|   |   | 
| Line 22: | Line 22: | 
|  |  |  |  | 
|  | ===Example 2=== |  | ===Example 2=== | 
| − | <p>Evaluate \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).</p> | + | <p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\dx </math>.</p> | 
|  |  |  |  | 
|  | <p><strong>Solution</strong></p> |  | <p><strong>Solution</strong></p> | 
| Line 28: | Line 28: | 
|  | <p>This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:</p> |  | <p>This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:</p> | 
|  |  |  |  | 
| − | <p>$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\ dx.$$</p> | + | <p><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\dx </math>/p> | 
|  |  |  |  | 
| − | <p>The first integral is handled using a straightforwardapplication of Theorem \(\PageIndex{2}\); the second integral is handled by substitution, with\(u = 16-x^2\). We handle each separately.</p> | + | <p>The first integral is handled straightforward; the second integral is handled by substitution, with <math>u = 16-x^2</math>. We handle each separately. </p> | 
|  |  |  |  | 
| − | <p>\(\displaystyle \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.\)</p> | + | <p><math>\int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.</math></p> | 
|  |  |  |  | 
| − | <p>\(\displaystyle \int\frac{x}{\sqrt{16-x^2}}\ dx\): Set\(u = 16-x^2\), so\(du = -2xdx\) and\(xdx = -du/2\). We have</p> | + | <p><math>\int\frac{x}{\sqrt{16-x^2}}\ dx</math>: Set <math>u = 16-x^2</math>, so <math>du = -2xdx<math> and <math>xdx = -du/2</math>. We have</p> | 
|  |  |  |  | 
| − | <p>\[\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &= \int\frac{-du/2}{\sqrt{u}}\\ &= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &= - \sqrt{u} + C\\ &= -\sqrt{16-x^2} + C.\end{align}\]</p> | + | <p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &= \int\frac{-du/2}{\sqrt{u}}\\ &= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &= - \sqrt{u} + C\\ &= -\sqrt{16-x^2} + C.\end{align}</math></p> | 
|  |  |  |  | 
|  | <p>Combining these together, we have</p> |  | <p>Combining these together, we have</p> | 
|  |  |  |  | 
| − | <p>$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.$$</p> | + | <p><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.</math></p> | 
|  |  |  |  | 
|  | ==Resources== |  | ==Resources== | 
		Revision as of 14:30, 28 October 2021
 
 
 
Example 1
Evaluate the integral

Solution
Substitute  . Then
. Then  and we have
 and we have

Applying the formula with  we obtain
 we obtain

Example 2
Evaluate Failed to parse (unknown function "\dx"): {\displaystyle  \int \frac{4-x}{\sqrt{16-x^2}}\dx }
.
Solution
This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:
Failed to parse (unknown function "\dx"): {\displaystyle  \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\dx }
/p>
The first integral is handled straightforward; the second integral is handled by substitution, with  . We handle each separately.
. We handle each separately. 

 : Set
: Set  , so
, so  . We have
. We have

Combining these together, we have

Resources
Integration into Inverse trigonometric functions using Substitution by The Organic Chemistry Tutor
Integrating using Inverse Trigonometric Functions by  patrickJMT