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Notes for Math 220, Calculus 2

Brenton LeMesurier

Section 3.3 Trigonometric Substitution

References.

Substitution Revisited.

So far, substitution has been done by rearranging the integrand in \(\int f(x) \, dx\) as a product \(\displaystyle f(x) = a(g(x)) g'(x)\text{,}\) so that the substitution \(u=g(x)\text{,}\) \(du=g'(x) dx\) gives
\begin{equation*} \int f(x) \, dx = \int a(g(x)) g'(x) \, dx = \int a(u) \frac{du}{dx} \, dx = \int a(u) \, du. \end{equation*}
The main challenge is finding a choice of \(g\) that allows everything to be put in terms of the new variable \(u\text{,}\) with no remaining appearance of the original variable \(x\text{.}\)

Inverse Substitution.

Often, a better approach is “inverse substitution”: instead of choosing \(u = g(x)\text{,}\) we specify \(x=h(t)\text{;}\) the original variable as a function of a new one. (The function \(h\) is the inverse of the function \(g\) used in normal substitution.)
Then \(dx = h'(t) dt\text{,}\) and inserting this and \(x=h(t)\) automatically converts everything to the new variable \(t\text{:}\)
\begin{equation} \int f(x) \, dx = \int f(h(t)) h'(t) \, dt\tag{3.3.1} \end{equation}
An important question is (as always with substitution, and with integration by parts) Is this new integral easier to evaluate that the original one?

Inverse Substitution for Definite Integrals.

For definite integrals, there is an extra step of solving for the new limits of integration:
\begin{equation} \int_{x=a}^b f(x) \, dx = \int_ {t=c}^d f(h(t)) h'(t) \, dt\tag{3.3.2} \end{equation}
where \(a = h(c)\) and \(b = h(d)\text{,}\) so some equation solving is required.

Example: The Area of a Circle.

An example of how this can be useful is computing the area of the circle radius \(R\) by evaluating
\begin{equation*} A = \displaystyle \int_{-R} ^R 2 \sqrt{1-x^2} \, dx \end{equation*}
This will be done in two parts: first the indefinite integral, and then showing two ways to insert the limits of integration and evaluate the definite integral.
Solution.
A substitution that works is \(u = \arcsin(x/R)\text{,}\) but that is far from obvious!
The inverse substitution form \(x = R \sin \theta\text{,}\) \(-\pi/2 \leq \theta \leq \pi/2\) makes things easier:
\begin{equation*} dx = R \cos \theta \, d\theta \end{equation*}
and so
\begin{equation*} \int \sqrt{R^2-x^2} \, dx = \int \sqrt{R^2 - R^2 \sin^2 \theta} R \cos \theta \; d \theta = R^2 \int \sqrt{1 - \sin^2 \theta} \cos \theta \; d \theta = R^2 \int \cos^2 \theta \, d\theta \end{equation*}
(Here we used the domain of \(\theta\) to ensure that \(\sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos\theta\text{;}\) no minus sign.)
This integral is one that can be done with the methods of the previous section:
\begin{equation*} R^2 \int \cos^2 \theta \, d\theta = R^2 \int \frac12 + \frac12 \cos 2\theta \, d\theta = R^2 \left( \frac12\theta + \frac14 \sin 2\theta \right) + C. \end{equation*}
Next we need to express everything in terms of the original variable \(x\text{,}\) which requires inverting the substitution. In this case we can solve \(x = R\sin\theta\) to get \(\theta = \arcsin (x/R)\text{,}\) so
\begin{equation*} \int \sqrt{R-x^2} \, dx = R^2 \left( \frac12 \arcsin(x/R) + \frac14 \sin (2\arcsin x) \right) +C. \end{equation*}
However there is a far nicer form, using the identity \(\sin(2\theta) = 2 \sin\theta \cos\theta\) and the facts that
\begin{equation*} \sin\theta = x/R \text{ and } \cos\theta = \sqrt{R^2 - x^2}/R \text{ as seen above.} \end{equation*}
That gives \(\sin(2\theta) = 2 x \sqrt{R^2 - x^2}/R^2\) so
\begin{equation*} \int \sqrt{R-x^2} \, dx = \frac{R^2}{2} \arcsin(x/R) + \frac12 x \sqrt{R^2 - x^2} + C. \end{equation*}
One method would be to use the above indefinite integral with the Fundamental Theorem of Calculus to insert the limits of integration. However an alernative approach is to work directly with the definite integral, avoiding the need to convert back to the original variable \(x\text{.}\)
\begin{equation*} A = 2\int_{x=-R}^R \sqrt{R^2-x^2} \, dx = 2 R^2 \left[ \frac12\theta + \frac14 \sin(2\theta) \right]_{\theta=-\pi/2}^{\theta=\pi/2} = \pi R^2. \end{equation*}

Three Square Root Forms and Recommended Substitutions.

We have seen one of three basic square root forms that can occur in integrals, and for each there is an inverse substitution that often helps.
The three cases are where the integrand is a product of an integer power of \(x\) with an integer power of one of the following square root forms:
Table 3.3.3.
With \(\sqrt{a^2-x^2}\) \(\sqrt{a^2+x^2}\) \(\sqrt{x^2-a^2}\)
use \(x=a\sin\theta\) \(x=a\tan\theta\) \(x=a\sec\theta\)
giving \(\sqrt{a^2-x^2}=a\cos\theta\) \(\sqrt{a^2+x^2}=a\sec\theta\) \(\sqrt{x^2-a^2}=a\tan\theta\)
and \(dx=a\cos\theta \, d\theta\) \(dx=a\sec^2\theta \, d\theta\) \(dx=a\tan\theta\sec\theta \, d\theta\)
The new integrals will be expressible as products of powers of \(\sin\theta\) and \(\cos\theta\text{,}\) or alternatively of \(\tan\theta\) and \(\sec\theta\text{;}\) the methods of the previous section are then useful.

Converting the Integral Back From \(\theta\) to \(x\).

After using these substitutions and integrating, the answer is in terms of \(\theta\) and various trig. functions of \(\theta\text{.}\) This needs to be converted back into terms of the original variable.
Using \(x=a\sin\theta\) for integrals involving \(\sqrt{a^2-x^2}\text{:}\)
\(\cos\theta = \displaystyle\frac{\sqrt{a^2-x^2}}{a}\text{,}\) \(\sin\theta = \displaystyle\frac{x}{a}\text{,}\) \(\tan\theta = \displaystyle\frac{x}{\sqrt{a^2-x^2}}\text{,}\) \(\theta = \sin^{-1}\displaystyle\frac{x}{a}\text{.}\)
Using \(x=a\tan\theta\) for integrals involving \(\sqrt{a^2+x^2}\text{:}\)
\(\cos\theta = \displaystyle\frac{a}{\sqrt{a^2+x^2}}\text{,}\) \(\sin\theta = \displaystyle\frac{x}{\sqrt{a^2+x^2}}\text{,}\) \(\tan\theta = \displaystyle\frac{x}{a}\text{,}\) \(\theta = \tan^{-1}\displaystyle\frac{x}{a}\text{.}\)
Using \(x=a\sec\theta\) for integrals involving \(\sqrt{x^2-a^2}\text{:}\)
\(\cos\theta = \displaystyle\frac{a}{x}\text{,}\) \(\sin\theta = \displaystyle\frac{\sqrt{x^2-a^2}}{x}\text{,}\) \(\tan\theta = \displaystyle\frac{\sqrt{x^2-a^2}}{a}\text{,}\) \(\theta = \sec^{-1}\displaystyle\frac{x}{a}\text{.}\)
It is often easiest to work these formulas out from drawing an appropriate right triangle with angle \(\theta\text{,}\) so draw the following three right triangles
  1. For \(x=a\sin\theta\text{:}\) opposite side \(x\text{,}\) hypotenuse \(a\text{,}\) giving \(\sin\theta=x/a\text{.}\)
  2. For \(x=a\tan\theta\text{:}\) opposite side \(x\text{,}\) adjacent side \(a\text{,}\) giving \(\tan\theta = x/a\text{.}\)
  3. For \(x=a\sec\theta\text{:}\) hypotenuse \(x\text{,}\) adjacent side \(a\text{,}\) giving \(\sec\theta = x/a\text{.}\)
Use the Pythagorean Theorem to get the third side,
and then read off the sine, cosine, etc.
Hint.
There can be more than one way to approach a problem, so always try the easiest methods first!
Never neglect simplifications: after a trig.{} substitution, explore trig. identities to simplify the new integrand.
More on simplification: after a trig. substitution, expressing in terms of sines and cosines often helps.
Hint.
What is \(a\text{?}\)
Integer powers of the square root terms can also be handled; particularly odd powers of a root, since an even power eliminates the root, giving just a polynomial or rational function. (The latter case will be handled in Section 3.4.)

Roots of quadratics: eliminating the \(bx\) term.

Roots of other quadratics \(ax^2+bx+c\) need one more simplification first: complete the square to get a form like
\begin{equation*} ax^2+bx+c = \pm a(x-b/(2a)) \pm d^2, \end{equation*}
and then use the substitution
\begin{equation*} u=x-b/(2a) \quad \text{(or )} \end{equation*}

Section Study Guide.

Study Calculus Volume 2, Section 3.3 2 ; in particular
  • learn the right-triangle diagrams for each of the three main cases
  • review the three Problem Solving Strategies, one for each of those three cases
  • study all Examples and Checkpoints
  • and do one or several exercises from each of the groups 131–133, 134–142, 146 & 147, and 160–164.
openstax.org/books/calculus-volume-2/pages/3-3-trigonometric-substitution
openstax.org/books/calculus-volume-2/pages/3-3-trigonometric-substitution
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