We discuss ways of computing indefinite integrals involving trigonometric functions.
Subsection3.2.1Products of Trigonometric Functions
We first study how to integrate functions of the form \(\sin(mx)\sin(nx)\text{,}\)\(\cos(mx)\cos(nx)\) and \(\sin(mx)\cos(nx)\text{.}\) The idea for computing such integrals is to convert the integrand, which is a product of trigonometric functions, into a sum of trigonometric functions. This can be done using trigonometric identities in Appendix A. Here we offer another way of deducing them. First, it follows from Euler’s formula that
\begin{align*}
\int \sin(mx) &\cos(nx)dx =
\int\dfrac{1}{2}(\sin(m+n)(x)
+ \sin(m-n)(x))dx\\
&= \begin{cases} -\dfrac{1}{2(m+n)}\cos(m+n)(x)
- \dfrac{1}{2(m-n)}\cos(m-n)(x) + C & m \neq n \\ -
\dfrac{1}{4m}\cos(2mx) + C. & m = n
\end{cases}
\end{align*}
If either \(m\) or \(n\) is odd then the suggested strategy turns the differential form to be integrated into a form \(p(u) du\) where \(p(u)\) is a polynomial in \(u\text{.}\) This last form can be integrated easily solving the original integral. For instance,
\begin{align*}
\sin^{2k+1}(x)\cos^m(x)dx &=
(\sin^{2}(x))^k\cos^m(x)\sin(x)dx\\
&=-(1-u^2)^k u^m du.
\end{align*}
Certainly, both \(m\) and \(n\) can be odd, then either the first or the second substitution will work.
If both \(m\) and \(n\) are even, then an application of (A.14) will halve the powers involved. Repeat the process until an odd power of either sine or cosine appears. That will bring us back to either the first or the second case. The following examples illustrate how to apply these strategies.
Just like before, if \(m\text{,}\) the power of tangent (equivalently the power of sine), is odd, one makes the substitution \(u=\cos(x)\) and turn the integrand into sum of negative powers of \(u\text{.}\)
If the power of tangent is even, then one turns the integrand into a sum of powers of \(\sec(x)\) and uses the reduction formula for secant Item 3.5.3.d to handle the rest. This reduces the problem of finding the integral to integrating a single power of \(\sec(x)\) which we have found earlier (Example 2.7, see also half-angle substitution 3.2.3).
The strategies above express the integral in terms of integrals of power of secants. Even though they cover all cases, they may not be the most efficient ways in handling high even power of tangents. In those cases, the substitution \(u=\tan(x)\) leads to a simpler expression of the antiderivatives as that avoids expressing a high even power of tangent in terms of secant. Let us illustrate this point with a couple examples.
Example3.19.
Let us compute \(\int \tan^{10}(x)\sec^2(x) dx\text{.}\) Instead of expression the integrand in terms of secant and use the secant reduction formula, let \(u = \tan(x)\text{.}\) Then \(du = \sec^2(x)dx\) and the integral turns into
\begin{gather*}
\int u^{10} du = \frac{1}{11}u^{11} +C =
\frac{1}{11}\tan^{11}(x) + C.
\end{gather*}
Example3.20.
Compute \(\int \tan^{8}(x)dx\text{.}\) First, turn a \(\tan^2(x)\) into \(\sec^2(x)-1\text{.}\) That is write \(\tan^{8}(x)\) as
There is a way to convert integrals of rational functions in sine and cosine into integrals of rational functions of a single variable. It comes from the following parametrization of the unit circle.
Figure3.23.Half-Angle Substitution
By intersecting the line \(\ell\) with the unit circle, we get
The form on the right-hand side is a differential form involving a rational function in \(t\) which we will study systematically in Section 3.4. For now let us apply this technique to compute the indefinite integral of \(\sec(x)\text{.}\)
