Objectives
- Apply strategies for evaluating integrals of the forms for integers \(m,n\geq 0\text{,}\) potentially using trigonometric identities:
- \(\displaystyle \ds \int \sin^mx\cos^nx\, dx\)
- \(\displaystyle \ds \int \sec^mx\tan^nx\, dx\)
Section 7.2 Trigonometric Integrals
Subsection 7.2.1 Before Class
https://mymedia.ou.edu/media/7.2-1/1_asqv45iuSubsubsection 7.2.1.1 Trigonometric Identities
Here are some identities from trigonometry which will be helpful as we move through this section:
| \(\sin^2\theta + \cos^2\theta = 1\) | \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(\cos 2\theta = 1-2\sin^2\theta\) | \(\cos 2\theta = 2\cos^2\theta - 1\) |
| \(\tan^2\theta + 1 = \sec^2\theta\) | \(1+\cot^2\theta = \csc^2\theta\) |
| \(\sin (\theta + \phi) = \sin\theta\cos\phi + \sin\phi\cos\theta\) | \(\sin(\theta - \phi) =\sin\theta\cos\phi -\sin\phi\cos\theta\) |
| \(\cos (\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi\) | \(\cos(\theta - \phi) = \cos\theta\cos\phi + \sin\theta\sin\phi\) |
Occasionally, you’ll want to rearrange some of these identities to arrive at new ones:
| \(\sin\theta\cos\phi = \dfrac{1}{2}[\sin(\theta - \phi) + \sin (\theta + \phi)]\) | \(\sin\theta\sin\phi = \dfrac{1}{2}[\cos(\theta - \phi) - \cos(\theta + \phi)]\) |
| \(\cos\theta\cos\phi =\dfrac{1}{2}[\cos(\theta-\phi) + \cos (\theta + \phi)]\) |
Subsubsection 7.2.1.2 Trigonometric Integrals
Example 7.2.1.
Compute \(\ds \int \sin^3 x\, dx\)
Solution.
\(\ds \int \sin^3 x\, dx = -\cos x + \dfrac{1}{3}\cos^3x + C\)
Example 7.2.2.
Compute \(\ds \int \sin^5x\cos^2x\, dx\)
Solution.
\(\ds \int \sin^5x\cos^2x\, dx = -\dfrac{1}{7}\cos^7x + \dfrac{2}{5}\cos^5x-\dfrac{1}{3}\cos^3x + C\)
Example 7.2.3.
Find the area under the curve of \(f(x) = \sin^2x\) from \(0\) to \(\pi\)
Solution.
\(\ds \int_0^{\pi} \sin^2x\, dx = \dfrac{\pi}{2}\)
Example 7.2.4.
Compute \(\ds \int \cos^4x\, dx\)
Solution.
\(\ds \int \cos^4x\, dx = \dfrac{3}{8}x + \dfrac{1}{5}\sin 2x + \dfrac{1}{32}\sin 4x + C\)
Subsection 7.2.2 Pre-Class Activities
Example 7.2.5.
Use this space to write any questions you might have from the videos.
Solution.
Answers vary
Example 7.2.6.
Evaluate \(\ds \int \sin^2 x\cos^3x\, dx\)
Solution.
\(\ds \int \sin^2 x\cos^3x\, dx = \dfrac{1}{3}\sin^3x - \dfrac{1}{5}\sin^5x + C\)
Example 7.2.7.
Evaluate \(\ds \int \sin x\cos x\, dx\) in four ways: (1) using the substitution \(u = \cos x\text{;}\) (2) using the substitution \(u = \sin x\text{;}\) (3) using the double-angle identity for sine; (4) using integration by parts. Compare your work between the four methods.
Solution.
- Using the cosine substitution, we get \(-\dfrac{1}{2}\cos^2x+C\)
- Using the sine substitution, we get \(\dfrac{1}{2}\sin^2x + C\)
- Using the double-angle identity, we get \(-\dfrac{1}{4}\cos 2x + C\)
- If we set \(u=\sin x\) and \(dv = \cos x\, dx\text{,}\) we get\begin{equation*} \ds \int \sin x\cos x\, dx = \dfrac{1}{2}\sin^2x + C \end{equation*}However, if we set \(u=\cos x\) and \(dv = \sin x\, dx\text{,}\) we get\begin{equation*} \ds \int \sin x\cos x\, dx = -\dfrac{1}{2}\cos^2x + C \end{equation*}
It seems as though we should get the same answer for all four integrals, but we don’t. However, each of these answers can be transformed into the other by using the trigonometric identities listed in the identity table earlier.
Subsection 7.2.3 In Class
Subsubsection 7.2.3.1 Strategies for Trig Integrals
Evaluating \(\ds \int \sin^mx\cos^nx\, dx\).
- If the power of cosine is odd:\begin{align*} \ds \int \sin^mx\cos^{2k+1}x\, dx \amp = \int\sin^m x\cos^{2k}x\cdot \cos x\, dx\\ \amp = \ds \int \sin^m x (\cos^2x)^k\cos x\, dx\\ \amp = \ds \int \sin^m x (1-\sin^2x)^k\cos x\, dx \end{align*}Now set \(u = \sin x\text{,}\) simplify, and integrate.
- If the power of sine is odd:\begin{align*} \ds \int \sin^{2k+1}x\cos^n x\, dx \amp = \int (\sin^2 x)^k\cdot \sin x\cdot \cos^n x\, dx\\ \amp = \ds \int (1-\cos^2x)^k\cdot \cos^n x\cdot \sin x\, dx \end{align*}Now set \(u = \cos x\text{,}\) simplify, and integrate.
- If the powers of sine and cosine are even: use the power reducing identities \(\sin^2x = \dfrac{1}{2}(1-\cos 2x)\) or \(\cos^2x = \dfrac{1}{2}(1+\cos 2x)\)
Example 7.2.8.
Evaluate \(\ds \int \sin^3\theta\cos^4\theta \, d\theta\)
Solution.
\(\ds \int \sin^3\theta\cos^4\theta \, d\theta = \dfrac{1}{7}\cos^7x - \dfrac{1}{5}\cos^5x + C\)
Example 7.2.9.
Evaluate \(\ds \int \tan x \sec x\, dx\)
Solution.
\(\ds \int \tan x \sec x\, dx = \sec x + C\)
Evaluating \(\ds \int \tan^mx\sec^nx\, dx\).
- If the power of secant is even:\begin{align*} \ds \int \tan^mx\sec^{2k}x\, dx \amp = \int \tan^m x\sec^{2k-2}\cdot \sec^2x\, dx\\ \amp = \ds \int \tan^m x (\sec^2x)^{k-1}\cdot \sec^2 x\, dx\\ \amp = \ds \int \tan^m x (1+\tan^2x)^{k-1}\cdot \sec^2 x\, dx \end{align*}Now set \(u = \tan x\text{,}\) simplify, and integrate.
- If the power of tangent is odd:\begin{align*} \ds \int \tan^{2k+1}x \sec^n x\, dx \amp = \int \tan^{2k} x\cdot \tan x \cdot \sec^{n-1} x\cdot \sec x\, dx\\ \amp = \ds \int (\tan^2x)^k\cdot\sec^{n-1}x\cdot \sec x\tan x\, dx\\ \amp = \ds \int (\sec^2x-1)^k\cdot \sec^{n-1}x\cdot \sec x\tan x\, dx \end{align*}Now let \(u = \sec x\text{,}\) simplify, and integrate.
Example 7.2.10.
Evaluate \(\ds \int \tan^3\theta\sec^5\theta\, d\theta\)
Solution.
\(\ds \int \tan^3\theta\sec^5\theta\, d\theta = \dfrac{1}{7}\sec^7 x - \dfrac{1}{5}\sec^5 x + C\)
Example 7.2.11.
Show that \(\ds \int \sec x\, dx = \ln |\sec x + \tan x| + C\)
Solution.
Multiply the integrand by a form of one:
\begin{equation*}
\sec x = \sec x\cdot \dfrac{\sec x + \tan x}{\sec x + \tan x}
\end{equation*}
So that we have
\begin{equation*}
\ds \int \sec x\, dx = \int \dfrac{\sec^2x + \sec x\tan x}{\sec x + \tan x}\, dx
\end{equation*}
Let \(u = \sec x + \tan x\text{.}\) So, \(du = (\sec x\tan x + \sec^2x)\, dx\text{.}\) Rewriting gives
\begin{equation*}
\ds \int \dfrac{\sec^2x + \sec x\tan x}{\sec x + \tan x}\, dx = \int \dfrac{1}{u}\, du = \ln |u| + C
\end{equation*}
Replacing gives
\begin{equation*}
\ln |\sec x + \tan x| + C
\end{equation*}
Example 7.2.12.
Evaluate \(\ds \int \tan^3x\, dx\)
Solution.
\(\ds \int \tan^3x\, dx = \dfrac{1}{2}\sec^2 x + \ln |\cos x| + C\)
Example 7.2.13.
Find \(\ds \int \sec^3x\, dx\)
Solution.
\(\ds \int \sec^3x\, dx = \dfrac{1}{2}(\sec x\tan x + \ln |\sec x + \tan x|) + C\)
Example 7.2.14.
Evaluate \(\ds \int \sin 3x\cos 4x\, dx\)
Solution.
\(\ds \int \sin 3x\cos 4x\, dx = \dfrac{1}{2}\cos x - \dfrac{1}{14}\cos 7x + C\)
Example 7.2.15.
Evaluate \(\ds \int \cos 5x\cos 7x\, dx\)
Solution.
\(\ds \int \cos 5x\cos 7x\, dx = \dfrac{1}{4}\sin 2x + \dfrac{1}{24}\sin 12x + C\)
Example 7.2.16.
Evaluate \(\ds \int_{\pi/4}^{\pi/2} \csc^4\theta\cot^4\theta\, d\theta\)
Solution.
\(\ds \int_{\pi/4}^{\pi/2} \csc^4\theta\cot^4\theta\, d\theta = \dfrac{12}{35}\)
Example 7.2.17.
Find \(\ds \int_0^{\pi/2} \sin^2x\cos^2x\, dx\)
Solution.
\(\ds \int_0^{\pi/2} \sin^2x\cos^2x\, dx = \dfrac{\pi}{16}\)
Example 7.2.18.
Evaluate \(\ds \int t\sin^2 t\, dt\)
Solution.
\(\ds \int t\sin^2 t\, dt = \dfrac{1}{4}t^2-\dfrac{1}{4}t\sin 2t - \dfrac{1}{8}\cos 2t + C\)
Subsection 7.2.4 After Class Activities
Example 7.2.19.
Explain why \(\ds \int_{-\pi}^{\pi} \sin mx \cos nx = 0\) for any positive integers \(m\) and \(n\text{.}\) There are several ways of coming to this answer!
Solution.
Answers vary
Example 7.2.20.
Evaluate \(\ds \int_0^\pi \cos^4(2t)\, dt\)
Solution.
\(\ds \int_0^\pi \cos^4(2t)\, dt = \dfrac{3\pi}{8}\)
Example 7.2.21.
Evaluate \(\ds \int \sqrt{\cos\theta}\sin^3\theta \, d\theta\)
Solution.
\(\ds \int \sqrt{\cos\theta}\sin^3\theta \, d\theta = \dfrac{2}{7}\cos^{7/2}\theta - \dfrac{2}{3}\cos^{3/2}\theta + C\)
Example 7.2.22.
Find \(\ds \int \tan^2\theta \sec^4\theta\, d\theta\)
Solution.
\(\ds \int \tan^2\theta \sec^4\theta\, d\theta = \dfrac{1}{3}\tan^3 \theta + \dfrac{1}{5}\tan^5\theta + C\)
Example 7.2.23.
Evaluate \(\ds \int_0^{\pi/2} \cos 5t\cos 10t\, dt\)
Solution.
\(\ds \int_0^{\pi/2} \cos 5t\cos 10t\, dt = \dfrac{1}{10}\sin 5t + \dfrac{1}{30}\sin 15t + C\)
Example 7.2.24.
Evaluate \(\ds \int \dfrac{dx}{\cos x - 1}\)
Solution.
\(\ds \int \dfrac{dx}{\cos x - 1} = \csc x + \cot x + C\)
