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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

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

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

## 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

=-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

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)]

=1.5670

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):

(3pi)/2

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

## Exercises

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

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 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:

"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.

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