Trigonometric Substitution

When an integrand contains \(\sqrt{a^2-x^2}\), \(\sqrt{a^2+x^2}\), or \(\sqrt{x^2-a^2}\), swapping \(x\) for a trig function of a new angle \(\theta\) makes the radical collapse into a single trig factor — and a right triangle carries you back to \(x\) at the end.

By the end you'll be able to match each of the three radical forms to its substitution, convert \(dx\) along with the radical, and use a reference triangle to translate the answer back into \(x\).

Predict: as \(x \to a\), what happens to the \(\sqrt{a^2-x^2}\) leg and to \(\theta\)? Drag the x slider toward a to check.

This is the reference triangle for \(x = a\sin\theta\): the opposite side is x, the hypotenuse is a, so \(\sin\theta = x/a\) is built in — and the Pythagorean theorem forces the adjacent side to be \(\sqrt{a^2-x^2}\), which equals \(a\cos\theta\). Hover or tab to any side for its exact value.

θ = 30.0° · sin θ = x/a = 0.500 · √(a²−x²) = √(9.00 − 2.25) = 2.60 = a·cos θ ✓

Reference triangle for x = a sin θ — the radical is the adjacent leg
opposite = x adjacent = √(a²−x²) = a cos θ hypotenuse = a

Trigonometric substitution evaluates integrals containing \(\sqrt{a^2-x^2}\), \(\sqrt{a^2+x^2}\), or \(\sqrt{x^2-a^2}\) by substituting \(x = a\sin\theta\), \(a\tan\theta\), or \(a\sec\theta\), turning the radical into a single trig factor and requiring back-substitution via a reference triangle.

One radical, one substitution, one identity

The sign pattern inside the radical tells you everything. Each of the three forms pairs with exactly one substitution, chosen so a Pythagorean identity collapses it:

Radical form Substitution Identity used Radical becomes
\(\sqrt{a^2-x^2}\) \(x = a\sin\theta\) \(1-\sin^2\theta = \cos^2\theta\) \(a\cos\theta\)
\(\sqrt{a^2+x^2}\) \(x = a\tan\theta\) \(1+\tan^2\theta = \sec^2\theta\) \(a\sec\theta\)
\(\sqrt{x^2-a^2}\) \(x = a\sec\theta\) \(\sec^2\theta - 1 = \tan^2\theta\) \(a\tan\theta\)
Visual

Every row of that table is a right triangle. For \(x = a\sin\theta\) — the triangle in the demo above — the ratio \(\sin\theta = x/a\) fixes the opposite side at x and the hypotenuse at a, and the Pythagorean theorem hands you the third side, \(\sqrt{a^2-x^2}\). That's why the radical "collapses": it was never anything more mysterious than one leg of a right triangle. The same picture works in reverse at the end of the problem — any trig function of \(\theta\) you need is just a ratio of two labeled sides.

Algebraic

Substituting \(x = a\sin\theta\) gives \(a^2 - x^2 = a^2(1-\sin^2\theta) = a^2\cos^2\theta\), so \(\sqrt{a^2-x^2} = a\cos\theta\) (taking \(\theta\in[-\pi/2,\pi/2]\) so \(\cos\theta \ge 0\)). Two bookkeeping steps matter: convert \(dx\) too — here \(dx = a\cos\theta\,d\theta\) — and after integrating, back-substitute every trig function of \(\theta\) into an expression in \(x\). What's left in between is a pure trig integral, exactly the kind you handled in the last two lessons on powers of sine–cosine and secant–tangent.

Worked example

Evaluate \(\displaystyle\int \frac{dx}{\sqrt{9-x^2}}\). The radical matches \(\sqrt{a^2-x^2}\) with \(a = 3\), so substitute \(x = 3\sin\theta\), which gives \(dx = 3\cos\theta\,d\theta\) and \(\sqrt{9-x^2} = 3\cos\theta\). Then \[\int \frac{3\cos\theta\,d\theta}{3\cos\theta} = \int d\theta = \theta + C.\] Back-substitute: \(x = 3\sin\theta\) means \(\theta = \arcsin(x/3)\), so \[\int\frac{dx}{\sqrt{9-x^2}} = \arcsin\!\left(\frac{x}{3}\right) + C.\]

Your turn

Same pattern, new numbers: \(\displaystyle\int \frac{dx}{\sqrt{4-x^2}}\). Here \(a = 2\), so substitute \(x = 2\sin\theta\), giving \(dx = 2\cos\theta\,d\theta\) and \(\sqrt{4-x^2} = 2\cos\theta\). The integral becomes \(\int d\theta = \theta + C\). Now finish it — back-substitute to write \(\theta\) in terms of \(x\): \(\theta = \) ____

Reveal the answer

From \(x = 2\sin\theta\), \(\sin\theta = x/2\), so \(\theta = \arcsin(x/2)\) and \[\int\frac{dx}{\sqrt{4-x^2}} = \arcsin\!\left(\frac{x}{2}\right) + C.\] Set \(a = 2\) on the slider above and watch the triangle: \(\theta\) is literally the angle whose sine is \(x/2\).

Back-substitution: let the triangle do the algebra

The worked example only needed \(\theta\) itself, but most problems end with something like \(\sin\theta\) or \(\tan\theta\) when the original variable was \(x = a\tan\theta\). Don't try to simplify \(\sin(\arctan(x/a))\) symbolically — draw the triangle instead. Put the substitution's ratio into a right triangle (for \(x = a\tan\theta\): opposite \(x\), adjacent \(a\)), fill in the third side with the Pythagorean theorem (\(\sqrt{x^2+a^2}\)), and read off whatever trig function you need as a ratio of sides.

More info — why the restriction on θ makes the collapse legal

Strictly, \(\sqrt{a^2\cos^2\theta} = a\,|\cos\theta|\) — the square root of a square is an absolute value. The substitution \(x = a\sin\theta\) comes with the range \(\theta\in[-\pi/2,\pi/2]\) precisely so that \(\cos\theta \ge 0\) and the absolute value can be dropped. Each row of the table carries its own range for the same reason (\(\theta\in(-\pi/2,\pi/2)\) for tangent, and a range keeping \(\tan\theta \ge 0\) for secant). You can see it in the demo: as \(x\) sweeps from \(0\) to \(a\), \(\theta\) only ever moves from \(0°\) to \(90°\), staying where cosine is nonnegative. The OpenStax section in Dive deeper below states these ranges precisely for all three cases.

One last routing note: trig substitution is for these exact radical patterns. If the integrand is a ratio of polynomials with no radical, the next lesson's partial-fraction decomposition is the tool instead.

Check your understanding

Question 1 of 4

An integrand contains \(\sqrt{x^2-49}\). Which substitution turns the radical into a single trig factor?

Question 2 of 4

Evaluate \(\displaystyle\int \frac{dx}{\sqrt{25-x^2}}\).

Question 3 of 4

You substituted \(x = 6\tan\theta\) and your antiderivative in \(\theta\) contains \(\sin\theta\). Using a reference triangle, what is \(\sin\theta\) in terms of \(x\)?

Question 4 of 4

Substituting \(x = 3\sin\theta\) into \(\displaystyle\int \frac{x^2}{\sqrt{9-x^2}}\,dx\) produces \(9\int \sin^2\theta\,d\theta\). Which identity finishes the integration?

Recap

  • Match the radical to the row: \(\sqrt{a^2-x^2} \Rightarrow x = a\sin\theta\); \(\sqrt{a^2+x^2} \Rightarrow x = a\tan\theta\); \(\sqrt{x^2-a^2} \Rightarrow x = a\sec\theta\). Each collapses the radical to a single trig factor via a Pythagorean identity.
  • Convert \(dx\) along with the radical (e.g. \(dx = a\cos\theta\,d\theta\) for the sine substitution) — forgetting it leaves a wrong constant factor.
  • The result is a pure trig integral in \(\theta\), handled with the sine–cosine and secant–tangent techniques from the previous lessons.
  • Back-substitute with a reference triangle: build the substitution's ratio into a right triangle, fill the third side by the Pythagorean theorem, and read any trig function of \(\theta\) as a ratio of sides in \(x\).

Dive deeper

Sources

  • Trigonometric Substitution