# The Trigonometric Functions by Stefan Waner and Steven R. Costenoble

## This Section: 4. Integrals of Trigonometric Functions

4. Integrals of Trigonometric Functions

Recall from the definition of an antiderivative that, if

$\frac{d}{dx} f(x) = g(x),$

then

$∫ g(x) dx = f(x) + C.$

That is, every time we have a differentiation formula, we get an integration formula for nothing. Here is a list of some of them.

 Derivative Rule Antiderivative Rule $\frac{d}{dx} \sin x = \cos x$ $∫ \cos x dx = \sin x + C$ $\frac{d}{dx} \cos x = -\sin x$ $∫ \sin x dx = -\cos x + C$ $\frac{d}{dx} \tan x = \sec^2x$ $∫\sec^2x dx = \tan x + C$ $\frac{d}{dx} cotan x = -cosec^2 x$ $∫ cosec^2x dx = -cotan x + C$ $\frac{d}{dx} \sec x = \sec x \tan x$ $∫ (\sec x \tan x) dx = \sec x + C$ $\frac{d}{dx} cosec x = -cosec x cotan x$ $∫ (cosec x cotan x) dx = -cosec x + C$

Notice that, quite by chance, we have come up with formulas for the antiderivatives of $\sin x$ and $\cos x.$

Question

We shall obtain some of them below, and leave others to the exercise set. (Some of them have already appeared as derivatives in Exercise Set 3...).

Example 1

Compute the following.

 (a) $∫ (3\sin x - 4\sec^2x) dx$ (b) $∫ \cos(2x - 6) dx$ (c) $∫ \sin x \cos^2x dx$ (d) $∫ \tan x dx$

Solution

(a) Consulting the table above,

$(3\sin x - 4\sec^2x) dx$ $=$  ∫ $3\sin x dx -$ ∫ $4\sec^2x dx$
(properties of integrals)
$=$  $3$ ∫ $\sin x dx - 4$ ∫ $\sec^2x dx$
(properties of integrals)
$=$ $-3\cos x - 4\tan x + C$ (from the table)

(b) The calculation of $∫ \cos(2x - 6) dx$ requires a substitution:

 $\color{blue}{u = 2x-6}$ $\color{blue}{\frac{du}{dx}=2}$ $\color{blue}{dx = \frac{1}{2} du}$

We now have

 ∫ $\cos(2x - 6) dx$ $=$ $∫ \cos u$$\frac{1}{2} du (using the substitution) = \frac{1}{2} ∫ \cos u du (properties of integrals) = \frac{1}{2}$$ sen u + C$ (from the table) $=$ $\frac{1}{2}$$sen(2x-6) + C. (using the substitution) (c) This one can also be done using a substitution. The trick is to substitute for the term \cos x as follows:  \color{blue}{u = \cos x } \color{blue}{\frac{du}{dx} = -sen x} \color{blue}{dx = - \frac{1}{sen x} du} We now have  ∫ sen x \cos^2x dx = ∫(sen x)u^2 \left(- \frac{1}{sen x}\right) du (using the substitution) = - ∫u^2 du = - \frac{u^3}{3} + C = - \frac{\cos^3x}{3} + C. (using the substitution) (d) Write ∫ \tan x dx as ∫(\sin x / \cos x) dx, and use the same substitution as in part (c):  ∫ \tan x dx = ∫\frac{sen x}{\cos x} dx = ∫\frac{sen x}{u} \left(- \frac{1}{sen x} \right) du (using the substitution) = - ∫\frac{1}{u} du = - \ln \|u\| + C = - \ln \|\cos x\| + C. (using the substitution) Before we go on... The method in part (b) gives us the following more general formulas:  Integral Rule General Rule ∫\cos x dx = sen x + C ∫ \cos(ax+b) dx = \frac{1}{a} sen(ax+b) + C ∫sen x dx = -\cos x + C ∫sen(ax+b) dx = - \frac{1}{a} \cos(ax+b) + C Not to keep you in suspense, here are the antiderivatives of all six trigonometric functions. (You will obtain them in the exercises.)  Integral Rule General Rule ∫\cos x dx = sen x + C ∫\cos(ax+b) dx = \frac{1}{a} sen(ax+b) + C ∫sen x dx = -\cos x + C ∫sen(ax+b) dx = - \frac{1}{a} \cos(ax+b) + C ∫\tan x dx = - \ln \|\cos x\| + C ∫\tan(ax+b) dx = -\frac{1}{a} \ln \|\cos(ax+b)\| + C ∫cotan x dx = \ln \|sen x\| + C ∫cotan(ax+b) dx = \frac{1}{a} \ln \|sen(ax+b)\| + C ∫\sec x dx = \ln \|\sec x + \tan x\| + C ∫\sec(ax+b) dx = \frac{1}{a} \ln \|\sec(ax+b) + \tan(ax+b)\| + C ∫cosec x dx = -\ln \|cosec x + cotan x\| + C ∫ cosec(ax+b) dx = - \frac{1}{a} \ln \|cosec(ax+b) + cotan(ax+b)\| + C Example 2 Total Sales Monthly sales of Ocean King Boogie Boards are given by s(t) = 1\,500\sin(π(t-7)/6) + 2\,000, where t is time in months, and t = 0 represents January 1. Estimate total sales over the four-month period beginning March 1. Solution Since total sales are given by the definite integral of monthly sales for the given period (t = 2 to t = 6), we have, consulting the above table, Total Sales =  ∫ 6$$2$ $[1\,500sen(π (t -7)/6) + 2\,000] dt$
$=$  $1,500$ $($ $\frac{6}{π}$ $)$ $sen(π(t -7)/6) + 2\,000t$ $6$$2 =  3\,038 boogie boards. We can also use the tabular method of integration by parts discussed in Section 7.1 of Calculus Applied to the Real World, or Section 14.1 of Finite Mathematics and Calculus Applied to the Real World. Example 3 Evaluate the following integrals (a) _∫(3x^2-2x+1)sen(x/2) dx (b) _∫e^{2x}\cos(3x) dx Solution (a) Since repeated differentiation of the first term (3x^2-2x+1) results in zero, we place it in the "D" column:  D I + 3x^2−2x+1 sen(x/2) ­ 6x−2 −2\cos(x/2) + 6 −4sen(x/2) ­ 0 8\cos(x/2) This gives: _∫(3x^2−2x+1)sen(x/2) dx = −2(3x^2−2x+1)\cos(x/2) + 4(6x−2)sen(x/2) + 48\cos(x/2) + C (b) Repeated differentiation does not annihilate either term. Actually, it doesn't matter which term we place in the "D" column, so let us place the trigonometric function there: D I + \cos(3x) e^{2x} ­ − 3sen(3x)  \frac{1}{2} e^{2x} + −9\cos(3x)  \frac{1}{4} e^{2x} This gives _∫e^{2x}\cos(3x) dx =$$\frac{1}{2}$ $e^{2x}\cos(3x) +$$\frac{3}{4}$$e^{2x}sen(3x) −$ $\frac{9}{4}$$_∫ e^{2x}\cos(3x) dx (We will add the constant of integration after we are done.) Notice that we have ended up with the same integral on the right as the one we started with. Calling this integral I gives: \color{red}{I}=$$\frac{1}{2}$$e^{2x} \cos(3x) +$$\frac{3}{4}$$e^{2x}sen(3x) −$$\frac{9}{4}\color{red}{I}$

and we can now solve for I:

$I +$$\frac{9}{4}$$I =$$\frac{1}{2} e^{2x}\cos(3x) +$$\frac{3}{4}$$e^{2x}sen(3x), that is, \frac{13}{4}$$I =$$\frac{1}{2} e^{2x}\cos(3x) +$$\frac{3}{4}$$e^{2x}sen(3x), so that _∫e^{2x}\cos(3x) dx = I =$$\frac{4}{13}$$[$$\frac{1}{2}$$e^{2x}\cos(3x) +$$\frac{3}{4}$$e^{2x}sen(3x)$$]$$+ C.$

We would welcome comments and suggestions for improving this resource.

Mail us at:
 Stefan Waner (matszw@hofstra.edu) Steven R. Costenoble (matsrc@hofstra.edu)

Last Updated: March, 1997