Difference between revisions of "Integrals Resulting in Inverse Trigonometric Functions"
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− | ==Example | + | ===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=== | ||
+ | <p>Evaluate \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).</p> | ||
+ | |||
+ | <p><strong>Solution</strong></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> | ||
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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>\[\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &= \int\frac{-du/2}{\sqrt{u}}\\ &= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &= - \sqrt{u} + C\\ &= -\sqrt{16-x^2} + C.\end{align}\]</p> | ||
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+ | <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> | ||
==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