Caluclus: Table of contents

Integration of Trigonometric Functions

Integrals of the form \(\displaystyle \int \sin^m x\, \cos ^n x\,dx\)

The integrals of the form \(\displaystyle \int \sin^m x\, \cos ^n x\,dx\) are very easy to evaluate in the case when one (or both) of the numbers \(m\) and \(n\) is odd integer.

Example 1 Evaluate the integral \(\displaystyle \int \sqrt[3]{\cos x}\cdot \sin^3x\,dx\).

Example 2 Evaluate the integral \(\displaystyle \int \sec x\,dx=\int\frac1{\cos x}\,dx\).

If both \(m\) and \(n\) are even, then use the formulas \(\cos^2\theta=\frac{1+\cos(2\theta)}2\) and \(\sin^2\theta=\frac{1-\cos(2\theta)}2\) to reduce them to the previous case.

Example 3 Evaluate the integral \(\displaystyle \int \cos^4x\,\sin^2 x\,dx\).

Integrals of the form \(\displaystyle \int \frac{P(\tan x)}{Q(\tan x)}\,dx\)

If the integrand is a rational function of \(\tan x\) the substitution \(\tan x=t\) transforms the integral in a rational function because \(x=\tan^{-1}t\) and consequently \(dx=\frac1{1+t^2}\,dt\) which is rational function as well.

It is worth noting that many integrals that do not appear to be of this form can be re-written as rational functions of \(\tan\) by dividing both numerator and denominator by the appropriate factor.

Example 3 Evaluate the integral \(\displaystyle \int \frac{2\sin^3x+\sin^2x\cos x-4\sin x\cos^2x+3\cos^3x}{\left(\sin^2x-\sin x\cos x-2\cos^2x\right)\cos x}\,dx\).

Integrals of the form \(\displaystyle \int\frac{P(\sin x, \cos x)}{Q(\sin x, \cos x)}\,dx\)

The magical substitution \(u=\tan\frac{x}2\) can bring the most general trigonometric integral to an integral of a rational function.

The the substitution owes its power to the fact that \(dx\), \(\sin x\), and \(\cos x\) can be expressed in terms of \(u\) in a relatively simple manner. First of all, \(x=2\tan^{-1}u\) and \(dx=\frac{2du}{u^2+1}\). We need to express \(\sin x\) and \(\cos x\) in terms of \(u\). \begin{eqnarray*}\sin x&=&2\sin\frac{x}2\cos\frac{x}2=2\frac{\sin{x}2}{\cos\frac{x}2}\cdot\cos^2\frac{x}2=2\tan\frac{x}2\cdot \frac1{\frac1{\cos^2\frac{x}2}}=\frac{2u}{1+u^2}\\ \cos x&=&\cos^2\frac{x}2-\sin^2\frac{x}2=\cos^2\frac{x}2\left(1-\tan\frac{x}2\right)=\frac1{\frac1{\cos^2\frac{x}2}}\left(1-u^2\right)=\frac{1-u^2}{1+u^2}.\end{eqnarray*}

Let us summarize the substitution: \begin{eqnarray*} u&=&\tan\frac{x}2\\ dx&=&\frac{2\,du}{u^2+1}\\ \sin x&=&\frac{2u}{u^2+1}\\ \cos x&=&\frac{1-u^2}{1+u^2}.\end{eqnarray*}

Example 4 Evaluate the integral \(\displaystyle \int\frac{1}{1+\sin x}\,dx\).

In the beginning we learned how to use the substitutions \(u=\sin x\), \(u=\cos x\), and \(u=\tan x\). This paragraph showed that these substitutions are unnecessary, as the substitution \(u=\tan\frac{x}2\) is more powerful than any of the substitutions we covered. However, the integrals resulting from substituting \(u=\tan\frac{x}2\) may be very unpleasant. One should attempt to use \(u=\sin x\), \(u=\cos x\), or \(u=\tan x\) whenever possible.

Practice problems

Problem 1. Evaluate the integral \(\displaystyle \int \sin^{10}x\cos x \,dx\).

Problem 2. Evaluate the integral \(\displaystyle \int \sin^3 x\cos^{6}x\,dx\).

Problem 3. Evaluate the integral \(\displaystyle \int_0^{\frac{\pi}4}\frac{ 3\sin x +2 \cos x}{ 4\sin x+ 3\cos x} \,dx\).

Problem 4. Evaluate the integral \(\displaystyle \int_0^{\frac{\pi}2}\frac1{6+10\cos x}\,dx\).

Problem 5. Evaluate the integral \(\displaystyle \int_0^{\frac{\pi}2}\frac1{10-6\cos x}\,dx\).