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`
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`
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.
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`
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 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`
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:
VC = 583.6 sin 377t
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.
