MAT187 L17-18 - Trigonometric Substitution

#MAT187 PCE
Recall the following trigonometric identities from high school:

\cos^{2} (θ) + \sin^{2} (θ) = 1

\tan^{2} (θ) + 1 = \sec^{2} (θ)

We can use these to our advantage when solving integrals using trigonometric substitution.

To walk through this method, we will use an example, as that is what the PCE does and I don't want to deviate from that.

Identifying Pythagorean Theorem in an Integral

Consider the integral $\int \frac{1}{\sqrt{4 - x^{2}}} d x$ . This is a difficult integral to solve normally, but notice that the $\sqrt{4 - x^{2}}$ term looks almost like a rearranged Pythagorean formula equation. That is, if:
$a = \sqrt{2^{2} - x^{2}}$ , $b = x$ , and $c = 2$
Then:

\begin{array}{r} a^{2} + b^{2} = c^{2} \\ a^{2} = c^{2} - b^{2} \\ a = \sqrt{c^{2} - b^{2}} \\ = \sqrt{2^{2} - x^{2}} \end{array}

Reference Triangle

This is what it would the above would like as a triangle—what the PCE calls a "reference triangle".

From this, we can identify that:

\sin θ = \frac{x}{2} ⟹ x = 2 \sin θ ...A)

\cos θ = \frac{\sqrt{4 - x^{2}}}{2} ⟹ \sqrt{4 - x^{2}} = 2 \cos θ ...B)

Substitution and Resubstitution

If you take the derivative of both sides of relationship A) with respect to x, then you get:

\begin{aligned} \frac{d}{d x} (x) & = \frac{d}{d x} (2 \sin θ) \\ 1 & = (2 \cos θ) \frac{d θ}{d x} \\ d x & = 2 \cos θ d θ \end{aligned}

...note that this is just a result of implicit differentiation.
So, we can substitute this in for $d x$ , and $2 \cos θ$ for $\sqrt{4 - x^{2}}$ in the integral we're trying to solve.

\begin{aligned} \int \frac{1}{\sqrt{4 - x^{2}}} d x & = \int \frac{2 \cos θ}{2 \cos θ} d θ \\ = \int 1 d θ \\ = θ + C \end{aligned}

We can rearrange relationship C) to isolate for $θ$ , and thus bring the integral back into something in terms of $x$ .

\begin{aligned} x & = 2 \sin θ \\ θ & = \arcsin (\frac{x}{2}) \\ \int \frac{1}{\sqrt{4 - x^{2}}} d x & = \arcsin (\frac{x}{2}) + C \end{aligned}

Domain of Trig Sub

Note that since $\sin (x) \in [- 1, 1]$ in the above example, when we specify that $\sin θ = \frac{x}{2}$ , we know that $x \in [- 2, 2]$ , which is derived from:

\begin{aligned} - 1 \leq (\sin θ & = \frac{x}{2}) \leq 1 \\ ⟹ - 1 \leq \frac{x}{2} \leq 1 \\ ⟹ - 2 \leq x \leq 2 \end{aligned}

This makes sense because, for $\frac{1}{\sqrt{4 - x^{2}}}$ to be real, we need this same domain.

For other functions, it may not be so cut-and-dry, so you will want to check the bounds every time.

Domains/ranges of common trig functions

You will want to remember these :3

Function	Domain	Range
$\arcsin (x)$	[-1,-1]	$[- \frac{π}{2}, \frac{π}{2}]$
$\arccos (x)$	$[- 1, 1]$	$[0, π]$
$\arctan (x)$	$x \in R$	$[- \frac{π}{2}, \frac{π}{2}]$
$\sin (x)$	$x \in R$	$[- 1, 1]$
$\cos (x)$	$x \in R$	$[- 1, 1]$
$\tan (x)$	$x \neq \frac{π (2 n + 1)}{2}$ , $n \in R$	$y \in R$