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

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==Example Problem==
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===Example 1===
 
<p>Evaluate the integral</p>
 
<p>Evaluate the integral</p>
  
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<p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.</math></p>
 
<p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.</math></p>
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===Example 2===
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<p>Evaluate \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).</p>
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<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>
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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>
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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>
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<p>\(\displaystyle \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.\)</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>
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<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>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>
  
 
==Resources==
 
==Resources==

Revision as of 14:23, 28 October 2021


Example 1

Evaluate the integral

Solution

Substitute . Then and we have

Applying the formula with we obtain

Example 2

Evaluate \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).

Solution

This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:

$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\ dx.$$

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.

\(\displaystyle \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.\)

\(\displaystyle \int\frac{x}{\sqrt{16-x^2}}\ dx\): Set \(u = 16-x^2\), so \(du = -2xdx\) and \(xdx = -du/2\). We have

\[\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &= \int\frac{-du/2}{\sqrt{u}}\\ &= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &= - \sqrt{u} + C\\ &= -\sqrt{16-x^2} + C.\end{align}\]

Combining these together, we have

$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.$$

Resources

Integration into Inverse trigonometric functions using Substitution by The Organic Chemistry Tutor

Integrating using Inverse Trigonometric Functions by patrickJMT