Further Integration of Trigonometric Functions

Video lesson · from 1:04:293 worked examples

One lesson video covers all of Integration, so it opens at 1:04:29 for this topic — not from the beginning.

We have previously learned how to differentiate trigonometric functions of the form $\sin(ax+b)$ and $\cos(ax+b)$ .

When integrating these, we must divide by the derivative of the angle (which is $a$ ):

\int \cos(ax+b) \, dx = \frac{1}{a} \sin(ax+b) + C

\int \sin(ax+b) \, dx = -\frac{1}{a} \cos(ax+b) + C

The minus sign swaps sides: $\int \sin x\,dx = -\cos x + C$ , while $\int \cos x\,dx = \sin x + C$ . This is the reverse of differentiation, and mixing the two up is the most common error in the topic.
Divide by the coefficient of $x$ : $\int \cos(4x)\,dx = \frac{1}{4}\sin(4x) + C$ . The chain factor divides when integrating and multiplies when differentiating.
Work in radians: these results are only valid in radians, and definite integrals must have radian limits.
Check by differentiating: it takes seconds and catches both the sign and the coefficient.

Example 1

Find $\int \cos 4x \, dx$ .

\int \cos 4x \, dx = \frac{1}{4} \sin 4x + C

Example 2

Find $\int \sin\left(4x + \frac{3\pi}{4}\right) \, dx$ .

\int \sin\left(4x + \frac{3\pi}{4}\right) \, dx = -\frac{1}{4} \cos\left(4x + \frac{3\pi}{4}\right) + C

Example 3

Find the area enclosed between the graph of $y = \sin\left(3x + \frac{\pi}{6}\right)$ , the x-axis, and the lines $x=0$ and $x = \frac{\pi}{6}$ .

We set up a definite integral:

\text{Area} = \int_{0}^{\pi/6} \sin\left(3x + \frac{\pi}{6}\right) \, dx

\begin{aligned} &= \left[ -\frac{1}{3} \cos\left(3x + \frac{\pi}{6}\right) \right]_{0}^{\pi/6} \\ &= \left( -\frac{1}{3} \cos\left(3\left(\frac{\pi}{6}\right) + \frac{\pi}{6}\right) \right) - \left( -\frac{1}{3} \cos\left(3(0) + \frac{\pi}{6}\right) \right) \\ &= \left( -\frac{1}{3} \cos\left(\frac{\pi}{2} + \frac{\pi}{6}\right) \right) - \left( -\frac{1}{3} \cos\left(\frac{\pi}{6}\right) \right) \end{aligned}

Since $\frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3}$ :

\begin{aligned} &= \left( -\frac{1}{3} \left(-\frac{1}{2}\right) \right) - \left( -\frac{1}{3} \left(\frac{\sqrt{3}}{2}\right) \right) \\ &= \frac{1}{6} + \frac{\sqrt{3}}{6} \\ &= \frac{1 + \sqrt{3}}{6} \end{aligned}

The area is $\frac{1 + \sqrt{3}}{6}$ square units.