Difference between revisions of "Integrals Resulting in Inverse Trigonometric Functions"

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===Example 2===
 
===Example 2===
<p>Evaluate \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).</p>
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<p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\dx </math>.</p>
  
 
<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>
  
<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>
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<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 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>
  
<p>\(\displaystyle \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.\)</p>
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<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>
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<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>\[\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &amp;= \int\frac{-du/2}{\sqrt{u}}\\ &amp;= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &amp;= - \sqrt{u} + C\\ &amp;= -\sqrt{16-x^2} + C.\end{align}\]</p>
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<p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &amp;= \int\frac{-du/2}{\sqrt{u}}\\ &amp;= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &amp;= - \sqrt{u} + C\\ &amp;= -\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 and we have

Applying the formula with we obtain

Example 2

Evaluate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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