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`


`=-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|sec x + tan x| + K`


`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`.


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).