# 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 377*t,* 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).

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