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Calculus for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

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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?
  1. \(\displaystyle u=9-4x^2\)
  2. \(\displaystyle u=\sqrt{9-4x^2}\)
  3. \(\displaystyle u=3-2x\)
  4. 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?
  1. Let \(v\) satisfy \(9-4x^2=9-9e^{2v}=9e^{-2v}\text{.}\)
  2. Let \(\theta\) satisfy \(9-4x^2=9-9\sin^2\theta=9\cos^2\theta\text{.}\)
  3. Let \(w\) satisfy \(9-4x^2=4-8\ln|w|=4\ln|2w|\text{.}\)
  4. 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 \(\displaystyle \int\cos^2 \theta\,d\theta=\dfrac{1}{2}\theta+\dfrac{1}{2}\sin\theta\cos\theta+C\text{.}\)
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)+\dfrac{9}{4}(\unknown)(\unknown)+C \end{align*}

Activity 5.4.5.

Use similar reasoning to complete the following proof that \(\dfrac{d}{dx}\left[\arcsin(x)\right]=\dfrac{1}{\sqrt{1-x^2}}\text{.}\)
\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 \dfrac{1}{\sqrt{1-x^2}} \,dx&=\displaystyle \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?
  1. \(\displaystyle \tan^2(x)-1=\sec^2(x)\)
  2. \(\displaystyle \tan^2(x)+1=\sec^2(x)\)
  3. \(\displaystyle \sec^2(x)-1=\tan^2(x)\)
  4. \(\displaystyle \sec^2(x)+1=\tan^2(x)\)

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^2-a^2x^2&=b^2-b^2\sin^2(\theta)=b^2\cos^2(\theta)\\ \text{So } x&=\frac{b}{a}\sin(\theta) \end{align*}
\begin{align*} \text{Let } b^2+a^2x^2&=b^2+b^2\tan^2(\theta)=b^2\sec^2(\theta)\\ \text{So } x&=\frac{b}{a}\tan(\theta) \end{align*}
\begin{align*} \text{Let } a^2x^2-b^2&=b^2\sec^2(\theta)-b^2=b^2\tan^2(\theta)\\ \text{So } x&=\frac{b}{a}\sec(\theta) \end{align*}

Activity 5.4.8.

Complete the following trigonometric substitution to find \(\displaystyle\int\dfrac{3}{4+25x^2}\,dx\text{.}\)
\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\dfrac{3}{\unknown}\,(\unknown\,d\theta)\\ &= \int \unknown \, d\theta\\ &= \unknown + C\\ &= \dfrac{3}{10}\arctan\left(\dfrac{5}{2}x\right) + C \end{align*}

Activity 5.4.9.

Complete the following trigonometric substitution to find \(\displaystyle\int\dfrac{7}{x\sqrt{9x^2-16}}\,dx\text{.}\)
\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*} \displaystyle \int\dfrac{7}{x\sqrt{9x^2-16}}\,dx &=\int\dfrac{7}{\unknown\sqrt{\unknown}}\,(\unknown\,d\theta)\\ &= \int \unknown\, d\theta\\ &= \unknown + C\\ &= \frac{7}{4}\arcsec\left(\dfrac{3}{4}x\right) + C \end{align*}

Activity 5.4.10.

Use appropriate trigonometric 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.

(b)
How does this value compare to what we know about areas of circles?

Subsection 5.4.2 Videos

Figure 114. Video: Use trigonometric substitution to compute indefinite integrals

Subsection 5.4.3 Exercises

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