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 evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate $$\int_{-3}^3\sqrt{9-x^2}\ dx = \frac{9\pi}{2} \label{eq_trigsub1}\tag{6.4.1}$$ 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 (6.4.1). After the example, we will generalize the method and give more examples.

Example6.4.1Using Trigonometric Substitution

Evaluate $\ds \int_{-3}^3\sqrt{9-x^2}\ dx\text{.}$

Solution

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 Idea6.4.2Trigonometric Substitution
1. For integrands containing $\sqrt{a^2-x^2}\text{:}$ 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 $\sqrt{a^2-x^2} = a\cos(\theta)\text{.}$

2. For integrands containing $\sqrt{x^2+a^2}\text{:}$ 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, $\sec(\theta) > 0\text{,}$ so $\sqrt{x^2+a^2} = a\sec(\theta)\text{.}$

3. For integrands containing $\sqrt{x^2-a^2}\text{:}$ 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 $\sqrt{x^2-a^2} = a\tan(\theta)\text{.}$

We will now apply Key Idea 3 to a type of problem that we were able to solve using algebra and substitution in Section 6.1. Example 6.4.3 shows an alternate way to approach these problems.

Example6.4.3Using Trigonometric Substitution

Evaluate $\ds \int \frac{1}{\sqrt{x^2-4}}\ dx\text{.}$

Solution
Example6.4.4Using Trigonometric Substitution

Evaluate $\ds \int \frac{1}{\sqrt{5+x^2}}\ dx\text{.}$

Solution
Example6.4.5Using Trigonometric Substitution

Evaluate $\ds \int \sqrt{4x^2-1}\ dx\text{.}$

Solution
Example6.4.6Using Trigonometric Substitution

Evaluate $\ds \int \frac{\sqrt{4-x^2}}{x^2}\ dx\text{.}$

Solution

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.

Example6.4.7Using Trigonometric Substitution

Evaluate $\ds \int\frac1{x^2+1}\ dx\text{.}$

Solution

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.

Example6.4.8Using Trigonometric Substitution

Evaluate $\ds\int\frac1{(x^2+6x+10)^2}\ dx\text{.}$

Solution

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.

Example6.4.9Definite integration and Trigonometric Substitution

Evaluate $\ds\int_0^5\frac{x^2}{\sqrt{x^2+25}}\ dx\text{.}$

Solution

The following equalities are very useful when evaluating integrals using Trigonometric Substitution.

Key Idea6.4.10Useful Equalities with Trigonometric Substitution
1. $\sin(2\theta) = 2\sin(\theta) \cos(\theta)$

2. $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) -1 = 1-2\sin^2(\theta)$

3. $\ds \int \sec^3(\theta) \ d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C$

4. $\ds \int \cos^2(\theta) \ d\theta = \int \frac12\big(1+\cos(2\theta)\big)\ d\theta = \frac12\big(\theta+\sin(\theta) \cos(\theta) \big)+C\text{.}$

The next section introduces Partial Fraction Decomposition, which is an algebraic technique that turns “complicated” fractions into sums of “simpler” fractions, making integration easier.

Subsection6.4.1Exercises

Terms and Concepts

In the following exercises, apply Trigonometric Substitution to evaluate the indefinite integrals.