Objectives
- Use the basic trigonometric substitutions to evaluate definite and indefinite integrals, and use trigonometric identities to write solutions without using inverse trigonometric functions (when possible)
Section 7.3 Trigonometric Substitution
Subsection 7.3.1 Before Class
https://mymedia.ou.edu/media/7.3-1/1_ymu3wt20https://mymedia.ou.edu/media/7.3-2/1_j9pld8uqSubsubsection 7.3.1.1 The Substitutions
Example 7.3.1.
Use trigonometry to complete the following reference triangles:
Example 7.3.2.
Let \(f(x) = \dfrac{1}{16+x^2}\)
- Consider \(\ds \int f(x)\, dx\text{.}\) Why do none of our previous integration techniques work here?
- Refer to the example at the top of the page. Which triangle has an expression similar to the radicand in \(f(x)\text{?}\) Write the substitution below.
- Treat the substitution you chose in part (b) like a \(u-\)substitution. Make the appropriate substitutions into the integral \(\ds \int \dfrac{1}{16+x^2}\, dx\text{,}\) and complete the integration.
Solution.
- Answers vary, but the idea is that we don’t have an appropriate rule for the pattern
- \(\displaystyle x = a\tan \theta\)
-
If \(x = 4\tan \theta\text{,}\) then \(dx = 4\sec^2\theta\, d\theta\text{.}\) Anywhere we see \(x\text{,}\) we will replace with \(4\tan\theta\text{.}\)Now, we have\begin{equation*} \ds \int \dfrac{1}{16 + (4\tan\theta)^2}\, (4\sec^2\theta\, d\theta) \end{equation*}Simplifying, we have\begin{equation*} \ds \int \dfrac{4\sec^2\theta}{16(1+\tan^2\theta)}\, d\theta \end{equation*}Using identities, we arrive at\begin{equation*} \ds \dfrac{1}{4}\int \dfrac{\sec^2\theta}{\sec^2\theta}\, d\theta = \dfrac{1}{4}\int d\theta \end{equation*}Integrating gives\begin{equation*} \ds \dfrac{1}{4}\int d\theta = \dfrac{1}{4}\theta + C \end{equation*}However, our original variable was \(x\text{,}\) not \(\theta\text{;}\) since we declared \(x = 4\tan \theta\text{,}\) we rearrange to get \(\theta = \inv{\tan}\lrpar{\dfrac{x}{4}}\text{.}\) So, our answer becomes\begin{equation*} \dfrac{1}{4}\inv{\tan}\lrpar{\dfrac{x}{4}} + C \end{equation*}
Example 7.3.3.
Compute \(\ds \int \dfrac{\sqrt{9-x^2}}{x^2}\, dx\)
Solution.
\(\ds \int \dfrac{\sqrt{9-x^2}}{x^2}\, dx = -\dfrac{\sqrt{9-x^2}}{x} - \inv{\sin}\lrpar{\dfrac{x}{3}} + C\)
Example 7.3.4.
Evaluate \(\ds \int \dfrac{\sqrt{x^2-9}}{x}\, dx\)
Solution.
\(\ds \int \dfrac{\sqrt{x^2-9}}{x}\, dx = \sqrt{x^2-9} - 3\inv{\sec}\lrpar{\dfrac{x}{3}} + C\)
Subsection 7.3.2 Pre-Class Activities
Example 7.3.5.
Use this space to write any questions you have from the videos.
Solution.
Answers vary
Example 7.3.6.
Compute \(\ds \int_0^1 \dfrac{3}{x^2 + 4}\, dx\)
Solution.
\(\ds \int_0^1 \dfrac{3}{x^2 + 4}\, dx = \dfrac{3}{2}\inv{\tan}\lrpar{\dfrac{1}{2}}\)
Example 7.3.7.
Compute \(\ds \int \dfrac{x}{x^2 - 4}\, dx\text{.}\) Is a trigonometric substitution necessary here? Why or why not?
Solution.
\(\ds \int \dfrac{x}{x^2 - 4}\, dx = \dfrac{1}{2}\ln |x^2-4| + C\text{.}\) A trig substitution isn’t necessary here, as we can use a u-substitution.
Example 7.3.8.
Compute \(\ds \int \dfrac{1}{x^2\sqrt{4-x^2}}\, dx\) using the substitution \(x = 2\sin \theta\)
Solution.
\(\ds \int \dfrac{1}{x^2\sqrt{4-x^2}}\, dx = -\dfrac{\sqrt{4-x^2}}{4x} + C\)
Question 7.3.9.
Using the previous examples as guideposts, fill out the following table:
| If you see an integrand involving | Try a substitution using |
Solution.
| If you see an integrand involving | Try a substitution using |
| \(x^2 + a^2\) | \(x = a\tan\theta\) |
| \(a^2-x^2\) | \(x = a\sin\theta\) |
| \(x^2-a^2\) | \(x = a\sec\theta\) |
Subsection 7.3.3 In Class
Subsubsection 7.3.3.1 Examples
Example 7.3.10.
Compute \(\ds \int \dfrac{dx}{x^2\sqrt{x^2 + 9}}\)
Solution.
\(\ds \int \dfrac{dx}{x^2\sqrt{x^2 + 9}} = -\dfrac{1}{9}\lrpar{\dfrac{\sqrt{x^2+9}}{x}} + C\)
Example 7.3.11.
Evaluate \(\ds \int \dfrac{16x^3}{(4x^2+9)^{3/2}}\, dx\)
Solution.
\(\ds \int \dfrac{16x^3}{(4x^2+9)^{3/2}}\, dx = \sqrt{4x^2+9} + \dfrac{9}{\sqrt{4x^2+9}} + C\)
Example 7.3.12.
Compute \(\ds \int \dfrac{\sqrt{x^2-1}}{x^4}\, dx\)
Solution.
\(\ds \int \dfrac{\sqrt{x^2-1}}{x^4}\, dx = \dfrac{1}{3}\lrpar{\dfrac{\sqrt{x^2-1}}{x}}^3 + C\)
Example 7.3.13.
Find \(\ds \int \dfrac{x}{\sqrt{3-2x-x^2}}\, dx\)
Solution.
\(\ds \int \dfrac{x}{\sqrt{3-2x-x^2}}\, dx = -\sqrt{4-(x+1)^2} - \inv{\sin}\lrpar{\dfrac{x+1}{2}}+C\)
Example 7.3.14.
Evaluate \(\ds \int \dfrac{dx}{(x^2-1)^{3/2}}\)
Solution.
\(\ds \int \dfrac{dx}{(x^2-1)^{3/2}} = -\dfrac{x}{\sqrt{x^2-1}} + C\)
Example 7.3.15.
Show that \(\ds \int_0^a x^2\sqrt{a^2-x^2}\, dx = \dfrac{\pi}{16}a^4\)
Solution.
Set \(x = a\sin\theta\text{.}\) Then, \(dx = a\cos\theta\text{.}\) Making our substitutions, we get
\begin{equation*}
\ds \int_0^a a^2\sin^2\theta \sqrt{a^2-a^2\sin\theta}\, (a\cos\theta)\, d\theta
\end{equation*}
After some simplification, we get
\begin{equation*}
\ds \int_0^a a^4\sin^2\theta\cos^2\theta\, d\theta
\end{equation*}
Using the double angle identity for sine, we have
\begin{equation*}
\ds \int_0^a a^4 \lrpar{\dfrac{1}{2}\sin 2\theta}^2\, d\theta = \dfrac{a^4}{4}\int \sin^2 2\theta\, d\theta
\end{equation*}
Now use a power reducing identity:
\begin{equation*}
\ds \dfrac{a^4}{4}\int_0^a \dfrac{1}{2} (1-\cos 4\theta)\, d\theta
\end{equation*}
More simplification gives
\begin{equation*}
\ds \dfrac{a^4}{8} \int_0^a 1-\cos 4\theta\, d\theta
\end{equation*}
Integrating and using double angle identities gives
\begin{equation*}
\dfrac{a^4}{8}\lrpar{\theta - \dfrac{1}{4}\sin 4\theta}\bigg \rvert_{x = 0}^{x = a}
\end{equation*}
After reduction, it becomes clear that the sine term is zero at the endpoints, so the integral becomes
\begin{equation*}
\dfrac{a^4}{8}\lrpar{\inv{\sin}\lrpar{\dfrac{x}{a}}}\bigg\rvert_0^a
\end{equation*}
Evaluating gives our answer.
Example 7.3.16.
Evaluate \(\ds \int \sqrt{x^2 + 2x}\, dx\)
Solution.
\(\ds \int \dfrac{x}{\sqrt{x^2 + 1}}\, dx = \dfrac{1}{2}(x+1)\sqrt{(x+1)^2-1} - \dfrac{3}{2}\ln |x+1 + \sqrt{(x+1)^2-1}| + C\)
Example 7.3.17.
Compute \(\ds \int \dfrac{x}{\sqrt{x^2 + 1}}\, dx\)
Solution.
\(\ds \int \dfrac{x}{\sqrt{x^2 + 1}}\, dx = \sqrt{x^2+1} + C\)
Subsection 7.3.4 After Class Activities
Example 7.3.18.
Use trig substitution to prove that the area of the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) is \(\pi ab\text{,}\) where \(a\) is the length of the major axis and \(b\) is the length of the minor axis.
Solution.
Write the ellipse in terms of \(y\) alone first; this gives \(y = \pm \dfrac{b}{a}\sqrt{a^2-x^2}\text{.}\) We can take the positive root, integrate a quarter of the ellipse, and multiply the resulting area by 4.
Now, the area will be given by \(\ds \int_0^a \dfrac{b}{a}\sqrt{a^2-x^2}\, dx\text{.}\) Use the substitution \(x = a\sin\theta\text{;}\) then, \(dx = a\cos\theta\, d\theta\text{.}\) Then
\begin{align*}
\ds \int_0^a \dfrac{b}{a}\sqrt{a^2-x^2}\, dx \amp = \dfrac{b}{a}\int_0^a \sqrt{a^2-a^2\sin^2\theta} (a\cos\theta)\, d\theta\\
\amp = \ds ab\int_0^a \cos^2\theta\, d\theta\\
\amp = \ds \dfrac{ab}{2}\int_0^a 1 + \cos 2\theta\, d\theta\\
\amp = \dfrac{ab}{2}\lrpar{\theta + \dfrac{1}{2}\sin 2\theta}\bigg\rvert_0^a\\
\amp = \dfrac{ab}{2}\lrpar{\inv{\sin}\lrpar{\dfrac{x}{a} + \dfrac{x\sqrt{a^2-x^2}}{a^2}}}\bigg\rvert_0^a\\
\amp = \dfrac{\pi ab}{4}
\end{align*}
Multiplying by 4 gives the area, \(\pi ab\)
Example 7.3.19.
Evaluate \(\ds \int_{\sqrt{2}/3}^{2/3} \dfrac{dx}{x^5\sqrt{9x^2-1}}\)
Solution.
\(\ds \int_{\sqrt{2}/3}^{2/3} \dfrac{dx}{x^5\sqrt{9x^2-1}} = \dfrac{81}{64}(7\sqrt{3}-16 + 2\pi)\)
Example 7.3.20.
Compute \(\ds \int_0^1 \sqrt{x-x^2}\, dx\)
Solution.
\(\ds \int_0^1 \sqrt{x-x^2}\, dx = \dfrac{\pi}{8} \)
