Learning Outcomes
- Compute integrals involving products of trigonometric functions.
Section 13.3 Integration of Trigonometry (TI3)
Subsection 13.3.1 Activities
Activity 13.46.
Consider \(\displaystyle\int \sin(x)\cos(x) \, dx\text{.}\) Which substitution would you choose to evaluate this integral?
- \(\displaystyle u=\sin(x)\)
- \(\displaystyle u=\cos(x)\)
- \(\displaystyle u=\sin(x)\cos(x)\)
- Substitution is not effective
Activity 13.47.
Consider \(\displaystyle\int \sin^4(x)\cos(x) \, dx\text{.}\) Which substitution would you choose to evaluate this integral?
- \(\displaystyle u=\sin(x)\)
- \(\displaystyle u=\sin^4(x)\)
- \(\displaystyle u=\cos(x)\)
- Substitution is not effective
Activity 13.48.
Consider \(\displaystyle\int \sin^4(x)\cos^3(x) \, dx\text{.}\) Which substitution would you choose to evaluate this integral?
- \(\displaystyle u=\sin(x)\)
- \(\displaystyle u=\cos^3(x)\)
- \(\displaystyle u=\cos(x)\)
- Substitution is not effective
Activity 13.49.
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 13.50.
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?
- The even power of \(\sin^4(x)\)
- The odd power of \(\cos^3(x)\)
Activity 13.51.
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?
- \(u=\sin(x)\) due to \(\sin^5(x)\)’s odd power.
- \(u=\sin(x)\) due to \(\cos^2(x)\)’s even power.
- \(u=\cos(x)\) due to \(\sin^5(x)\)’s odd power.
- \(u=\cos(x)\) due to \(\cos^2(x)\)’s even power.
Observation 13.52.
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 13.53.
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 13.54.
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 13.55.
Consider \(\displaystyle\int \sin^4(x)\cos^4(x) \, dx\text{.}\) Which would be the most useful way to rewrite the integral?
- \(\displaystyle \displaystyle\int (1-\cos^2(x))^2\cos^4(x) \, dx\)
- \(\displaystyle \displaystyle\int \sin^4(x)(1-\sin^2(x))^2 \, dx\)
- \(\displaystyle \displaystyle\int \left(\frac{1-\cos(2x)}{2}\right)^2\left(\frac{1+\cos(2x)}{2}\right)^2 \, dx\)
Activity 13.56.
Consider \(\displaystyle\int \sin^3(x)\cos^5(x) \, dx\text{.}\) Which would be the most useful way to rewrite the integral?
- \(\displaystyle \displaystyle\int (1-\cos^2(x))\cos^5(x) \sin(x)\, dx\)
- \(\displaystyle \displaystyle\int \sin^3(x)\left(\frac{1+\cos(2x)}{2}\right)^2\cos(x) \, dx\)
- \(\displaystyle \displaystyle\int \sin^3(x)(1-\sin^2(x))^2\cos(x) \, dx\)
Remark 13.57.
We might also use some other trigonometric identities to manipulate our integrands, listed in Section B.2.
Activity 13.58.
Consider \(\displaystyle\int \sin(\theta)\sin(3\theta) \, d\theta\text{.}\)
(a)
Find an identity from Section B.2 which could be used to transform our integrand.
(b)
Rewrite the integrand using the selected identity.
(c)
Evaluate the integral.
Subsection 13.3.2 Videos
Subsection 13.3.3 Exercises
Exercises available at
https://tbil.org/preview/calculus/exercises/#/bank/TI3/
