Learning Outcomes
- Use trigonometric substitution to compute indefinite integrals.
Section 5.4 Trigonometric Substitution (TI4)
Subsection 5.4.1 Activities
Activity 5.4.1.
Consider \(\displaystyle \int \sqrt{9-4x^2} \,dx\text{.}\) Which substitution would you choose to evaluate this integral?
- \(\displaystyle u=9-4x^2\)
- \(\displaystyle u=\sqrt{9-4x^2}\)
- \(\displaystyle u=3-2x\)
- Substitution is not effective
Activity 5.4.2.
To find \(\displaystyle \int \sqrt{9-4x^2} \,dx\text{,}\) we will need a more advanced substitution. Which of these candidates is most reasonable?
- Let \(v\) satisfy \(9-4x^2=9-9e^{2v}=9e^{-2v}\text{.}\)
- Let \(\theta\) satisfy \(9-4x^2=9-9\sin^2\theta=9\cos^2\theta\text{.}\)
- Let \(w\) satisfy \(9-4x^2=4-8\ln|w|=4\ln|2w|\text{.}\)
- Let \(\phi\) satisfy \(9-4x^2=4-4\cos^2\phi=4\sin^2\phi\text{.}\)
Activity 5.4.3.
Fill in the missing \(\unknown\)s for the following calculation.
\begin{align*}
\text{Let }9-4x^2&=9-9\sin^2\theta=9\cos^2\theta\\
4x^2&=\unknown\\
x&=\unknown\\
dx&=\unknown\,d\theta
\end{align*}
\begin{align*}
\int\sqrt{9-4x^2} \,dx&=\int\sqrt{\unknown}\,(\unknown\,d\theta)\\
&= \int\frac{9}{2}\cos^2 \theta\,d\theta
\end{align*}
Activity 5.4.4.
From Section 5.3 we may find \(\int\cos^2 \theta\,d\theta=\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta+C\text{.}\)
Use this to continue your work in the previous activity and complete the integration by trigonometric substitution.
\begin{align*}
\sin(\theta)&=\unknown\\
\theta&=\arcsin(\unknown)\\
\cos(\theta)&=\unknown\sqrt{\unknown}
\end{align*}
\begin{align*}
\int\sqrt{9-4x^2} \,dx&= \cdots = \int\frac{9}{2}\cos^2 \theta\,d\theta\\
&= \frac{9}{2}\left(\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta\right)+C\\
&= \frac{9}{4}(\unknown)+\frac{9}{4}(\unknown)(\unknown)+C
\end{align*}
Activity 5.4.5.
Use similar reasoning to complete the following proof that \(\frac{d}{dx}\left[\arcsin(x)\right]=\frac{1}{\sqrt{1-x^2}}\text{.}\)
\begin{align*}
\text{Let }1-x^2&=1-\unknown\theta=\unknown\theta\\
x^2&=\unknown\\
x&=\unknown\\
dx&=\unknown\,d\theta\\
\theta&=\unknown
\end{align*}
\begin{align*}
\int \frac{1}{\sqrt{1-x^2}} \,dx&=\int\frac{1}{\sqrt{\unknown}}\,(\unknown\,d\theta)\\
&= \int d\theta\\
&= \unknown + C\\
&= \arcsin(x) + C
\end{align*}
Activity 5.4.6.
Substitutions of the form
\begin{equation*}
16-25x^2=16-16\sin^2x=16\cos^2x
\end{equation*}
are made possible due to the Pythagorean identity \(\sin^2(x)+\cos^2(x)=1\text{.}\)
Which two of these four identities can be obtained from dividing both sides of \(\sin^2(x)+\cos^2(x)=1\) by \(\cos^2(x)\) and rearranging?
- \(\displaystyle \tan^2(x)-1=\sec^2(x)\)
- \(\displaystyle \tan^2(x)+1=\sec^2(x)\)
- \(\displaystyle \sec^2(x)-1=\tan^2(x)\)
- \(\displaystyle \sec^2(x)+1=\tan^2(x)\)
Observation 5.4.7.
In summary, certain quadratic expressions inside an integral may be substituted with trigonometric functions to take advantage of trigonometric identities and simplify the integrand:
\begin{align*}
\text{Let } b-ax^2&=b-b\sin^2(\theta)=b\cos^2(\theta)\\
\text{So } x&=\frac{b}{a}\sin(\theta)
\end{align*}
\begin{align*}
\text{Let } b+ax^2&=b+b\tan^2(\theta)=b\sec^2(\theta)\\
\text{So } x&=\frac{b}{a}\tan(\theta)
\end{align*}
\begin{align*}
\text{Let } ax^2-b&=b\sec^2(\theta)-b=b\tan^2(\theta)\\
\text{So } x&=\frac{b}{a}\sec(\theta)
\end{align*}
Activity 5.4.8.
Complete the following trignometric substitution to find \(\displaystyle\int\frac{3}{4+25x^2}\,dx\text{.}\)
\begin{align*}
\text{Let }4+25x^2&=2+\unknown\theta=\unknown\theta\\
25x^2&=\unknown\\
x&=\unknown\\
dx&=\unknown\,d\theta\\
\theta&=\unknown
\end{align*}
\begin{align*}
\int\frac{3}{4+25x^2}\,dx &=\int\frac{3}{\unknown}\,(\unknown\,d\theta)\\
&= \int \unknown d\theta\\
&= \unknown + C\\
&= \frac{3}{10}\arctan(\frac{5}{2}x) + C
\end{align*}
Activity 5.4.9.
Complete the following trignometric substitution to find \(\displaystyle\int\frac{7}{x\sqrt{9x^2-16}}\,dx\text{.}\)
\begin{align*}
\text{Let }9x^2-16&=\unknown\theta-16=\unknown\theta\\
9x^2&=\unknown\\
x&=\unknown\\
dx&=\unknown\,d\theta\\
\theta&=\unknown
\end{align*}
\begin{align*}
\int\frac{7}{x\sqrt{9x^2-16}}\,dx &=\int\frac{7}{\unknown\sqrt{\unknown}}\,(\unknown\,d\theta)\\
&= \int \unknown d\theta\\
&= \unknown + C\\
&= \frac{7}{4}\arcsec(\frac{3}{4}x) + C
\end{align*}
Activity 5.4.10.
Use appropriate trignometric substitutions and the given trigonometric integrals to find each of the following.
(a)
\begin{align*}
\int \frac{\sqrt{-9 \, x^{2} + 16}}{x^{2}}\,dx &= \cdots \\
&=\int \frac{3\cos^2\theta}{\sin^2\theta}\,d\theta\\
&=-3\theta-3\frac{\cos\theta}{\sin\theta}+C\\
&= - 3 \, \arcsin\left(\unknown\right)-\frac{\sqrt{\unknown}}{\unknown} +C
\end{align*}
(b)
\begin{align*}
\int \frac{2 \, \sqrt{9 \, x^{2} - 16}}{x}\,dx &= \cdots \\
&=\int 8\tan^2\theta\,d\theta\\
&=8\tan\theta-8\theta+C\\
&= \unknown \, \sqrt{\unknown} - 8 \, \operatorname{arcsec}\left(\unknown\right) +C
\end{align*}
(c)
\begin{align*}
\int \frac{1}{\sqrt{81 \, x^{2} + 4}}\,dx&= \cdots \\
&=\int \frac{1}{9}\sec\theta\,d\theta\\
&=\frac{1}{9}\log|\sec\theta+\tan\theta|+C\\
&= \frac{1}{9} \, \log\left| \unknown + \frac{1}{2} \, \sqrt{\unknown} \right| +C
\end{align*}
Activity 5.4.11.
Consider the unit circle \(x^2+y^2=1\text{.}\) Find a function \(f(x)\) so that \(y=f(x)\) is the graph of the upper-half semicircle of the unit circle.
Activity 5.4.12.
(a)
Find the area under the curve \(y=f(x)\) from Activity 5.4.11.
(b)
How does this value compare to what we know about areas of circles?
