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Trigonometric Substitution

Calculus 2 · Chapter 1: Review & Advanced Integration TechniquesSection 1.4 · lesson 4 of 5411 min · free

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In this lesson

Turning stubborn radicals into clean trigonometric integrals.

  • Match a radical to one of three template substitutions
  • Use a reference right triangle to return cleanly to x
  • Transform limits so definite integrals need no triangle at all
  • Complete the square to reach standard form

Transcript

Everything said in the video, section by section (1,594 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 1.4

Trigonometric Substitution

A structural engineer modeling a suspension cable. A physicist computing the electric field along the axis of a charged ring. Both of them keep running into the same shape of integral. A square root of a sum or a difference of squares. Root of a squared minus x squared. Root of x squared plus a squared. And ordinary u-substitution stalls completely on these. Why? Because no part of the integrand is the derivative of what's sitting under the root. There's nothing to absorb into d u. So the breakthrough comes from a genuinely different idea. Replace x itself with a trigonometric function of a new angle theta — chosen so the radical collapses through a Pythagorean identity.

The Problem

When u-Substitution Stalls

Let me make sure the reversal here is clear, because it catches people out. In section one point one, you substituted to remove complexity. You spotted an inner function and swapped it out. Here you're doing the opposite. You're deliberately introducing a trig function into an integral that didn't have one. That sounds like it should make things worse. It doesn't, and here's why. Trading an algebraic problem for a trigonometric one is a trade worth making — because you already know how to handle trigonometric integrals. You spent all of section one point three learning exactly that.

The Templates

Three Cases, Three Identities

There are three template cases, and each one is engineered around a Pythagorean identity. The two you need are: one minus sine squared theta equals cosine squared theta. And one plus tangent squared theta equals secant squared theta. Multiply either one by a squared, factor it under a root, and the radical vanishes entirely. Your first practical step is always identifying a. It's the positive square root of the constant term. So root of nine minus x squared has a equals three. Root of x squared plus four has a equals two. Case one. For root a squared minus x squared, substitute x equals a sine theta. Case two. For root a squared plus x squared, substitute x equals a tangent theta. Case three. For root x squared minus a squared, substitute x equals a secant theta. Now — here's a memory aid that actually sticks, because it explains the choice rather than just naming it. A difference with x squared second pairs with sine, because one minus sine squared stays under one. A sum pairs with tangent, because one plus tangent squared grows without bound. And a difference with x squared first pairs with secant, because secant squared minus one is only non-negative when secant is at least one in size. The signs tell you which function can actually live there.

Figure 1.4

Draw the Triangle, Do Not Memorise the Conversions

After you integrate in theta, you have to get back to x. And this is where most of the errors happen. So here's the technique. Don't memorize inverse-trig conversions. Draw a triangle. Take Case one, where x equals a sine theta — equivalently, sine theta is x over a. Sine is opposite over hypotenuse. So label the side opposite theta as x, and the hypotenuse as a. The Pythagorean theorem then forces the third side to be root a squared minus x squared. And now every trig function of theta can be read straight off the picture. Cosine theta is root a squared minus x squared, over a. Tangent theta is x over that same radical. The same construction works for all three cases. The single rule to remember is this. The defining substitution is always a basic ratio. Sine in Case one, tangent in Case two, secant in Case three. Place x and a on the two sides that ratio names. Let Pythagoras give you the third. Read off whatever you need. Under exam pressure, this is dramatically less error-prone than trying to recall a table of identities.

Worked Example 1

Case 1 - the Sine Substitution

Let's run Case one all the way through. Evaluate the integral of d x over root nine minus x squared. Match the form first. a squared is nine, so a is three. The pattern is root a squared minus x squared, which calls for x equals three sine theta. That gives d x equals three cosine theta d theta. And the radical becomes three cosine theta. Now substitute. And watch what happens. The three cosine theta on top and the three cosine theta on the bottom cancel completely. You're left with the integral of d theta. Which is just theta, plus C. That's the whole point of the method. A radical you couldn't touch became the easiest integral in calculus. Now back-substitute. From x equals three sine theta, sine theta is x over three, so theta is arcsine of x over three. And notice what you've just done. That's one of the standard inverse-trig antiderivatives — derived from scratch, not memorized. You can check it. Differentiate arcsine of x over three and you get one over root nine minus x squared.

Worked Example 2

Case 2 - the Tangent Substitution

Now Case two, and this one shows you why the triangle matters. Evaluate the integral of d x over the quantity x squared plus four, to the three halves. First, rewrite that power. x squared plus four to the three halves is just root x squared plus four, cubed. With a equals two, use x equals two tangent theta. Then d x is two secant squared theta d theta, and the radical is two secant theta. So the denominator becomes two secant theta, cubed — that's eight secant cubed theta. Substitute and simplify. The secant squared over secant cubed leaves one over secant, which is cosine. So you get one quarter the integral of cosine theta, which is one quarter sine theta. But you can't stop there. The answer has to be in x. So draw the Case two triangle. Tangent theta is x over two — opposite is x, adjacent is two. Pythagoras gives the hypotenuse: root x squared plus four. Read off sine theta: opposite over hypotenuse, which is x over root x squared plus four. Final answer: x over four root x squared plus four, plus C. And this exact integral is what appears when you compute the on-axis magnetic field of a current loop — the Biot-Savart law. Case two earns its keep in electromagnetism.

Definite Integrals

Transform the Limits and Skip the Triangle

Now, what changes with a definite integral? Something rather nice. You can skip the triangle entirely. The efficient route is to convert the x-limits into theta-limits at the moment you substitute, then evaluate the whole thing in theta. The rule is the same as any substitution. Each limit is an x-value, so each has to be replaced by the corresponding theta-value. Let's evaluate the integral from zero to two of root four minus x squared. With a equals two, use x equals two sine theta. Transform the limits. When x is zero, theta is zero. When x is two, sine theta is one, so theta is pi over two. The integral becomes four times the integral of cosine squared theta, from zero to pi over two. And now you use the power-reduction identity from the last section. Cosine squared is one plus cosine two theta, over two. Evaluate, and the answer is pi. Now — that answer is not an accident, and you should notice why. This integral is the area of a quarter of a circle of radius two. One quarter times pi times two squared. Which is pi. Keep that geometric check in your pocket. It's a free safeguard on any Case one problem with symmetric limits.

Widening the Net

Complete the Square to Reach Standard Form

One last extension, and it widens the net considerably. So far every radical has been a clean difference or sum of squares. But lots of integrals hide that pattern inside a general quadratic. Completing the square exposes it. The procedure: factor the leading coefficient out of the x-terms, add and subtract the square of half the linear coefficient, and regroup. Try the integral of d x over root x squared plus six x plus thirteen. Complete the square. x squared plus six x plus nine, plus four. Which is x plus three, squared, plus four. Now shift. Let u be x plus three, so d u equals d x. And look at what you have. The integral of d u over root u squared plus four. That's a clean Case two form with a equals two. Substitute, and the integral reduces to the integral of secant theta — which is log of absolute secant plus tangent. Back-substitute from the Case two triangle, and you land on log of absolute root x plus three squared plus four, plus x plus three. One detail worth noting. Dividing inside the logarithm produces a minus log two, and the constant of integration simply absorbs it. Completing the square extends this technique to any integrand built on an irreducible quadratic under a root.

Key Takeaways

What to Carry Forward

Five things to take away. Match the radical to a case. Root a squared minus x squared takes sine. Root a squared plus x squared takes tangent. Root x squared minus a squared takes secant. Each substitution uses a Pythagorean identity to make the square root disappear, leaving a trigonometric integral — exactly the kind section one point three taught you to handle. Build a reference triangle from the substitution to convert your theta-answer back into x. That's for indefinite integrals. For definite integrals, transform the limits to theta-values and evaluate in theta. No triangle required at all. And complete the square first whenever the radicand is a general quadratic, then shift to reach a standard form. Next up is section one point five, Partial Fraction Decomposition — which breaks a rational function into pieces you already know how to integrate.

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FAQ

Questions students ask us

About Trigonometric Substitution

What is trigonometric substitution in calculus?
Trigonometric substitution is an integration technique that replaces x with a trig function of a new angle \theta so that a radical like \sqrt{a^2-x^2} collapses through a Pythagorean identity. Use it when u-substitution stalls because nothing in the integrand is the derivative of the expression under the root. If the radicand is a general quadratic, such as x^2+6x+13, complete the square first to get (x+3)^2+4.
Which trig substitution do I use for each type of radical?
For \sqrt{a^2-x^2} use x=a\sin\theta, for \sqrt{a^2+x^2} use x=a\tan\theta, and for \sqrt{x^2-a^2} use x=a\sec\theta. Here a is the positive square root of the constant term, so \sqrt{9-x^2} has a=3. Each choice works because 1-\sin^2\theta=\cos^2\theta and 1+\tan^2\theta=\sec^2\theta, or equivalently \sec^2\theta-1=\tan^2\theta, turn the radicand into a perfect square.
How do you convert a trig substitution answer back to x?
Draw a reference right triangle from the substitution instead of memorizing conversion identities. For x=2\tan\theta, you have \tan\theta=x/2, so label the opposite side x and the adjacent side 2. The Pythagorean theorem gives the hypotenuse \sqrt{x^2+4}. Then read off whatever you need, such as \sin\theta=x/\sqrt{x^2+4}. This turns \frac{1}{4}\sin\theta+C into \frac{x}{4\sqrt{x^2+4}}+C.
Do you need a reference triangle for definite integrals with trig substitution?
No, because you can convert the x-limits into \theta-limits when you substitute and then evaluate entirely in \theta. For \int_0^2\sqrt{4-x^2}\,dx, let x=2\sin\theta. Then x=0 gives \theta=0, and x=2 gives \theta=\pi/2. The integral becomes 4\int_0^{\pi/2}\cos^2\theta\,d\theta, and power reduction yields \pi. This matches the area of a quarter circle of radius 2.

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