IN3.4: Integration of trigonometric functions

The integration of trigonometric functions in solving problems is related to oscillatory motion, waves and other periodic phenomena in physics and engineering.

This module deals with integration of trigonometric functions. These include: \[\begin{align*} & \int\sin\left(2x+3\right)dx\\ & \int\cos\left(5x\right)dx\\ & \int_{1}^{2}\sec^{2}\left(x-2\right)dx. \end{align*}\]

Indefinite integral (antiderivative) of a trigonometric function

Recall that:

\[\begin{align*} \frac{d}{dx}\cos\left(x\right) & =-\sin\left(x\right)\\ \frac{d}{dx}\sin\left(x\right) & =\cos\left(x\right)\\ \frac{d}{dx}\tan\left(x\right) & =\sec^{2}\left(x\right). \end{align*}\] It follows that indefinite integrals (or antiderivatives) of \(\sin\left(x\right),\ \cos\left(x\right)\) and \(\sec^{2}\left(x\right)\) are of the form \[\begin{align*} \int\sin\left(x\right)dx & =-\cos\left(x\right)+c\\ \int\cos\left(x\right)dx & =\sin\left(x\right)+c\\ \int\sec^{2}\left(x\right)dx & =\tan\left(x\right)+c \end{align*}\] where \(c\) is a constant.

More general forms are: 1 These forms may be derived using integration by substitution. For example, let \(u=ax+b\) then \(du/dx=a.\) Noting \(\sin\left(ax+b\right)=1\times\sin\left(ax+b\right)\) and using substitution, \[\begin{align*} \int\sin\left(ax+b\right)dx & =\int\frac{1}{a}\sin\left(u\right)\frac{du}{dx}dx\\ & =\frac{1}{a}\int\sin\left(u\right)du\\ & =\frac{1}{a}\left(-\cos\left(u\right)\right)+c\\ & =-\frac{1}{a}\cos\left(ax+b\ \right)+c. \end{align*}\] \[\begin{align*} \int\sin\left(ax+b\right)dx & =-\frac{1}{a}\cos\left(ax+b\right)+c\\ \int\cos\left(ax+b\right)dx & =\frac{1}{a}\sin\left(ax+b\right)+c\\ \int\sec^{2}\left(ax+b\right)dx & =\frac{1}{a}\tan\left(ax+b\right)+c \end{align*}\] where \(a,\,b\) and \(c\) are constants.

Examples

1. \(\int5\cos\left(x\right)dx=5\sin\left(x\right)+c,\ \left(a=1,\,b=0\right)\)
2. \(\int3\cos\left(3-2x\right)dx=-\frac{3}{2}\sin\left(3-2x\right)+c\ \left(a=-2,\;b=3\right)\)
3. \(\int3\sin\left(3-x\right)dx=3\cos\left(3-x\right)+c,\ \left(a=-1,\;b=3\right)\)
4. \(\int\sec^{2}\left(\frac{x}{2}\right)dx=2\tan\left(\frac{x}{2}\right)+c\ \left(a=\frac{1}{2},\;b=0\right)\).

Definite integral of a trigonometric function

Now that we know how to get an indefinite integral (or antiderivative) of a trigonometric function we can consider definite integrals. To evaluate a definite integral we determine an antiderivative and calculate the difference of the values of the antiderivatve at the limits defined in the definite integral. For example consider

\[\begin{align*} \int_{0}^{\pi/2}3\cos\left(x\right)dx. \end{align*}\] From the previous section we know an antiderivative is \(3\sin\left(x\right)+c\) where \(c\) is a constant. The limits of the integral are \(0\) and \(\pi/2\). So we have \[\begin{align*} \int_{0}^{\pi/2}3\cos\left(x\right)dx & =\left[3\sin\left(x\right)+c\right]_{x=0}^{x=\pi/2} & \left(1\right)\\ & =\left(3\sin\left(\frac{\pi}{2}\right)+c\right)-\left(3\sin\left(0\right)+c\right) & \left(2\right)\\ & =3+c-0-c & \left(3\right)\\ & =3. & \left(4\right) \end{align*}\]

Note that the notation in line \(\left(1\right)\) \[\begin{align*} \left[3\sin\left(x\right)+c\right]_{x=0}^{x=\pi/2} \end{align*}\] means substitute \(x=\pi/2\) in the expression in brackets and subtract the expression in brackets evaluated at \(x=0.\)

Note also that the constant \(c\) in lines \(\left(1\right)\) to \(\left(3\right)\) has no effect when evaluating a definite integral. Consequently we usually leave it out and write \[\begin{align*} \int_{0}^{\pi/2}3\cos\left(x\right)dx & =\left[3\sin\left(x\right)\right]_{x=0}^{x=\pi/2} & \left(1\right)\\ & =\left(3\sin\left(\frac{\pi}{2}\right)\right)-\left(3\sin\left(0\right)\right) & \left(2\right)\\ & =3-0 & \left(3\right)\\ & =3. & \left(4\right) \end{align*}\]

Examples

1. Evaluate \(\int_{0}^{\,\pi/4}5\cos\left(x\right)dx.\)
   Solution: 2 Remember that \(\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}.\) \[\begin{align*} \int_{0}^{\,\pi/4}5\cos\left(x\right)dx & =\left[5\sin\left(x\right)\right]_{x=0}^{x=\pi/4}\\ & =5\sin\left(\frac{\pi}{4}\right)-5\sin\left(0\right)\\ & =\frac{5}{\sqrt{2}} \end{align*}\]
2. Evaluate \(\int_{-\pi/4}^{\pi/4}3\cos\left(\pi-2x\right)dx\).
   Solution: 3 Remember that \(\sin\left(\pi/2\right)=1\) and \(\sin\left(3\pi/2\right)=-1.\) \[\begin{align*} \int_{-\pi/4}^{\pi/4}3\cos\left(\pi-2x\right)dx & =\left[-\frac{3}{2}\sin\left(\pi-2x\right)\right]_{x=-\pi/4}^{x=\pi/4}\\ & =\left(-\frac{3}{2}\sin\left(\pi-\frac{\pi}{2}\right)\right)-\left(-\frac{3}{2}\sin\left(\pi-2\left(-\frac{\pi}{4}\right)\right)\right)\\ & =\left(-\frac{3}{2}\sin\left(\frac{\pi}{2}\right)\right)-\left(-\frac{3}{2}\sin\left(\pi+\frac{\pi}{2}\right)\right)\\ & =-\frac{3}{2}-\left(-\frac{3}{2}\left(-1\right)\right)\\ & =-\frac{3}{2}-\frac{3}{2}\\ & =-3. \end{align*}\]
3. Evaluate \(\int_{0}^{\pi/6}3\sin\left(\pi-2x\right)dx\).
   Solution: 4 Remember that \(\cos\left(2\pi/3\right)=-1/2\) and \(\cos\left(\pi\right)=-1.\) \[\begin{align*} \int_{0}^{\pi/6}3\sin\left(\pi-2x\right)dx & =\left[\frac{3}{2}\cos\left(\pi-2x\right)\right]_{x=0}^{x=\pi/6}\\ & =\left(\frac{3}{2}\cos\left(\pi-2\left(\frac{\pi}{6}\right)\right)\right)-\left(\frac{3}{2}\cos\left(\pi\right)\right)\\ & =\left(\frac{3}{2}\cos\left(\pi-\frac{\pi}{3}\right)\right)-\left(\frac{3}{2}\left(-1\right)\right)\\ & =\left(\frac{3}{2}\cos\left(\frac{2\pi}{3}\right)\right)+\frac{3}{2}\\ & =\frac{3}{2}\left(-\frac{1}{2}\right)+\frac{3}{2}\\ & =\frac{3}{4}\\ & =0.75 \end{align*}\]
4. Evaluate \(\int_{0}^{2\pi/3}\sec^{2}\left(\frac{x}{2}\right)dx.\)
   Solution: 5 Remember that \(\tan\left(0\right)=0\) and \(\tan\left(\pi/3\right)=\sqrt{3}\). \[\begin{align*} \int_{0}^{2\pi/3}\sec^{2}\left(\frac{x}{2}\right)dx & =\left[2\tan\left(\frac{x}{2}\right)\right]_{x=0}^{x=2\pi/3}\\ & =\left(2\tan\left(\frac{1}{2}\times\frac{2\pi}{3}\right)\right)-\left(2\tan\left(0\right)\right)\\ & =2\tan\left(\frac{\pi}{3}\right)\\ & =2\sqrt{3}. \end{align*}\]

Exercises

1. Calculate:
   \(\begin{aligned}a) & \int\sec^{2}\left(4x\right)dx & b) & \int2\cos\left(1-x\right)dx & c) & 2\sin\left(\frac{5-3x}{4}\right)dx\text{ }\end{aligned}\)
2. Evaluate:
   \(\begin{aligned}a) & \intop_{0}^{\pi/2}3\cos\left(2x+\pi\right)dx & b) & \int_{-\pi}^{0}5\sin\left(x/2\right)dx & c) & \int_{0}^{\pi/2}2\sec^{2}\left(x/3\right)\end{aligned} dx\)