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| ===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> |
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| <p><strong>Solution</strong></p> | | <p><strong>Solution</strong></p> |
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| <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> |
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− | <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> |
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− | <p>The first integral is handled using a straightforward application 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> |
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− | <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> |
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− | <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> |
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− | <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> |
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| <p>Combining these together, we have</p> | | <p>Combining these together, we have</p> |
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− | <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> |
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| ==Resources== | | ==Resources== |
Revision as of 14:30, 28 October 2021
Example 1
Evaluate the integral
Solution
Substitute . Then and we have
Applying the formula with 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.
: Set , so . 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