Trigonometric Substitution

The process of finding antiderivatives can feel a bit frightening as integrands become more complicated. For instance, consider the following integral:

$I = \int \frac{x}{\sqrt{3x^2 + 2x - 1}} \ dx$How to solve it? You can use integration by parts or u-substitution after doing some algebraic manipulation. However, these methods can be inefficient because of their trial-and-error nature. Fortunately, there are techniques to identify and solve indefinite integrals like this one with less guesswork.

In this topic, you'll take a look at the Trigonometric substitution technique and how it can help solve integrals with integrands that contain rational functions with quadratic trinomial functions.

Rational quadratic trinomials

Before describing the actual technique, let's take a look at a few expressions:

$\frac{1}{x^2 - 2x + 5} \ ,\quad \frac{1}{x \cdot \sqrt{-x^2 + 5x - 6}} \ , \quad x \sqrt{x^2+4} \ , \quad \frac{x}{(x-1) \cdot \sqrt{x^2 - 1}}$All of them have the following elements in common: they can all be expressed as rational functions, $R(x) = \frac{P(x)}{Q(x)}$; they contain a quadratic trinomial factor:$\ ax^2 + bx + c$; and they may contain a function of $x$ as another factor.

As such, all the expressions from the beginning can be represented as follows:

$R(x, ax^2 +bx +c) \quad \text{or} \quad R\left(x, \sqrt{ax^2 +bx +c}\right)$

Trigonometric link

These rational functions that contain quadratic trinomials have a special relationship with trigonometric functions. It can be described in two facts:

1. Completing the square allows you to convert expressions of the form $ax^2+bx+c$ to any of the three forms:

   • $1+g^2(x)$

   • $1-g^2(x)$, or

   • $g^2(x) - 1$

2. Substituting $g(x)$ for the right trigonometric function allows you to make use of well-known trigonometric identities:

   • $\sec^2(x) = 1 + \tan^2(x)$

   • $\cos^2(x) = 1 - \sin^2(x)$, and

   • $\tan^2(x) = \sec^2(x) - 1$

Now, when dealing with integrands of the form:

$\int R(x, ax^2 +bx +c) \ dx \quad \text{or} \quad \int R\left(x, \sqrt{ax^2 +bx +c}\right) \ dx$Another helpful fact:

• taking the derivative of the trigonometric functions on one side of the identities mentioned in (2) results in expressions that include functions from the other side of the equation. This can potentially lead to canceling factors after a change of variables.

An example

Let's consider the integral from before:

$I = \int \frac{x}{\sqrt{3x^2 + 2x - 1}} \ dx$First, let's complete the square for the argument of the radical function in the denominator,

You should be able to see how

$g^2(x) - 1 = \left( \frac{3x+1}{2} \right) ^2 - 1 \quad \text{if} \quad g(x) = \frac{3x+1}{2}$

Then, let's define the following substitution:

$\frac{3x+1}{2} = \sec(\theta)$And solve for $x$ to find $dx$,

$x =\frac{2 \sec(\theta) - 1}{3} \implies dx = \frac{2}{3} \sec(\theta) \tan(\theta) \ d\theta$Now, let's perform the change of variables by substituting our new expressions in the original integral,

$I = \int \frac{x}{\sqrt{3x^2 + 2x - 1}} \ dx = \int \frac{x}{\sqrt{\frac{4}{3} \cdot \left(\left( \frac{3x+1}{2} \right)^2 - 1 \right)}} \ dx \implies I = \int \frac{\frac{1}{3} \cdot (2 \sec(\theta) - 1)}{\sqrt{\frac{4}{3} \cdot \left(\sec^2(\theta) - 1 \right) }} \cdot \frac{2}{3} \cdot \sec(\theta) \cdot \tan(\theta) \ d\theta$. Simplifying,

[ALERT]The $\sqrt{\tan^2(\theta)}$ factor in the denominator canceled out one of the factors in the numerator, simplifying the integrand expression as a whole.[/ALERT]

From previous topics, you should already know the antiderivatives for these two integrals you are left with:$I = \frac{2}{3\sqrt{3}} \int \sec^2(\theta) \ d\theta - \frac{1}{3\sqrt{3}} \int \sec(\theta) \ d\theta = \frac{2}{3\sqrt{3}} \tan(\theta) - \frac{1}{3\sqrt{3}} \ln| \tan(\theta) + \sec(\theta) | + C$The change of variables still needs to be reversed, but first let's recap on the technique.

Trigonometric substitution

The Trigonometric substitution technique can be summarized as follows:

• Complete the square to reach any of the following expressions

$a) \quad 1+g^2(x)$

$b) \quad 1-g^2(x)$

$c) \quad g^2(x) - 1$

• Depending on the form of the resulting expression, substitute $g(x)$ with the correct trig function:

$1) \quad 1+g^2(x) \quad \text{and} \quad g(x) = \tan(\theta) \implies 1+ \tan^2(\theta) = \sec^2(\theta)$
$2) \quad 1-g^2(x) \quad \text{and} \quad g(x) = \sin(\theta) \implies 1 - \sin^2(\theta) = \cos^2(\theta)$
$3) \quad 1-g^2(x) \quad \text{and} \quad g(x) = \cos(\theta) \implies 1 - \cos^2(\theta) = \sin^2(\theta)$
$4) \quad g^2(x) - 1 \quad \text{and} \quad g(x) = \sec(\theta) \implies \sec^2(\theta) - 1 = \tan^2(\theta)$

• Solve for $x$ and determine $dx$ as a function of $\theta$

• Simplify and solve easier integrals

• Reverse all changes of variable

Reversing changes of variable

In order to reverse the change of variable from $\theta$ back to $x$, construct a right triangle:

If $a$, $b$, and $c$ can be expressed as functions of $x$; any of the six fundamental trigonometric functions can be expressed as a function of $x$ as well, based on their right-angled triangle definition:

Let's resume the previous example:

From the original change of variables, you know that

$\frac{3x+1}{2} = \sec(\theta)$

From the right-triangle definition of the secant function, you should know also that

$\sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{c}{a}$

Let's define $a$ and $c$ as follows:

$\sec(\theta) = \frac{3x+1}{2} \implies \begin{cases} c = 3x+1 \\ a = 2 \end{cases}$

With a little help from Pythagoras' theorem, $b$ can be found

$a^2 + b^2 = c^2 \implies (2)^2 + b^2 = (3x+1)^2 \implies b = \sqrt{ (3x+1)^2 - 4}$Thus, the length of all sides of the triangle can be defined like so

Therefore:

Substituting our newfound expressions:

Finally,

$I =\int \frac{x}{\sqrt{3x^2 + 2x - 1}} \ dx = \frac{\sqrt{3x^2+2x-1}}{3} - \frac{1}{3\sqrt{3}} \ln \left| \frac{\sqrt{9x^2+6x-3}+3x+1}{2} \right| + C$

Conclusion

In this topic, you have learned about the trigonometric substitution integration method, which can be summarized as follows:

• First, complete the square if necessary.

• Perform the appropriate trigonometric substitution:

• Solve an easier integral!

$\int R(x, \ \sqrt{ax^2 + bx + c}) dx \ \ \text{or} \ \int R(x, ax^2 + bx + c) dx = \int \frac{P(x)}{Q(x)} dx \implies\int \frac{U(\theta)}{\cancel{V(\theta)}} \cdot \cancel{V(\theta)} d\theta = \int U(\theta) d\theta$

• Finally, reverse a change of variables.

In addition to looking for the three main substitution cases, you have learned how to use Pythagoras' triangle to perform a change of variables for functions involving trigonometric expressions.