Learning Lab

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: 11 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.

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 \[\begin{align*} \int_{0}^{\pi/2}3\cos\left(x\right)dx & =\left[3\sin\left(x\right)\right]_{x=0}^{x=\pi/2} & \left(1\right)\\ & =\left(3\sin\left(\frac{\pi}{2}\right)\right)-\left(3\sin\left(0\right)\right) & \left(2\right)\\ & =3-0 & \left(3\right)\\ & =3. & \left(4\right) \end{align*}\]

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:
   \(\begin{aligned}a) & \intop_{0}^{\pi/2}3\cos\left(2x+\pi\right)dx & b) & \int_{-\pi}^{0}5\sin\left(x/2\right)dx & c) & \int_{0}^{\pi/2}2\sec^{2}\left(x/3\right)\end{aligned} dx\)