \\ &= a^{2}\cos^{2}\theta, \quad\text{so}\quad \sqrt{a^{2} - x^{2}} \\ &= a\cos\theta, \end{aligned}$$
because $\cos\theta \ge 0$ on that range. The root is gone. The same trick works with $1 + \tan^{2}\theta = \sec^{2}\theta$ and $\sec^{2}\theta - 1 = \tan^{2}\theta$.
| The integral has | Substitute | Identity | The root becomes |
|---|---|---|---|
| $\sqrt{a^{2} - x^{2}}$ | $x = a\sin\theta$ | $1 - \sin^{2}\theta = \cos^{2}\theta$ | $a\cos\theta$ |
| $\sqrt{a^{2} + x^{2}}$ | $x = a\tan\theta$ | $1 + \tan^{2}\theta = \sec^{2}\theta$ | $a\sec\theta$ |
| $\sqrt{x^{2} - a^{2}}$ | $x = a\sec\theta$ | $\sec^{2}\theta - 1 = \tan^{2}\theta$ | $a\tan\theta$ |
Key idea
Substitute to remove the root, integrate in $\theta$, then go back to $x$. Draw a right triangle from the substitution ($\sin\theta = \frac{x}{a}$, $\tan\theta = \frac{x}{a}$ or $\sec\theta = \frac{x}{a}$) and read every other trig function off it.
Worked example: Back to x with a triangle
Find $\int \frac{dx}{x^{2}\sqrt{9 - x^{2}}}$.
- Let $x = 3\sin\theta$: $dx = 3\cos\theta\,d\theta$ and $\sqrt{9 - x^{2}} = 3\cos\theta$.
- The integral becomes $$\begin{aligned} &\int \frac{3\cos\theta\,d\theta}{9\sin^{2}\theta\cdot 3\cos\theta} \\ &= \frac{1}{9}\int \csc^{2}\theta\,d\theta \\ &= -\frac{1}{9}\cot\theta + C. \end{aligned}$$
- Triangle: $\sin\theta = \frac{x}{3}$, so the opposite side is $x$, the hypotenuse is 3 and the adjacent side is $\sqrt{9 - x^{2}}$. Then $\cot\theta = \frac{\sqrt{9 - x^{2}}}{x}$.
- Answer: $-\frac{\sqrt{9 - x^{2}}}{9x} + C$.
Answer: $-\frac{\sqrt{9 - x^{2}}}{9x} + C$
Common mistake
Stopping in $\theta$, or mixing variables. An answer like $-\frac{1}{9}\cot\theta + C$ is not an antiderivative of a function of $x$ until you rewrite it in $x$. Writing $\cot\left(\arcsin\frac{x}{3}\right)$ is correct but unfinished; the triangle turns it into $\frac{\sqrt{9 - x^{2}}}{x}$. Check: differentiate your final answer and compare with the integrand.
Worked example: A definite integral: change the limits
Evaluate $\int_0^{3}\sqrt{9 - x^{2}}\,dx$.
- Let $x = 3\sin\theta$. When $x = 0$, $\theta = 0$; when $x = 3$, $\theta = \frac{\pi}{2}$.
- The integral becomes $$\begin{aligned} &\int_0^{\pi/2} 9\cos^{2}\theta\,d\theta \\ &= 9\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2} \\ &= \frac{9\pi}{4}. \end{aligned}$$
- With new limits there is no need to go back to $x$. Check: this is a quarter of a disk of radius 3, area $\frac{1}{4}\pi\cdot 9$.
Answer: $\frac{9\pi}{4}$
Worked example: The tangent case
Find $\int \frac{dx}{\sqrt{x^{2} + 4}}$.
- Let $x = 2\tan\theta$: $dx = 2\sec^{2}\theta\,d\theta$, $\sqrt{x^{2} + 4} = 2\sec\theta$.
- The integral becomes $\int \sec\theta\,d\theta = \ln\left|\sec\theta + \tan\theta\right| + C$.
- Triangle: $\tan\theta = \frac{x}{2}$, hypotenuse $\sqrt{x^{2} + 4}$, so $\sec\theta = \frac{\sqrt{x^{2} + 4}}{2}$.
- $$\begin{aligned} &\ln\left|\frac{x + \sqrt{x^{2} + 4}}{2}\right| + C \\ &= \ln\left|x + \sqrt{x^{2} + 4}\right| + C, \end{aligned}$$ since $-\ln 2$ joins the constant.
Answer: $\ln\left|x + \sqrt{x^{2} + 4}\right| + C$