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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 Section A.2.

Activity 5.3.13.

Consider \(\displaystyle\int \sin(\theta)\sin(3\theta) \, d\theta\text{.}\)
(a)
Find an identity from Section A.2 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 113. Video: Compute integrals involving products of trigonometric functions

Subsection 5.3.3 Exercises

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