Special Integrals - Part 3

Draw the upper half of a circle of radius 3 on graph paper and shade the area under it from x = 0 to x = 2. Can you write that area as a number?

The curve is y = \sqrt{9 - x^2}, so the question is just

The substitutions from Part 1 do not work here. You could try x = 3\sin\theta, and that will eventually give an answer, but you end up integrating \cos^2\theta, which is its own little detour. The Part 2 recipe — completing the square — does not apply, because there is no square to complete; the quadratic inside the radical is already a^2 - x^2.

What you need is a direct formula for \int \sqrt{a^2 - x^2}\,dx, the same way Part 1 gave you direct formulas for \int \frac{dx}{x^2 + a^2} and its relatives. That formula exists, it is short, and once you know how it is derived you never have to re-derive it.

The three radicals

There are only three canonical square-root quadratics, and they are exactly the three from Part 1 with the reciprocal removed:

Geometrically, each of these is the area under a curve: the first is a piece of a circle, the second and third are pieces of hyperbolas. The formulas all have the same shape: a polynomial piece (always \frac{x}{2} times the original radical) plus an inverse-trig or logarithm piece.

Here are the three results up front. The rest of the article is about where they come from and how to use them.

Notice the symmetry. The first term is the same in all three: half of x multiplied by the original radical. The second term is what changes: \sin^{-1} for the circle, \ln for the two hyperbolas (with a sign flip in the last one).

Where the first formula comes from

The fastest derivation is a guess-and-solve. You will see the same move in integration by parts later — this is it in miniature.

Start by asking: what would you guess the antiderivative of \sqrt{a^2 - x^2} to look like? It should involve \sqrt{a^2 - x^2} itself, because differentiating anything else is unlikely to produce that radical. Try differentiating x\sqrt{a^2 - x^2} and see what comes out.

Put the two terms over a common denominator:

Now split that fraction:

The second piece is almost 2\sqrt{a^2 - x^2}, except with a flipped sign: \sqrt{a^2 - x^2} = \frac{a^2 - x^2}{\sqrt{a^2 - x^2}}, so \frac{2x^2}{\sqrt{a^2 - x^2}} = \frac{2a^2 - 2(a^2 - x^2)}{\sqrt{a^2 - x^2}} = \frac{2a^2}{\sqrt{a^2 - x^2}} - 2\sqrt{a^2 - x^2}. Substituting:

Now integrate both sides:

The first integral on the right is a Part 1 standard: \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(x/a) + C. Solve for the integral you want:

That is the first formula. The other two come from exactly the same recipe: differentiate x\sqrt{x^2 \pm a^2}, collect the pieces, and solve for the integral you want. The only thing that changes is the standard integral you call on at the end — \int \frac{dx}{\sqrt{x^2 + a^2}} or \int \frac{dx}{\sqrt{x^2 - a^2}}, both of which are Part 1 logarithm formulas.

The geometric meaning

You should not memorise these formulas as bare symbols. Each of them has a picture.

Check it. Plug x = 3 and x = -3 into \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}(x/3):

The formula and the half-circle area agree. That is a strong reason to trust the formula.

First worked example

Time to use the formula on the question that opened this article.

The sanity-check paragraph: the full right half of the circle (from x = 0 to x = 3) has area \frac{9\pi}{4} \approx 7.07. Your answer of 5.52 is smaller than that, as it must be, because the interval [0, 2] is a subset of [0, 3]. The picture confirms the number.

Second worked example

The second canonical formula is the \sqrt{x^2 + a^2} one. Unlike the first, this radical has no upper bound — the curve y = \sqrt{x^2 + a^2} is the upper half of a hyperbola and it grows without bound as x moves away from zero.

Notice that the antiderivative has a non-zero value at x = 0: F(0) = 0 + 2\ln(\sqrt{4}) = 2\ln 2 \approx 1.386. That is not a mistake — antiderivatives are only determined up to a constant, and when you subtract F values to compute a definite integral, any constant cancels.

The general case: \int \sqrt{ax^2 + bx + c}\,dx

Once you have the three canonical radicals, the general case follows the same recipe as Part 2. Complete the square inside the radical, shift to a new variable, and recognise which of the three forms you have.

The coefficient a can be positive or negative — the two cases produce different radicals.

Case 1: a > 0. Factor a out, complete the square:

Set t = x + \frac{b}{2a} and pull \sqrt{a} out of the radical. Two sub-cases:

• If 4ac - b^2 > 0: the inside is t^2 + k^2. Use the \sqrt{x^2 + a^2} formula.
• If 4ac - b^2 < 0: the inside is t^2 - k^2. Use the \sqrt{x^2 - a^2} formula.

Case 2: a < 0. Write a = -|a|, factor the minus out, and complete the square on what is left. The cleanest way is to just run it by hand on each specific problem. The result will be of the form k^2 - t^2, and you use the \sqrt{a^2 - x^2} formula. (If it comes out the other way around, the integrand is imaginary and the integral is not defined over the reals.)

Here is a full example of the a < 0 case. Take \int \sqrt{3 - 2x - x^2}\,dx. Complete the square inside the radical:

Set t = x + 1, so dt = dx, and

Substituting back t = x + 1:

The integrand is defined only when 3 - 2x - x^2 \geq 0, which is -3 \leq x \leq 1 — exactly the interval where the shifted radical \sqrt{4 - t^2} makes sense.

Common confusions

A few things students reliably get wrong the first time they meet these formulas.

• "The \sqrt{a^2 - x^2} formula and the \frac{1}{\sqrt{a^2 - x^2}} formula are almost the same." They share a \sin^{-1} term and they share a factor of a^2, but the polynomial piece \frac{x}{2}\sqrt{a^2 - x^2} is only in the first one. The Part 3 formulas are longer than the Part 1 formulas because the integrand has higher algebraic weight.

• "I can just use x = a\sin\theta substitution every time." You can, and it always works, but you end up integrating \cos^2\theta or \sec^3\theta, which needs its own round of techniques. The direct formula is faster and less error-prone.

• "The formula with \ln has no domain restriction." It does. The argument x + \sqrt{x^2 - a^2} is only positive for x \geq a, and the integrand \sqrt{x^2 - a^2} is only real for |x| \geq a. On x \leq -a, the formula is still valid but the log argument becomes negative, which is why the absolute-value bars are in the boxed result.

• "The middle term should be \frac{a}{2}, not \frac{a^2}{2}." Careful with units. The coefficient is a^2/2 because it is the a^2 that appears inside the radical. Getting it wrong gives an answer that differs from the correct one by a factor of a — dimensionally wrong.

Going deeper

The trig-substitution route

There is a second way to derive the \int \sqrt{a^2 - x^2}\,dx formula, and it generalises to the hyperbolic cases cleanly. Set x = a\sin\theta, so dx = a\cos\theta\,d\theta and \sqrt{a^2 - x^2} = a\cos\theta. Then

Use the double-angle identity \cos^2\theta = \frac{1 + \cos 2\theta}{2}:

Now unravel. \theta = \sin^{-1}(x/a). And \sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\frac{x}{a}\cdot\frac{\sqrt{a^2 - x^2}}{a} = \frac{2x\sqrt{a^2 - x^2}}{a^2}. Substituting:

Same answer. The guess-and-solve route and the trig-substitution route are both valid; the guess-and-solve route is shorter because it uses the Part 1 result directly, without detouring through \cos^2\theta.

The hyperbolic route

For \int \sqrt{x^2 + a^2}\,dx, the natural substitution is x = a\sinh\theta, so \sqrt{x^2 + a^2} = a\cosh\theta and dx = a\cosh\theta\,d\theta. Then

Since \theta = \sinh^{-1}(x/a) and \sinh^{-1}(u) = \ln(u + \sqrt{u^2 + 1}), this becomes

The first term simplifies: \ln(\frac{x + \sqrt{x^2 + a^2}}{a}) = \ln(x + \sqrt{x^2 + a^2}) - \ln a, and the \ln a is absorbed into the constant. Final answer:

The logarithm in the Part 3 formula is exactly \sinh^{-1}(x/a) up to an additive constant. When you see \ln(x + \sqrt{x^2 + a^2}) in an answer, you can read it as an inverse hyperbolic sine.

The connection to Part 1 and Part 2

Look at the three families side by side. Part 1 handles integrands like \frac{1}{\text{quadratic}} and \frac{1}{\sqrt{\text{quadratic}}}. Part 2 extends those to general ax^2 + bx + c by completing the square. Part 3 handles \sqrt{\text{quadratic}} — one power of the radical higher than Part 1's reciprocal-root forms.

The connection between them is direct: if you differentiate the Part 3 formulas, you land back on the Part 1 results. Specifically,

and a similar identity for the other two. The three families are a tightly linked ladder: each rung is one power of the quadratic above the previous rung.

Why the \frac{x}{2} shows up in every formula

This is worth understanding, because it is not an accident. The radical \sqrt{p(x)}, where p(x) is a quadratic, is a curve whose anti-derivative should have the units of "area." Near any point x_0, the area under the curve from 0 to x_0 should scale roughly like x_0 times the height of the curve at x_0 — which is x_0 \cdot \sqrt{p(x_0)}. The factor of \frac{1}{2} accounts for the fact that the curve is rising or falling over the interval, so on average the height is somewhere between zero and the maximum.

That is the same geometric fact that gives a triangle an area of \frac{1}{2} \cdot \text{base} \cdot \text{height}. The Part 3 formulas are the calculus echo of that half.

Where this leads next

• Special Integrals - Part 1 — the reciprocal forms that these formulas reduce to after differentiation.
• Special Integrals - Part 2 — the general ax^2 + bx + c machinery that turns any quadratic into a canonical form.
• Integration by Parts — the technique behind the "guess and solve" derivation, formalised as a general rule.
• Area Under Curve — where these integrals actually get used: computing the area bounded by circles, ellipses, and hyperbolas.
• Trigonometric Substitutions — the alternative route, when a problem is best attacked by changing to an angle variable first.