Integration Using Completing the Square (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

Integration using completing the square

What is completing the square?

Completing the square is a way of rewriting quadratic expressions
- A quadratic expression is of the form $a x^{2} + b x + c$
- Rewriting can make them easier to use for various purposes
The simple version of completing the square is
- $x^{2} + b x + c = {(x + \frac{b}{2})}^{2} + c - {(\frac{b}{2})}^{2}$
  - E.g. $x^{2} - 6 x + 2 = {(x - 3)}^{2} + 2 - {(- 3)}^{2} = {(x - 3)}^{2} - 7$
When there is a coefficient in front of the $x^{2}$ this becomes
- $a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + c - a {(\frac{b}{2 a})}^{2}$
  - E.g. $3 x^{2} + 12 x - 7 = 3 {(x + 2)}^{2} - 7 - 3 {(2)}^{2} = 3 {(x + 2)}^{2} - 19$

How can I integrate using completing the square?

Recall the standard integrals
- $\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C, - 1 < x < 1$
- $\int \frac{1}{1 + x^{2}} d x = arc \tan x + C$
- See the 'Derivatives & Antiderivatives' study guide
Completing the square can be used to get integrals into a form where these results can be used
- The methods from the 'Integrals of Composite Functions' study guide may be needed as well
E.g. $\int \frac{1}{x^{2} - 6 x + 13} d x$
- Complete the square on the denominator
  - $x^{2} - 6 x + 13 = {(x - 3)}^{2} + 4$
- Rewrite the function to be integrated
  - $\frac{1}{x^{2} - 6 x + 13} = \frac{1}{4 + {(x - 3)}^{2}} = \frac{1}{4} \cdot \frac{1}{1 + {(\frac{x - 3}{2})}^{2}}$
- That is now a multiple of $\frac{1}{1 + {(\dots)}^{2}}$
  - This tells us that $\arctan (\dots)$ is going to be be our likely answer
- To see if 'adjust and compensate' is needed, use the chain rule to differentiate $\arctan (\frac{x - 3}{2})$
  - $\frac{d}{d x} (\arctan (\frac{x - 3}{2})) = \frac{1}{1 + {(\frac{x - 3}{2})}^{2}} \cdot 1 = \frac{1}{1 + {(\frac{x - 3}{2})}^{2}}$
- Integrate
  - $\int \frac{1}{x^{2} - 6 x + 13} d x = \frac{1}{4} \int \frac{1}{1 + {(\frac{x - 3}{2})}^{2}} d x = \frac{1}{4} \arctan (\frac{x - 3}{2}) + C$

Worked Example

Find the indefinite integral $\int \frac{1}{\sqrt{21 - x^{2} + 4 x}} d x$ .

Answer:

Complete the square on the expression in the square root

$21 - x^{2} + 4 x = - {(x - 2)}^{2} + 21 - (- {(2)}^{2}) = 25 - {(x - 2)}^{2}$

Rewrite the function to be integrated

$\frac{1}{\sqrt{21 - x^{2} + 4 x}} = \frac{1}{\sqrt{25 - {(x - 2)}^{2}}} = \frac{1}{5} \cdot \frac{1}{\sqrt{1 - {(\frac{x - 2}{5})}^{2}}}$

That is now a multiple of $\frac{1}{\sqrt{1 - {(\dots)}^{2}}}$

This tells us that $arc \sin (\dots)$ is going to be be our likely answer

To see if 'adjust and compensate' is needed, use the chain rule to differentiate $\arcsin (\frac{x - 2}{5})$

$\frac{d}{d x} (arc \sin (\frac{x - 2}{5})) = \frac{1}{\sqrt{1 - {(\frac{x - 2}{5})}^{2}}} \cdot \frac{1}{5}$

Integrate

$\int \frac{1}{\sqrt{21 - x^{2} + 4 x}} d x = \int \frac{1}{5} \cdot \frac{1}{\sqrt{1 - {(\frac{x - 2}{5})}^{2}}} d x = arc \sin (\frac{x - 2}{5}) + C$

$\int \frac{1}{\sqrt{21 - x^{2} + 4 x}} d x = arc \sin (\frac{x - 2}{5}) + C$

Last updated: 13 August 2024

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