Trig Substitutions

A trig substitution is simply a change of variable where you replace $x$ with a trig function, like $x = a\sin\theta$, so that a square root disappears. The name is literal: you substitute a trig function for $x$. It's good for integrals with $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$ or $\sqrt{x^2 - a^2}$, which show up in areas of circles, arc lengths and electric fields. All we're really saying is that such a root is secretly a side of a right triangle: in $\sqrt{a^2 - x^2}$, $a$ is the hypotenuse, $x$ is one leg, and the root is the other leg (by Pythagoras), so with $x = a\sin\theta$ the root becomes plain $a\cos\theta$. The table of three cases looks intimidating, but it's the same triangle drawn three ways; bear with it.

Try $\int \sqrt{1 - x^2}\,dx$ with the tools so far. A $u$-substitution needs an $x$ outside the root to absorb $du$, and there isn't one. Parts doesn't obviously help either. Yet this is just the area under a semicircle, so it had better have an answer.

Here's the idea. You know $1 - \sin^2\theta = \cos^2\theta$. So if $x$ were $\sin\theta$, the thing under the root would be a perfect square, and $\sqrt{\cos^2\theta}$ is just $\cos\theta$. The root is gone. Ordinary substitution names an inner piece $u$; this kind runs the other way and replaces $x$ with a trig function chosen so that the root collapses.

Why it works. Each identity in the table rewrites "one minus a square," "one plus a square" or "a square minus one" as a single square, so the root disappears. The three substitutions are chosen to match those three shapes. Geometrically, $x$, $a$ and the root are the three sides of a right triangle, and $\theta$ is one of its angles.

The recipe:

1. Pick the substitution from the table and compute $dx$. For $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$.
2. Simplify the root: $\sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{\cos^2\theta} = a\cos\theta$.
3. Integrate in $\theta$, usually with the previous lesson's tricks.
4. Convert back to $x$ with a right triangle. If $\sin\theta = x/a$, draw a triangle with opposite side $x$ and hypotenuse $a$. The adjacent side is $\sqrt{a^2 - x^2}$, and you can read every other trig function off the triangle.

For a definite integral, change the limits to $\theta$ values and skip step 4.

If the root holds a general quadratic like $\sqrt{x^2 + 4x + 13}$, complete the square first: it becomes $\sqrt{(x+2)^2 + 9}$, which is the $a^2 + u^2$ case with $u = x + 2$. There's more on that in the lesson on quadratics later in this unit.

A common mistake. At the end, it's tempting to undo something like $\sin 2\theta$ by writing $\sin(2\arcsin x)$ and stopping there. That's correct but unfinished. Use $\sin 2\theta = 2\sin\theta\cos\theta$ first, then read $\sin\theta$ and $\cos\theta$ off the triangle, so the answer comes out in plain algebra in $x$.

The electric field of a charged wire, the gravitational pull of a rod, and the length of a cable all reduce to these roots. The integral $\int \frac{dx}{(x^2 + a^2)^{3/2}}$ is exactly the field of a line charge.