Integration of trigonometric functions

How do you integrate a trigonometic function? This skill is important for solving problems involving wave patterns in physics, analysing alternating current circuits in engineering and determining structural loads with curved geomoetries in architecture. Use this resource to learn how.

You already learnt about the general forms of Circular functions. Examples include \(\sin(2x+3)\), \(\cos(5x)\) and \(\sec^{2}(x-2)\).

Remember:
\[\dfrac{d}{dx}(\cos(x))dx=-\sin(x)\] \[\dfrac{d}{dx}(\sin(x))dx=\cos(x)\] \[\dfrac{d}{dx}(\tan(x))dx=\sec^{2}(x)\]

This means that the antiderivatives are:

\[\int\sin(x)dx=-\cos(x)+c\] \[\int\cos(x)dx=\sin(x)+c\] \[\int\sec^{2}(x)dx=\tan(x)+c\]

where \(c\) is a constant.

The more general forms are:

\[\begin{align*} \int\sin\left(ax+b\right)dx & =-\frac{1}{a}\cos\left(ax+b\right)+c\\ \int\cos\left(ax+b\right)dx & =\frac{1}{a}\sin\left(ax+b\right)+c\\ \int\sec^{2}\left(ax+b\right)dx & =\frac{1}{a}\tan\left(ax+b\right)+c \end{align*}\]

where \(a\), \(b\) and \(c\) are constants.

Example 1 – integrating trigonometric functions

Find \(\int5\cos(x)dx\).

Here, \(a=1\) and \(b=0\).
\[\int5\cos(x)dx=5\sin(x)+c\]

Integrate \(3\cos(3-2x)\) with respect to \(x\).

Here, \(a=-2\) and \(b=3\).
\[\int3\cos(3-2x)dx=-\dfrac{3}{2}\sin(3-2x)+c\]

Find \(\int3\sin(3-x)dx\).

Here, \(a=-1\) and \(b=3\).
\[\int3\sin(3-x)dx=3\cos(3-x)+c\]

Evaluate \(\int_{0}^{\frac{\pi}{2}}3\cos(x)dx\).
\[\begin{align*} \int_{0}^{\frac{\pi}{2}}3\cos(x)dx & = \left[3\sin(x)+c\right]_{x=0}^{x=\frac{\pi}{2}}\\
& = \left(3\sin\left(\frac{\pi}{2}\right)+c\right)-\left(3\sin\left(0\right)+c\right)\\
& =3+c-0-c\\
& =3
\end{align*}\]

Evaluate \(\int_{0}^{\frac{\pi}{4}}5\cos(x)dx\).
\[\begin{align*} \int_{0}^{\frac{\pi}{4}}5\cos(x)dx & = \left[5\sin(x)\right]_{x=0}^{x=\frac{\pi}{4}}\\
& = 5\sin\left(\frac{\pi}{4}\right)-5\sin(0)\\
& = \frac{5}{\sqrt{2}}
\end{align*}\]

Evaluate \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}3\cos(\pi-2x)dx\).

Remember that \(\sin\left(\dfrac{\pi}{2}\right)=1\) and \(\sin\left(\dfrac{3\pi}{2}\right)=-1\).
\[\begin{align*} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}3\cos(\pi-2x)dx & = \left[-\frac{3}{2}\sin(\pi-2x)\right]_{x=-\frac{\pi}{4}}^{x=\frac{\pi}{4}}\\
& = \left(-\frac{3}{2}\sin\left(\pi-\frac{\pi}{2}\right)\right)-\left(-\frac{3}{2}\sin\left(\pi-2\left(-\frac{\pi}{4}\right)\right)\right)\\
& = \left(-\frac{3}{2}\sin\left(\frac{\pi}{2}\right)\right)-\left(-\frac{3}{2}\sin\left(\pi+\frac{\pi}{2}\right)\right)\\
& = -\frac{3}{2}-\left(-\frac{3}{2}\left(-1\right)\right)\\
& = -\frac{3}{2}-\frac{3}{2}\\
& = -3
\end{align*}\]

Exercises

Calculate the following.

\(\int\sec^{2}\left(\dfrac{x}{2}\right)dx\)
\(\int\sec^{2}(4x)dx\)
\(\int2\cos(1-x)dx\)
\(2\sin\left(\dfrac{5-3x}{4}\right)dx\)

Evaluate the following.

\(\int_{0}^{\frac{\pi}{2}}3\cos(2x+\pi)dx\)
\(\int_{-\pi}^{0}5\sin\left(\dfrac{x}{2}\right)dx\)
\(\int_{0}^{\frac{\pi}{2}}2\sec^{2}\left(\dfrac{x}{3}\right)dx\)

Evaluate the folowing.

\(\int_{0}^{\frac{\pi}{6}}3\sin(\pi-2x)dx\)
\(\int_{0}^{\frac{2\pi}{3}}\sec^{2}\left(\dfrac{x}{2}\right)dx\)

1. \(2\tan\left(\dfrac{x}{2}\right)+c\)
2. \(\dfrac{1}{4}\tan(4x)+c\)
3. \(-2\sin(1-x)+c\)
4. \(\dfrac{8}{3}\cos\left(\dfrac{5-3x}{4}\right)+c\)
1. \(0\)
2. \(-10\)
3. \(2\sqrt{3}\)
1. \(0.75\)
2. \(2\sqrt{3}\)