# 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`

Answer

`inte^xcsc^2(e^x)dx`

Let `u=e^x`, then `du=e^xdx`.

`int e^x csc^2(e^x)dx=int csc^2u\ du`

`=-cot u+K`

`=-cot(e^x)+K`

### Example 2

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

Answer

`int(sin(1/x))/(x^2)dx`

Let `u=1/x`, then `du=-1/x^2dx`.

`int(sin(1/x))/(x^2)dx=-int sin u\ du`

`=cos u+K`

`=cos(1/x)+K`

Continues below ⇩

## 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`

Answer

`intcsc\ 2x\ cot 2x\ dx`

Let `u=2x`, then `du=2\ dx`.

`int csc\ 2x\ cot 2x\ dx =int csc\ u\ cot u(du)/2`

`=1/2int csc\ u\ cot u\ du`

`=1/2(-csc\ u)+K`

`=-1/2csc\ u+K`

`=-1/2csc\ 2x+K`

## 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`

Answer

`int x^2\ cot x^3dx`

Let `u=x^3`, then `du=3x^2\ dx`.

`intx^2\ cot x^3dx=intcot u(du)/3`

`=1/3intcot u\ du`

`=1/3ln\ |sin u|+K`

`=1/3ln\ |sin x^3|+K`

Easy to understand math videos:

MathTutorDVD.com

### Example 5

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

Answer

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

Let `u=x/2`, then `du=1/2\ dx`.

`6int_0^1tan {:x/2:}dx =6(-2)[ln|cos {:x/2:}|]_0^1`

`=-12[ln(cos {:1/2:})-ln\ (cos 0)]`

`=1.5670`

Of course, `x` is in radians.

Easy to understand math videos:

MathTutorDVD.com

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`**.**

Answer

We sketch the curve first to see what is going on.

This is the curve `y=sin(x)`:

The shaded region represents the integral we need to find.

We need to split the integration into 2 portions because one part of the curve is above the `x`-axis (the part from `0` to `pi`), and the rest of it is below the `x`-axis (the part from `pi` to `(3pi)/2`, and we'll need to take the absolute value).

`"Area" =int_0^pi sin x\ dx+|int_pi^(3pi//2)sin x\ dx|`

`=[-cos x]_0^pi+` `|-cos x|_pi^(3pi//2)`

`=[-cos pi-(-cos 0)]+` `|-cos (3pi)/2-(-cos pi)|`

`=[1+1]+|0-1|`

`=3\ "units"^2`

Get the Daily Math Tweet!

IntMath on Twitter

## Exercises

Integrate each of the given functions:

### Exercise 1

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

Answer

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

Recall that `sin 2x = 2\ sin x\ cos x`

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

`=2int(sin x)/(cos x)dx`

`=2int tan x\ dx`

`= -2\ ln\ |cos x|+K`

### Exercise 2

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

Answer

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

Now

`(1+sec x)^2` `=1+2\ sec x+sec^2x`

So

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

`=[x+2\ ln\ |sec x+tan x|+ ` `{:tan x ]_(pi//4)^(pi//3)`

`=[(pi/3+2(1.31696)+1.7321)-` `{:(pi/4+2(0.88137)+1)]`

`=1.8651`

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`

Answer

Note that μ = 1 millionth, or 10^{-6}.

`V_C =1/Cinti\ dt`

`=1/(500xx10^-6)int110\ cos 377t\ dt`

`=220000/377sin 377t+K`

`=583.6\ sin 377t+K`

When `t = 0`, `V_C= 0`, so `K = 0`.

So we have:

V= 583.6 sin 377_{C}t

Now, using cos(*a* − *b*) = cos *a* cos *b* + sin *a* sin *b*, we
have:

`cos(377t-pi/2) =cos 377t\ cos (pi/2)+` `sin 377t\ sin (pi/2)`

`=sin 377t `

This shows that `583.6 sin 377t` and `110 cos 377t` are `90^"o"=pi/2` out of phase.

### 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.

Answer

Recall from the previous section Work by a Variable Force, that:

`"Work"=intFdx`

So we need to calculate

`W=int(2+tan x)/(cos x)dx`

Since

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

`=2\ sec x+tan x\ sec x`,

we have, using the given formulas:

`W =int(2\ sec x+tan x\ sec x)dx`

`=2\ ln\|sec x+tan x|+sec x+K`

Since `W = 0` when `x = 0`, we have:

`0=2\ ln\ |sec 0+` `tan 0|+sec 0+K`

`0=2(0)+1+K`

So `K = -1`. So we have:

`W=2\ ln\ |sec x+tan x|+` `sec x-1`

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

Realistically, we can only define work for continuous values of *x*, so we would need to restrict the domain, something like:

The work done as a function of *x* for continuous *x*.

### Search IntMath, blog and Forum

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!