Computing Integrals by Completing the Square
We will review the method of completing the square in the context of
evaluating integrals.
Example
Let's start by evaluating
The denominator does not factor with rational coefficients, so
partial fractions is not a viable option. There is also no obvious
substitution to make. Instead, we will complete the square in the
denominator to get a recognizable form for the integral.
Now
| 2x2-12x+26 |
= |
2[x2-6x+13] |
Factor out the
coefficient of x2 |
|
|
= |
2[(x2-6x+9)+4] |
1/2(-6) = -3
(-3)2 = 9
so group x2-6x+9 |
|
|
= |
2[(x-3)2+4]. |
Factor your
perfect square. |
|
Returning to the integral,
|
|
= |
|
|
= |
|
|
|
⌠
⌡ |
du
u2 + a2
|
= |
1
a |
arctan |
/
|
\ |
u
a
|
\
|
/ |
+C |
|
|
|
|
= |
1
2
|
/
|
\ |
1
2
|
arctan((x-3)/2) |
\
|
/ |
+C |
|
|
= |
1
4
|
arctan |
/
|
\ |
x-3
2
|
\
|
/ |
+C. |
|
Certain other types of integrals can be evaluated by this method as
well:
Example
Consider
⌠
⌡
|
dx
________
√21-4x-x2 |
.
|
Now
| 21-4x-x2 |
= |
21-[x2+4x] |
Factor out the
coefficient (-1) of x2 |
|
|
= |
21+4-[x2+4x+4] |
1/2(4) = 2
22 = 4
so group x2+4x+4 |
|
|
= |
25-(x+2)2. |
Factor your
perfect square |
|
Returning to the integral,
|
|
=
|
⌠
⌡
|
dx
________
√25-(x+2)2 |
|
|
⌠
⌡
|
du
______
√ a2 - u2 |
= |
arcsin |
/
|
\ |
u
a
|
\
|
/ |
+C |
|
|
|
|
= |
| arcsin |
/
|
\ |
x+2
5
|
\
|
/ |
+C. |
|
Completing the square is a powerful method that is used to derive the
quadratic formula:
We will find the roots of ax2+bx+c = 0.
|
ax2 + bx + c |
= |
0 |
|
|
= |
0 |
|
|
|
= |
|
| Move the constant |
c
a | to the right) |
|
|
|
|
= |
|
Add the square of
1/2[coefficient of x]
to both sides) |
|
|
|
= |
|
Factor the left as a
perfect square;
simplify the right |
|
|
|
= |
|
Take the square root
of both sides) |
|
| x |
= |
|
|
which is the familiar quadratic formula!