In Section 5.2 we defined the definite integral as the “signed area under the curve.” In that section we had not yet learned the Fundamental Theorem of Calculus, so we only evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate
as we recognized that \(f(x) = \sqrt{9-x^2}\) described the upper half of a circle with radius 3.
We have since learned a number of integration techniques, including Substitution and Integration by Parts, yet we are still unable to evaluate the above integral without resorting to a geometric interpretation. This section introduces Trigonometric Substitution, a method of integration that fills this gap in our integration skill. This technique works on the same principle as Substitution as found in Section 6.1, though it can feel “backward.” In Section 6.1, we set \(u=f(x)\text{,}\) for some function \(f\text{,}\) and replaced \(f(x)\) with \(u\text{.}\) In this section, we will set \(x=f(\theta)\text{,}\) where \(f\) is a trigonometric function, then replace \(x\) with \(f(\theta)\text{.}\)
We start by demonstrating this method in evaluating the integral in Equation (6.4.1). After the example, we will generalize the method and give more examples.
We begin by noting that \(9\left(\sin^2(\theta) + \cos^2(\theta)\right) = 9\text{,}\) and hence \(9\cos^2(\theta) = 9-9\sin^2(\theta)\text{.}\) If we let \(x=3\sin(\theta)\text{,}\) then \(9-x^2 = 9-9\sin^2(\theta) = 9\cos^2(\theta)\text{.}\)
Setting \(x=3\sin(\theta)\) gives \(dx = 3\cos(\theta) \, d\theta\text{.}\) We are almost ready to substitute. We also wish to change our bounds of integration. The bound \(x=-3\) corresponds to \(\theta = -\pi/2\) (for when \(\theta = -\pi/2\text{,}\)\(x=3\sin(\theta) = -3\)). Likewise, the bound of \(x=3\) is replaced by the bound \(\theta = \pi/2\text{.}\) Thus
We now describe in detail Trigonometric Substitution. This method excels when dealing with integrands that contain \(\sqrt{a^2-x^2}\text{,}\)\(\sqrt{x^2-a^2}\) and \(\sqrt{x^2+a^2}\text{.}\) The following Key Idea outlines the procedure for each case, followed by more examples. Each right triangle acts as a reference to help us understand the relationships between \(x\) and \(\theta\text{.}\)
Key Idea6.4.3.Trigonometric Substitution.
Integrands containing \(\sqrt{a^2-x^2}\).
Let \(x=a\sin(\theta)\text{,}\)\(dx = a\cos(\theta) \, d\theta\text{.}\) Thus \(\theta = \sin^{-1}(x/a)\text{,}\) for \(-\pi/2\leq \theta\leq \pi/2\text{.}\) On this interval, \(\cos(\theta) \geq 0\text{,}\) so \(\sqrt{a^2-x^2} = a\cos(\theta)\text{.}\)
Integrands containing \(\sqrt{x^2+a^2}\).
Let \(x=a\tan(\theta)\text{,}\)\(dx = a\sec^2(\theta) \, d\theta\text{.}\) Thus \(\theta = \tan^{-1}(x/a)\text{,}\) for \(-\pi/2 \lt \theta \lt \pi/2\text{.}\) On this interval, \(\sec(\theta) \gt 0\text{,}\) so \(\sqrt{x^2+a^2} = a\sec(\theta)\text{.}\)
Integrands containing \(\sqrt{x^2-a^2}\).
Let \(x=a\sec(\theta)\text{,}\)\(dx = a\sec(\theta) \tan(\theta) \, d\theta\text{.}\) Thus \(\theta = \sec^{-1}(x/a)\text{.}\) If \(x/a\geq 1\text{,}\) then \(0\leq\theta\lt \pi/2\text{;}\) if \(x/a \leq -1\text{,}\) then \(\pi/2\lt \theta\leq \pi\text{.}\) We restrict our work to where \(x\geq a\text{,}\) so \(x/a\geq 1\text{,}\) and \(0\leq\theta\lt \pi/2\text{.}\) On this interval, \(\tan(\theta) \geq 0\text{,}\) so \(\sqrt{x^2-a^2} = a\tan(\theta)\text{.}\)
Using Item 2 in Key Idea 6.4.3, we recognize \(a=\sqrt{5}\) and set \(x= \sqrt{5}\tan(\theta)\text{.}\) This makes \(dx = \sqrt{5}\sec^2(\theta) \, d\theta\text{.}\) We will use the fact that \(\sqrt{5+x^2} = \sqrt{5+5\tan^2(\theta) } = \sqrt{5\sec^2(\theta) } = \sqrt{5}\sec(\theta)\text{.}\) Substituting, we have:
While the integration steps are over, we are not yet done. The original problem was stated in terms of \(x\text{,}\) whereas our answer is given in terms of \(\theta\text{.}\) We must convert back to \(x\text{.}\)
The reference triangle given in Figure 6.4.5 helps. With \(x=\sqrt{5}\tan(\theta)\text{,}\) we have
where the \(\ln\big(1/\sqrt{5}\big)\) term is absorbed into the constant \(C\text{.}\) (In Section 6.6 we will learn another way of approaching this problem.)
Solution2.Video solution
Example6.4.8.Using Trigonometric Substitution.
Evaluate \(\ds \int \sqrt{4x^2-1}\, dx\text{.}\)
Solution1.
We start by rewriting the integrand so that it looks like \(\sqrt{x^2-a^2}\) for some value of \(a\text{:}\)
So we have \(a=1/2\text{,}\) and following Part 3 of Key Idea 6.4.3, we set \(x= \frac12\sec(\theta)\text{,}\) and hence \(dx = \frac12\sec(\theta) \tan(\theta) \, d\theta\text{.}\) We now rewrite the integral with these substitutions:
We are not yet done. Our original integral is given in terms of \(x\text{,}\) whereas our final answer, as given, is in terms of \(\theta\text{.}\) We need to rewrite our answer in terms of \(x\text{.}\) With \(a=1/2\text{,}\) and \(x=\frac12\sec(\theta)\text{,}\) the reference triangle in Figure 6.4.6 shows that
We use Part 1 of Key Idea 6.4.3 with \(a=2\text{,}\)\(x=2\sin(\theta)\text{,}\)\(dx = 2\cos(\theta)\) and hence \(\sqrt{4-x^2} = 2\cos(\theta)\text{.}\) This gives
We need to rewrite our answer in terms of \(x\text{.}\) Using the reference triangle found in Figure 6.4.4, we have \(\cot(\theta) = \sqrt{4-x^2}/x\) and \(\theta = \sin^{-1}(x/2)\text{.}\) Thus
Trigonometric Substitution can be applied in many situations, even those not of the form \(\sqrt{a^2-x^2}\text{,}\)\(\sqrt{x^2-a^2}\) or \(\sqrt{x^2+a^2}\text{.}\) In the following example, we apply it to an integral we already know how to handle.
Example6.4.10.Using Trigonometric Substitution.
Evaluate \(\ds \int\frac1{x^2+1}\, dx\text{.}\)
Solution.
We know the answer already as \(\tan^{-1}(x) +C\text{.}\) We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function.
Using Part 2 of Key Idea 6.4.3, let \(x=\tan(\theta)\text{,}\)\(dx=\sec^2(\theta) \, d\theta\) and note that \(x^2+1 = \tan^2(\theta) +1 = \sec^2(\theta)\text{.}\) Thus
Since \(x=\tan(\theta)\text{,}\)\(\theta = \tan^{-1}(x)\text{,}\) and we conclude that \(\ds \int\frac1{x^2+1}\, dx = \tan^{-1}(x) +C\text{.}\)
The next example is similar to the previous one in that it does not involve a square-root. It shows how several techniques and identities can be combined to obtain a solution.
We start by completing the square, then make the substitution \(u=x+3\text{,}\) followed by the trigonometric substitution of \(u=\tan(\theta)\text{:}\)
We need to return to the variable \(x\text{.}\) As \(u=\tan(\theta)\text{,}\)\(\theta = \tan^{-1}(u)\text{.}\) Using the identity \(\sin(2\theta) = 2\sin(\theta) \cos(\theta)\) and using the reference triangle found in Figure 6.4.5, we have
Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substitution, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of \(x\) to one in terms of \(\theta\text{,}\) then converting back to \(x\)) and then evaluate using the original bounds. It is much more straightforward, though, to change the bounds as we substitute.
Example6.4.12.Definite integration and Trigonometric Substitution.
Using Part 2 of Key Idea 6.4.3, we set \(x=5\tan(\theta)\text{,}\)\(dx = 5\sec^2(\theta) \, d\theta\text{,}\) and note that \(\sqrt{x^2+25} = 5\sec(\theta)\text{.}\) As we substitute, we can also change the bounds of integration.
The lower bound of the original integral is \(x=0\text{.}\) As \(x=5\tan(\theta)\text{,}\) we solve for \(\theta\) and find \(\theta = \tan^{-1}(x/5)\text{.}\) Thus the new lower bound is \(\theta = \tan^{-1}(0) = 0\text{.}\) The original upper bound is \(x=5\text{,}\) thus the new upper bound is \(\theta = \tan^{-1}(5/5) = \pi/4\text{.}\)
The next section introduces Partial Fraction Decomposition, which is an algebraic technique that turns “complicated” fractions into sums of “simpler” fractions, making integration easier.
ExercisesExercises
Terms and Concepts
1.
Trigonometric Substitution works on the same principles as Integration by Substitution, though it can feel “ ”.
2.
If one uses Trigonometric Substitution on an integrand containing \({\sqrt{16-x^{2}}}\text{,}\) then one should set \(x={}\).
3.
Consider the Pythagorean Identity \(\sin^2(\theta) +\cos^2(\theta) = 1\text{.}\)
What identity is obtained when both sides are divided by \(\cos^2(\theta)\text{?}\)
Use the new identity to simplify \(9\tan^2(\theta) + 9\text{.}\)
4.
Why does Part 1 of Key Idea 6.4.2 state that \(\sqrt{a^2-x^2} = a\cos(\theta)\text{,}\) and not \(\abs{a\cos(\theta) }\text{?}\)
Problems
Exercise Group.
Apply Trigonometric Substitution to evaluate the indefinite integral.
5.
\(\ds \int \sqrt{x^2+1}\, dx\)
6.
\(\ds \int \sqrt{x^2+4}\, dx\)
7.
\(\ds \int \sqrt{1-x^2}\, dx\)
8.
\(\ds \int \sqrt{9-x^2}\, dx\)
9.
\(\ds \int \sqrt{x^2-1}\, dx\)
10.
\(\ds \int \sqrt{x^2-16}\, dx\)
11.
\(\ds \int {\sqrt{16x^{2}+1}}\, dx\)
12.
\(\ds \int {\sqrt{1-25x^{2}}}\, dx\)
13.
\(\ds \int {\sqrt{36x^{2}-1}}\, dx\)
14.
\(\ds \int {\frac{6}{\sqrt{x^{2}+2}}}\, dx\)
15.
\(\ds \int {\frac{7}{\sqrt{13-x^{2}}}}\, dx\)
16.
\(\ds \int {\frac{8}{\sqrt{x^{2}-6}}}\, dx\)
Exercise Group.
Evaluate the indefinite integral. Trigonometric Substitution may not be required.
Evaluate the definite integral by making the proper trigonometric substitution and changing the bounds of integration. (Note: the corresponding indefinite integrals appeared previously in the Section 6.4 exercises.)