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

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Section 5.3 Integration of Trigonometry (TI3)

Subsection 5.3.1 Activities

Activity 5.3.1.

Consider \(\displaystyle\int \sin(x)\cos(x) \, dx\text{.}\) Which substitution would you choose to evaluate this integral?
  1. \(\displaystyle u=\sin(x)\)
  2. \(\displaystyle u=\cos(x)\)
  3. \(\displaystyle u=\sin(x)\cos(x)\)
  4. Substitution is not effective

Activity 5.3.2.

Consider \(\displaystyle\int \sin^4(x)\cos(x) \, dx\text{.}\) Which substitution would you choose to evaluate this integral?
  1. \(\displaystyle u=\sin(x)\)
  2. \(\displaystyle u=\sin^4(x)\)
  3. \(\displaystyle u=\cos(x)\)
  4. Substitution is not effective

Activity 5.3.3.

Consider \(\displaystyle\int \sin^4(x)\cos^3(x) \, dx\text{.}\) Which substitution would you choose to evaluate this integral?
  1. \(\displaystyle u=\sin(x)\)
  2. \(\displaystyle u=\cos^3(x)\)
  3. \(\displaystyle u=\cos(x)\)
  4. Substitution is not effective

Activity 5.3.4.

It’s possible to use substitution to evaluate \(\displaystyle\int \sin^4(x)\cos^3(x) \, dx\text{,}\) by taking advantage of the trigonometric identity \(\sin^2(x)+\cos^2(x)=1\text{.}\)
Complete the following substitution of \(u=\sin(x),\, du=\cos(x)\,dx\) by filling in the missing \(\unknown\)s.
\begin{align*} \int \sin^4(x)\cos^3(x)\,dx &=\int\sin^4(x)(\,\unknown\,)\cos(x)\,dx\\ &=\int\sin^4(x)(1-\unknown)\cos(x)\,dx\\ &= \int\unknown(1-\unknown)\,du\\ &= \int (u^4-u^6)\,du\\ &= \frac{1}{5}u^5-\frac{1}{7}u^7+C\\ &= \unknown \end{align*}

Activity 5.3.5.

Trying to substitute \(u=\cos(x),du=-\sin(x)\,dx\) in the previous example is less successful.
\begin{align*} \int \sin^4(x)\cos^3(x)\,dx &=-\int\sin^3(x)\cos^3(x)(-\sin(x)\,dx)\\ &=-\int\sin^3(x)u^3\,du\\ &= \cdots? \end{align*}
Which feature of \(\sin^4(x)\cos^3(x)\) made \(u=\sin(x)\) the better choice?
  1. The even power of \(\sin^4(x)\)
  2. The odd power of \(\cos^3(x)\)

Activity 5.3.6.

Try to show
\begin{equation*} \int \sin^5(x)\cos^2(x)\,dx= -\frac{1}{7} \, \cos^{7}\left(x\right) + \frac{2}{5} \, \cos^{5}\left(x\right) - \frac{1}{3} \, \cos^{3}\left(x\right)+C \end{equation*}
by first trying \(u=\sin(x)\text{,}\) and then trying \(u=\cos(x)\) instead.
Which substitution worked better and why?
  1. \(u=\sin(x)\) due to \(\sin^5(x)\)’s odd power.
  2. \(u=\sin(x)\) due to \(\cos^2(x)\)’s even power.
  3. \(u=\cos(x)\) due to \(\sin^5(x)\)’s odd power.
  4. \(u=\cos(x)\) due to \(\cos^2(x)\)’s even power.

Observation 5.3.7.

When integrating the form \(\displaystyle \int \sin^m(x)\cos^n(x)\,dx\text{:}\)
  • If \(\sin\)’s power is odd, rewrite the integral as \(\displaystyle \int g(\cos(x))\sin(x)\,dx\) and use \(u=\cos(x)\text{.}\)
  • If \(\cos\)’s power is odd, rewrite the integral as \(\displaystyle \int h(\sin(x))\cos(x)\,dx\) and use \(u=\sin(x)\text{.}\)

Activity 5.3.8.

Let’s consider \(\displaystyle\int \sin^2(x) \, dx\text{.}\)
(a)
Use the fact that \(\sin^2(\theta)=\displaystyle\frac{1-\cos(2\theta)}{2}\) to rewrite the integrand using the above identities as an integral involving \(\cos(2x)\text{.}\)
(b)
Show that the integral evaluates to \(\dfrac{1}{2} \, x - \dfrac{1}{4} \, \sin\left(2 \, x\right)+C\text{.}\)

Activity 5.3.9.

Let’s consider \(\displaystyle\int \sin^2(x)\cos^2(x) \, dx\text{.}\)
(a)
Use the fact that \(\cos^2(\theta)=\displaystyle\frac{1+\cos(2\theta)}{2}\) and \(\sin^2(\theta)=\displaystyle\frac{1-\cos(2\theta)}{2}\) to rewrite the integrand using the above identities as an integral involving \(\cos^2(2x)\text{.}\)
(b)
Use the above identities to rewrite this new integrand as one involving \(\cos(4x)\text{.}\)
(c)
Show that integral evaluates to \(\dfrac{1}{8} \, x - \dfrac{1}{32} \, \sin\left(4 \, x\right)+C\text{.}\)

Activity 5.3.10.

Consider \(\displaystyle\int \sin^4(x)\cos^4(x) \, dx\text{.}\) Which would be the most useful way to rewrite the integral?
  1. \(\displaystyle \displaystyle\int (1-\cos^2(x))^2\cos^4(x) \, dx\)
  2. \(\displaystyle \displaystyle\int \sin^4(x)(1-\sin^2(x))^2 \, dx\)
  3. \(\displaystyle \displaystyle\int \left(\frac{1-\cos(2x)}{2}\right)^2\left(\frac{1+\cos(2x)}{2}\right)^2 \, dx\)

Activity 5.3.11.

Consider \(\displaystyle\int \sin^3(x)\cos^5(x) \, dx\text{.}\) Which would be the most useful way to rewrite the integral?
  1. \(\displaystyle \displaystyle\int (1-\cos^2(x))\cos^5(x) \sin(x)\, dx\)
  2. \(\displaystyle \displaystyle\int \sin^3(x)\left(\frac{1+\cos(2x)}{2}\right)^2\cos(x) \, dx\)
  3. \(\displaystyle \displaystyle\int \sin^3(x)(1-\sin^2(x))^2\cos(x) \, dx\)

Remark 5.3.12.

We might also use some other trigonometric identities to manipulate our integrands, listed in Appendix B.

Activity 5.3.13.

Consider \(\displaystyle\int \sin(\theta)\sin(3\theta) \, d\theta\text{.}\)
(a)
Find an identity from Appendix B which could be used to transform our integrand.
(b)
Rewrite the integrand using the selected identity.
(c)
Evaluate the integral.

Subsection 5.3.2 Videos

Figure 108. Video: Compute integrals involving products of trigonometric functions

Subsection 5.3.3 Exercises

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