Trigonometric substitution

When you reach for it

u-substitution works when the derivative of an inside expression sits in the integrand as a factor. In \displaystyle \int \sqrt{49 - x^2}\,dx the inside is 49 - x^2, its derivative -2x is nowhere to be seen, and no choice of u helps. Yet roots of this kind are everywhere: \displaystyle \sqrt{49 - x^2} is a half circle, the area of an ellipse and the length of a parabola’s arc both lead to such roots, and partial fractions leaves behind powers of x^2 + a^2 that call for the same move. Riemann sums can only approximate the area under a half circle; the exact value needs an antiderivative of the root.

The idea is to rename x itself as a trig function of a new angle \theta, chosen so that a Pythagorean identity turns the expression under the root into a perfect square. Reach for it when an integrand holds \displaystyle \sqrt{a^2 - x^2}, \displaystyle \sqrt{a^2 + x^2} or \displaystyle \sqrt{x^2 - a^2} with a > 0, a power of a^2 \pm x^2 or x^2 - a^2, or any quadratic that completing the square brings to one of these shapes, and a plain substitution does not already do the job.

The procedure

Day 1: substituting for x itself

Ordinary substitution names a new variable as a function of x. Here x is named as a function of the new variable, x = g(\theta), and the question is why that is allowed.

Theorem (inverse substitution). Let g have a continuous derivative on an interval J with g'(\theta) \ne 0 inside J, so that g is one-to-one there and has an inverse \theta = g^{-1}(x) on the interval I of values g takes. If G'(\theta) = f\big(g(\theta)\big)\,g'(\theta) on J, then G\big(g^{-1}(x)\big) is an antiderivative of f on I.

The proof is the chain rule together with the derivative of an inverse function, \displaystyle \frac{d\theta}{dx} = \frac{1}{g'(\theta)}, which is how Derivatives of inverse trigonometric functions found the derivative of \arcsin x:

In practice: write x = g(\theta) and dx = g'(\theta)\,d\theta, integrate in \theta, and return to x through \theta = g^{-1}(x). The theorem also says why g must be one-to-one: without an inverse there is no way back to x.

The three choices. Each shape is matched to an identity from the Pythagorean identity and an interval on which the substitution is one-to-one.

The identities used are 1 - \sin^2\theta = \cos^2\theta, 1 + \tan^2\theta = \sec^2\theta and \sec^2\theta - 1 = \tan^2\theta. The intervals for \theta are \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} for the sine, \displaystyle -\frac{\pi}{2} < \theta < \frac{\pi}{2} for the tangent, and \displaystyle 0 \le \theta < \frac{\pi}{2} or \displaystyle \frac{\pi}{2} < \theta \le \pi for the secant.

Take the first row. On \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] the sine rises from -1 to 1, so each x in [-a, a] comes from exactly one \theta, namely \displaystyle \theta = \arcsin\frac{x}{a}. The root becomes

and the last step holds because \cos\theta \ge 0 on that interval. The square root of a square is an absolute value; the interval is chosen so the bars can be dropped. In the second row the tangent takes every real value once on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), so \displaystyle \theta = \arctan\frac{x}{a}, and \sec\theta > 0 there. At \displaystyle \theta = \pm\frac{\pi}{2} the sine row has g'(\theta) = a\cos\theta = 0; the theorem asks for g' \ne 0 only inside the interval, so the answer holds on (-a, a) and extends to the endpoints by continuity where the integrand allows.

Getting back to x. After integrating, the answer is in \theta and must be rewritten in x. For x = a\sin\theta, draw a right triangle with angle \theta, hypotenuse a and opposite side x, since \displaystyle \sin\theta = \frac{x}{a}. The Pythagorean theorem gives the adjacent side \displaystyle \sqrt{a^2 - x^2}, and then \displaystyle \cos\theta = \frac{\sqrt{a^2 - x^2}}{a} and \displaystyle \tan\theta = \frac{x}{\sqrt{a^2 - x^2}} can be read off (Figure 1). A triangle needs \displaystyle 0 < \theta < \frac{\pi}{2}, but the ratios hold on the whole interval, because they also follow from algebra: \displaystyle \cos\theta = \sqrt{1 - \sin^2\theta} there, since \cos\theta \ge 0. The triangle is a memory aid for identities that are true throughout.

Definite integrals. As with u-substitution, the limits can move with the variable: each x-limit becomes \theta = g^{-1}(x), and the \theta-integral is evaluated with no return to x (Example 3).

Day 2: the secant case and shifted squares

Two pieces. The root \displaystyle \sqrt{x^2 - a^2} is defined for x \ge a and for x \le -a, two separate intervals. The secant is one-to-one on \displaystyle \left[0, \frac{\pi}{2}\right), giving every x \ge a, and on \displaystyle \left(\frac{\pi}{2}, \pi\right], giving every x \le -a. In both, \displaystyle \theta = \arccos\frac{a}{x}. The root is a\lvert\tan\theta\rvert, which is a\tan\theta on the first piece and -a\tan\theta on the second, where the tangent is negative. Each piece is worked separately (Example 4).

Shifted squares. A quadratic such as x^2 - 4x + 13 is not yet one of the three shapes. Completing the square writes it as (x - h)^2 + k, and the substitution is made for x - h instead of x. Since x - h is x moved by a constant, dx is unchanged. A leading coefficient other than 1 is handled the same way: in \displaystyle \sqrt{4x^2 - 9} the square is (2x)^2, so 2x = 3\sec\theta.

Powers with no root. (x^2 + a^2)^2 has no root, but x = a\tan\theta still turns it into a^4\sec^4\theta, and the integral becomes a power of cosine (Example 6). This is the form partial fractions leaves behind for a repeated quadratic factor.

First look for something simpler. In \displaystyle \int \frac{x}{\sqrt{x^2 + 16}}\,dx the factor x is half the derivative of x^2 + 16, and a plain substitution finishes in one line. The trig substitution would work too, but it is the longer road.

Worked examples

Boundaries

• The substitution must be one-to-one on the interval used, and the sign of the root must be read on that interval. The table’s intervals make the root come out positive. If \theta ran over \displaystyle \left[-\frac{\pi}{2}, \pi\right] in the sine row, each x in (0, a) would come from two angles, and on \displaystyle \frac{\pi}{2} < \theta < \pi the root would be -a\cos\theta.
• \displaystyle \sqrt{x^2 - a^2} = a\lvert\tan\theta\rvert, and for x \le -a this is -a\tan\theta. An antiderivative found on x \ge a need not hold on x \le -a (Example 4).
• Each answer holds only on an interval where the integrand is defined: the formula for \displaystyle \int \frac{dx}{x^2\sqrt{x^2 + 9}} holds on x < 0 and on x > 0 separately, with constants that need not match, and nothing involving \displaystyle \sqrt{a^2 - x^2} holds outside [-a, a].
• A definite integral whose integrand is unbounded at an endpoint, such as \displaystyle \int_0^5 \frac{dx}{\sqrt{25 - x^2}}, is improper and needs a limit, as in Improper integrals with an infinite integrand.
• The method needs a sum or difference of two squares, one of them constant, after completing the square. It does nothing for roots such as \displaystyle \sqrt{1 + x^3}.

Where this goes next

The \theta-integrals that appear are the work of Trigonometric integrals, and \int \sec^3\theta\,d\theta, which arrives from \displaystyle \int \sqrt{x^2 + a^2}\,dx, comes from integration by parts. Partial fractions reduces a rational function to pieces, and the piece \displaystyle \frac{1}{(x^2 + a^2)^n} is a tangent substitution. The Arc length produces \displaystyle \sqrt{1 + f'(x)^2}: for the parabola y = x^2 that is \displaystyle \sqrt{1 + 4x^2}, a tangent substitution with 2x = \tan\theta. Choosing an integration technique gathers all of these into one way of reading an integrand.

Exercises

Day 1: sine and tangent

1. Find \displaystyle \int \dfrac{dx}{x^2\sqrt{16 - x^2}}.
2. Find \displaystyle \int \sqrt{9 - x^2}\,dx.
3. Find \displaystyle \int \dfrac{x^3}{\sqrt{4 - x^2}}\,dx.
4. Find \int \dfrac{dx}{\left(x^2 + 16\right)^{3/2}}.
5. Find \displaystyle \int \dfrac{dx}{\sqrt{x^2 + 25}}.
6. Find \displaystyle \int \sqrt{x^2 + 4}\,dx. (Use \displaystyle \int \sec^3\theta\,d\theta = \frac{1}{2}\sec\theta\tan\theta \displaystyle + \frac{1}{2}\ln\lvert \sec\theta + \tan\theta \rvert + C.)

Day 1: definite integrals and the window

7. Evaluate \int_0^{3/2} \dfrac{dx}{\left(9 - x^2\right)^{3/2}}.
8. Tahina cuts Fara’s window (Example 3) along the center line x = 0 and the line \displaystyle x = \frac{7}{2}. Find the area of glass between the two lines.
9. Find the area of the part of Fara’s window that lies above the line y = 2.

Day 1: reasoning

10. Explain why x = 5\sin\theta is used with \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} and not with \displaystyle \frac{\pi}{2} \le \theta \le \frac{3\pi}{2}, which also gives every x in [-5, 5].
11. Fara plans a tangent substitution for \displaystyle \int \dfrac{x}{\sqrt{x^2 + 16}}\,dx. Find the integral more quickly, and say what in the integrand signals the shorter route.
12. For x = a\tan\theta with \displaystyle -\frac{\pi}{2} < \theta < 0, no triangle has a negative side. Show that \displaystyle \sin\theta = \frac{x}{\sqrt{x^2 + a^2}} and \displaystyle \cos\theta = \frac{a}{\sqrt{x^2 + a^2}} still hold.

Day 2: the secant substitution

13. Find \displaystyle \int \dfrac{dx}{x^2\sqrt{x^2 - 9}}, and show the answer holds for x > 3 and for x < -3.
14. Find \displaystyle \int \dfrac{dx}{\sqrt{x^2 - 16}} for x > 4.
15. Evaluate \displaystyle \int_{-6}^{-2\sqrt{3}} \dfrac{dx}{x^2\sqrt{x^2 - 9}}, changing the limits to \theta.

Day 2: completing the square

16. Find \displaystyle \int \dfrac{dx}{\sqrt{x^2 + 6x + 13}}.
17. Find \displaystyle \int \sqrt{6x - x^2}\,dx.
18. Find \int \dfrac{dx}{\left(x^2 + 2x + 2\right)^2}.

Day 2: definite integrals

19. Evaluate \displaystyle \int_0^2 x^2\sqrt{4 - x^2}\,dx.
20. Evaluate \int_0^1 \dfrac{dx}{\left(1 + x^2\right)^{3/2}}.

Day 2: reasoning

21. Rivo says \displaystyle \ln\lvert x + \sqrt{x^2 - 16} \rvert is an antiderivative of \displaystyle \frac{1}{\sqrt{x^2 - 16}} on (-4, 4) as well. Explain why that cannot be.
22. Name the substitution for each and give the form it reaches: \displaystyle \sqrt{4x^2 - 9}; \left(x^2 + 2x + 5\right)^{-3/2}; \displaystyle \sqrt{3 - 2x - x^2}.
23. Use trigonometric substitutions to show that \displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C on (-a, a) and \displaystyle \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C, for a > 0.