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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`



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

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

`=-cot u+K`


Example 2

Integrate: `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`


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`


`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`


`=-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`


`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`

Example 5

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


`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)]`


Of course, `x` is in radians.

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`.


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

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

π 1 -1 x y

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)|`


`=3\ "units"^2`


Integrate each of the given functions:

Exercise 1

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


`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`


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


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


`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)]`


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`


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:

VC = 583.6 sin 377t

Now, using cos(ab) = 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.


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


So we need to calculate

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


`(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`


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.

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