# 4. Integration: Basic Trigonometric Forms

by M. Bourne

We obtain the following integral formulas by reversing the formulas for differentiation of trigonometric functions that we met earlier:

int sin\ u\ du=-cos u+K

int cos u\ du=sin\ u+K

int sec^2u\ du=tan u+K

int csc^2u\ du=-cot\ u+K

We now apply the power formula to integrate some examples.

NOTE: All angles in this section are in radians. The formulas don't work in degrees.

### Example 1

Integrate: inte^xcsc^2(e^x)dx

### Example 2

Integrate: int(sin(1/x))/(x^2)dx

## Integrals of sec u tan u, and csc x cot u

These are obtained by simply reversing the differentiation process.

int sec u\ tan u\ du=sec u+K

int csc u\ cot\ u\ du=-csc u+K

### Example 3

Integrate: int csc 2x\ cot\ 2x\ dx

## Integrals of tan x, cot x

Now, if we want to find int tan x\ dx, we note that

int tan x\ dx=int(sin x)/(cos x)dx

Let u=cos x, then du=-sin x\ dx. Our integral becomes:

int tan x\ dx=int(sin x)/(cos x)dx

=-int(du)/u

=-ln |u|+K

=-ln |cos x|+K

Similarly, it can be shown that

intcot\ x\ dx=ln\ |sin\ x|+K

## Integrals of sec x, csc x

To integrate sec x , we need to use a trick. We multiply and divide by (sec x + tan x), as follows:

int sec x dx = int sec x(sec x + tan x)/(sec x + tan x)dx

= int (sec^2 x + sec x tan x)/(sec x + tan x)dx

We integrate this using a substitution. We put

u=sec x + tan x, which gives us

du = (sec x tan x + sec^2 x)dx

So our integral can be written:

int (sec x tan x + sec^2 x)/(sec x + tan x) dx =int (du)/u

=ln|u|

=ln|sec x + tan x| + K

Therefore

int sec x dx =ln |sec x + tan x|+K

Using a similar process with a substitution of u=csc x + cot x and multiplying top and bottom by csc x + cot x, we obtain:

int csc x dx =-ln |csc x + cot x|+K

## Summary of Integrals of Trigonometric Functions

We summarise the trigonometric integrals as follows:

inttan u\ du=-ln\ |cos u|+K

intcot\ u\ du=ln\ |\sin\ u|+K

intsec u\ du =ln\ |\sec u+tan u|+K

intcsc u\ du =ln\ |csc u-cot\ u|+K

### Example 4

Integrate: intx^2cot\ x^3dx

### Example 5

Integrate: 6int_0^1 tan{:x/2:}dx

This is the curve y=6 tan(x/2):

The shaded region represents the integral we needed to find.

### Example 6

Find the area under the curve of y = sin\ x from x = 0 to x=(3pi)/2.

## Exercises

Integrate each of the given functions:

### Exercise 1

int(sin\ 2x)/(cos^2x)dx

### Exercise 2

int_(pi//4)^(pi//3)(1+sec x)^2dx

This is the curve y=(1+sec x)^2:

The shaded region represents the integral we needed to find.

### Exercise 3

If the current in a certain electric circuit is i = 110 cos 377t, find the expression for the voltage across a 500-μF capacitor as a function of time. The initial voltage is zero. Show that the voltage across the capacitor is 90° out of phase with the current.

We need the following result from electronics, which gives the voltage across a capacitor, where C is the capacitance:

V_C=1/Cinti\ dt

### Exercise 4

A force is given as a function of the distance from the origin as

F=(2+tan x)/(cos x)

Express the work done by this force as a function of x if W = 0 for x = 0.

The work done as a function of x (the solution we just found).