Trigonometric Substitution

Overview

Central question: How can a triangle encode a radical identity? You will match the form under a square root to a Pythagorean identity, choose a substitution that removes the radical, and return to the original variable without losing the domain. Trigonometric substitution is structural rewriting: the point is not to introduce an angle arbitrarily, but to make a stubborn quadratic expression equal to a square.

Inspect $\sqrt{9-x^2}$. It matches $a^2-x^2$ with $a=3$. The identity $1-\sin^2\theta=\cos^2\theta$ suggests $x=3\sin\theta$. Before calculating, predict what happens to the radicand:

$$9-x^2=9-9\sin^2\theta=9\cos^2\theta.$$

On an interval where $\cos\theta\ge0$, the root becomes $3\cos\theta$. The interval condition matters because $\sqrt{9\cos^2\theta}=3|\cos\theta|$, not automatically $3\cos\theta$.

Match the radical to the identity

The three standard patterns are

$$\sqrt{a^2-x^2}:\quad x=a\sin\theta, \qquad \sqrt{a^2+x^2}:\quad x=a\tan\theta, \qquad \sqrt{x^2-a^2}:\quad x=a\sec\theta.$$

Each is chosen so the identity $1-\sin^2\theta=\cos^2\theta$, $1+\tan^2\theta=\sec^2\theta$, or $\sec^2\theta-1=\tan^2\theta$ collapses the radical. These are patterns to justify, not a list to apply by visual resemblance alone.

Work one complete example

Evaluate

$$\int\frac{dx}{\sqrt{9-x^2}}.$$

Put $x=3\sin\theta$. Then $dx=3\cos\theta\,d\theta$, and on the principal interval $-\frac\pi2\le\theta\le\frac\pi2$ the radical is $3\cos\theta$. Thus

$$\int\frac{dx}{\sqrt{9-x^2}} =\int\frac{3\cos\theta\,d\theta}{3\cos\theta} =\int d\theta =\theta+C.$$

The substitution has removed the radical, but the answer is not yet in the original variable. From $x=3\sin\theta$, draw a right triangle with opposite side $x$ and hypotenuse $3$. Then $\theta=\arcsin\left(\frac{x}{3}\right)$, so

$$\int\frac{dx}{\sqrt{9-x^2}} =\arcsin\left(\frac{x}{3}\right)+C.$$

Contrast cases and domain checks

The integral $\int\frac{dx}{\sqrt{x^2+9}}$ does not use sine because the plus sign requires $1+\tan^2\theta=\sec^2\theta$. The integral $\int\frac{dx}{\sqrt{x^2-9}}$ requires secant because $\sec^2\theta-1=\tan^2\theta$. The sign under the radical controls the substitution.

After any trig substitution, differentiate the returned answer or use the original triangle relationship to check it. For the example above, differentiating $\arcsin(x/3)$ produces $\frac{1}{\sqrt{9-x^2}}$ on $-3<x<3$, matching both the integrand and its natural domain.

Follow the substitution through a nontrivial integral

The integral $\int\sqrt{16-x^2}\,dx$ uses the same pattern but does not cancel immediately. Put $x=4\sin\theta$, so $dx=4\cos\theta\,d\theta$. On $-\frac\pi2\le\theta\le\frac\pi2$, the radical becomes $4\cos\theta$. Thus

$$\int\sqrt{16-x^2}\,dx =16\int\cos^2\theta\,d\theta.$$

The trigonometric substitution has done its job by removing the radical; a half-angle identity finishes the remaining trig integral:

$$16\int\cos^2\theta\,d\theta =8\int\left(1+\cos2\theta\right)\,d\theta =8\theta+4\sin2\theta+C.$$

To return, use $\theta=\arcsin\left(\frac{x}{4}\right)$. The triangle also gives $\sin\theta=\frac{x}{4}$ and $\cos\theta=\frac{\sqrt{16-x^2}}{4}$, so

$$\sin2\theta=2\sin\theta\cos\theta =\frac{x\sqrt{16-x^2}}{8}.$$

The antiderivative can therefore be written as

$$4\arcsin\left(\frac{x}{4}\right) +\frac{x\sqrt{16-x^2}}{2}+C.$$

This longer example explains why a triangle is a reasoning tool rather than a decorative picture. It translates the trigonometric answer back into the original variable and keeps the radical’s positive branch visible. Differentiate the final expression to check it; the product rule terms cancel, leaving $\sqrt{16-x^2}$.

Choosing a domain deliberately

The substitution $x=4\sin\theta$ is not one-to-one on all real angles. Restricting $\theta$ to $\left[-\frac\pi2,\frac\pi2\right]$ makes sine one-to-one and makes cosine nonnegative. This is what lets the square root simplify consistently. A substitution is incomplete until its interval or other condition makes each inverse and each square root unambiguous.

For definite integrals, you may either convert the $x$-bounds to angle bounds or return to $x$ before evaluating. Do not mix an angle integrand with the old $x$-bounds. The same variable-consistency rule from ordinary substitution still applies.

Compare the three radical families

The difference between $a^2-x^2$, $a^2+x^2$, and $x^2-a^2$ is not superficial. The first describes values of $x$ constrained by $|x|\le a$ when the radical is real. It naturally produces an inverse sine when the work is returned to the original variable. The second is positive for every real $x$ and typically leads to logarithmic or inverse-hyperbolic-looking expressions after a tangent substitution. The third is real only outside the interval from $-a$ to $a$ and requires separate attention to sign when returning through a secant triangle.

Before committing to a substitution, factor out any positive constant and complete the square if needed. For example, $x^2+6x+13$ becomes $(x+3)^2+4$, which reveals an $a^2+u^2$ pattern after the shift $u=x+3$. This preliminary algebra is part of recognition. A trigonometric substitution should emerge from a clear quadratic form, not from a vague impression that a square root is present.

Stop and check

A tempting but incorrect statement is $\sqrt{9\cos^2\theta}=3\cos\theta$ for every angle. The misconception is forgetting that a square root is nonnegative. The correction is to write $3|\cos\theta|$ first, then choose and state an angle interval where the absolute value simplifies legitimately.

Retrieval check

For $\int\sqrt{16-x^2}\,dx$, state the substitution, its differential, and the identity that removes the radical. Choose $x=4\sin\theta$, so $dx=4\cos\theta\,d\theta$ and $\sqrt{16-x^2}=4\cos\theta$ on the selected interval. Explain why the new integral contains $\cos^2\theta$ rather than a remaining radical.

The next lesson returns from trigonometric identities to algebraic structure. A rational function is often not integrated as written; it is decomposed into simpler fractions whose antiderivatives are visible.