Trigonometric Substitution

The area of an ellipse, the electric field around a charged wire, the length of a cable slung between two poles — each ends in an integral jammed shut by the square root of a quadratic. Trig substitution is the standard key to that lock, and it turns out to be Pythagoras in disguise.

Stare at \sqrt{a^2 - x^2} for a moment. Where have you seen "square, minus another square, under a root" before? Pythagoras. If a right triangle has hypotenuse a and one leg x, the other leg is exactly \sqrt{a^2 - x^2}. Every one of the three awkward roots — \sqrt{a^2 - x^2}, \sqrt{a^2 + x^2}, \sqrt{x^2 - a^2} — is a right triangle in disguise: it is the missing side, waiting to be found.

That matters because these roots are poison to ordinary substitution: setting u = a^2 - x^2 demands a spare factor of x in the integrand, and usually there isn't one. So we run the trick backwards. Instead of hunting for an inner function, we volunteer one: replace x by a trig function chosen so that a Pythagorean identity collapses the root into a single, root-free trig function.

Each substitution is engineered so the quadratic under the root becomes a perfect square and the root dissolves. What is left is a pure trig integral — exactly the kind you already know how to handle from trig integrals.

Three roots, three triangles

You never need to memorise the table blindly — draw the triangle and it rebuilds itself. Place the angle \theta, decide which two sides you know, and Pythagoras hands you the third — which is always the root you are trying to kill:

For an integrand containing a root of a quadratic (here a > 0):

A quadratic that isn't in one of these tidy shapes usually can be coaxed into one: \sqrt{x^2 + 4x + 13} = \sqrt{(x+2)^2 + 9} after completing the square, and the shift u = x + 2 hands you a clean \sqrt{u^2 + a^2} with a = 3.

Worked example: \displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}

The root is \sqrt{a^2 - x^2}, so we reach for x = a\sin\theta.

Step 1 — substitute x = a\sin\theta and differentiate. Then dx = a\cos\theta\, d\theta:

x = a\sin\theta, \qquad dx = a\cos\theta\, d\theta.

Step 2 — simplify the root. Factor out a^2 and apply 1 - \sin^2\theta = \cos^2\theta:

\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2(1 - \sin^2\theta)} = \sqrt{a^2\cos^2\theta} = a\cos\theta.

The root is gone — that is the whole point. (We take \theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}], where \cos\theta \ge 0, so \sqrt{\cos^2\theta} = \cos\theta with no absolute value.)

Step 3 — assemble the new integral. Put the dx from Step 1 over the simplified root from Step 2:

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

Step 4 — cancel. The a\cos\theta top and bottom annihilate, leaving the easiest integral there is:

\int \frac{a\cos\theta\, d\theta}{a\cos\theta} = \int 1\, d\theta = \theta + C.

Step 5 — back-substitute to x. From x = a\sin\theta we get \sin\theta = x/a, hence \theta = \arcsin(x/a):

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

This one finished politely — the answer was \theta itself, so one arcsine did the job. The next example is the more typical (and more famous) case, where the answer comes out tangled in \sin\theta and \cos\theta and the triangle has to untangle it.

The star example: \displaystyle\int \sqrt{a^2 - x^2}\, dx

Same root, but now it sits upstairs — this is the integrand whose graph y = \sqrt{a^2 - x^2} is the top half of a circle of radius a. Substitute x = a\sin\theta, dx = a\cos\theta\, d\theta, and use \sqrt{a^2 - x^2} = a\cos\theta from before:

\int \sqrt{a^2 - x^2}\, dx = \int (a\cos\theta)(a\cos\theta\, d\theta) = a^2 \int \cos^2\theta\, d\theta.

A \cos^2 integral is bread and butter: power-reduce with \cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta):

a^2 \int \frac{1 + \cos 2\theta}{2}\, d\theta = \frac{a^2}{2}\left(\theta + \frac{\sin 2\theta}{2}\right) + C = \frac{a^2}{2}\big(\theta + \sin\theta\cos\theta\big) + C,

where we expanded \sin 2\theta = 2\sin\theta\cos\theta — always do this before converting back, so every piece is a plain trig function of \theta. Now the reference triangle (hypotenuse a, opposite x) reads off everything: \theta = \arcsin(x/a), \sin\theta = x/a, and \cos\theta = \sqrt{a^2 - x^2}/a. Substitute all three:

\int \sqrt{a^2 - x^2}\, dx = \frac{a^2}{2}\arcsin\!\left(\frac{x}{a}\right) + \frac{x}{2}\sqrt{a^2 - x^2} + C.

The payoff. Run it across the whole semicircle, from -a to a. The \tfrac{x}{2}\sqrt{a^2 - x^2} term vanishes at both ends (the root is zero there), leaving only the arcsine term:

\int_{-a}^{a} \sqrt{a^2 - x^2}\, dx = \frac{a^2}{2}\Big[\arcsin(1) - \arcsin(-1)\Big] = \frac{a^2}{2}\left(\frac{\pi}{2} + \frac{\pi}{2}\right) = \frac{\pi a^2}{2}.

Half a circle of radius a has area \tfrac{1}{2}\pi a^2 — calculus agreeing perfectly with geometry. Double it and there it is: \pi a^2, the area of the full circle.

Pretty much, yes. At school the formula A = \pi r^2 is handed to you as a fact; the computation you just did is one of the standard ways mathematics earns it. Define \pi through the circle's circumference (or through the trig functions), write the circle as the graph y = \pm\sqrt{r^2 - x^2}, integrate — and out falls \pi r^2, with the trig substitution doing the heavy lifting. Archimedes got there 2,200 years earlier without calculus, by squeezing the circle between polygons with more and more sides — an argument that is secretly a limit, which is why he is often called the grandfather of integration. The circle even repays the favour: whenever an integrand is the top half of a circle, the known area \tfrac{1}{2}\pi a^2 is a free sanity check on your algebra.

A tangent case: \displaystyle\int \frac{dx}{(x^2 + a^2)^{3/2}}

No square root in sight? Look again: a power of \tfrac{3}{2} is a root cubed — (x^2 + a^2)^{3/2} = \left(\sqrt{x^2 + a^2}\right)^3 — so the sum of squares still calls for x = a\tan\theta. (This exact integral appears all over physics: it is the shape of the field of a charged wire and the gravity of a ring.)

Substitute. With x = a\tan\theta we get dx = a\sec^2\theta\, d\theta, and the identity 1 + \tan^2\theta = \sec^2\theta collapses the bracket:

x^2 + a^2 = a^2\tan^2\theta + a^2 = a^2\sec^2\theta, \qquad (x^2 + a^2)^{3/2} = a^3\sec^3\theta,

taking \theta \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2}) where \sec\theta > 0. The integral turns friendly:

\int \frac{a\sec^2\theta\, d\theta}{a^3\sec^3\theta} = \frac{1}{a^2}\int \frac{d\theta}{\sec\theta} = \frac{1}{a^2}\int \cos\theta\, d\theta = \frac{\sin\theta}{a^2} + C.

Back to x — draw the triangle. For the tangent case the legs are a (adjacent) and x (opposite), so the hypotenuse is \sqrt{x^2 + a^2}. Reading straight off it, \sin\theta = \dfrac{x}{\sqrt{x^2 + a^2}}, and therefore:

\int \frac{dx}{(x^2 + a^2)^{3/2}} = \frac{x}{a^2\sqrt{x^2 + a^2}} + C.

Notice what the triangle did: it converted a trig function of \theta back into pure algebra in x, with no inverse-trig gymnastics at all. That final triangle step is not optional decoration — it is the second half of the method.

Three classic ways this method goes wrong:

The identity \cosh^2 t - \sinh^2 t = 1 is Pythagoras's hyperbolic twin, and it powers a rival substitution kit: x = a\sinh t turns \sqrt{a^2 + x^2} into a\cosh t, and x = a\cosh t turns \sqrt{x^2 - a^2} into a\sinh t. Sometimes it is genuinely slicker:

\int \frac{dx}{\sqrt{x^2 + a^2}} = \int \frac{a\cosh t\, dt}{a\cosh t} = t + C = \operatorname{arsinh}\frac{x}{a} + C = \ln\!\left(x + \sqrt{x^2 + a^2}\right) + C',

with no secant integral and no absolute-value headaches (\cosh t is always positive). The same integral done with x = a\tan\theta forces you through \int \sec\theta\, d\theta — famously fiddly. UK exam boards love the hyperbolic route; US textbooks lean trig. Both are the same geometry: sine and cosine parametrise a circle, sinh and cosh parametrise a hyperbola.

The reference triangle, step by step

This is the triangle that turns \theta back into x for the sine case. From x = a\sin\theta, read \sin\theta = \tfrac{x}{a} = \tfrac{\text{opposite}}{\text{hypotenuse}}: hypotenuse a, opposite side x. Pythagoras fills in the adjacent side as \sqrt{a^2 - x^2} — and then every trig function of \theta can be read straight off the picture: \cos\theta = \tfrac{\sqrt{a^2 - x^2}}{a}, \tan\theta = \tfrac{x}{\sqrt{a^2 - x^2}}, and so on. That is exactly how the \tfrac{x}{2}\sqrt{a^2 - x^2} term re-materialised in the circle-area answer. Step through the construction:

See it explained