Difference between revisions of "Integrals Resulting in Inverse Trigonometric Functions"
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===Example 2=== | ===Example 2=== | ||
− | <p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\dx </math>.</p> | + | <p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\text{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><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><math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx} = \int \frac{4}{\sqrt{16-x^2}}\text{dx} - \int \frac{x}{\sqrt{16-x^2}}\text{dx} </math>/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>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><math>\int \frac{4}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac{x}{4} + C.</math></p> | + | <p><math>\int \frac{4}{\sqrt{16-x^2}}\text{dx} = 4\arcsin\frac{x}{4} + C.</math></p> |
− | <p><math>\int\frac{x}{\sqrt{16-x^2}}\ dx</math>: Set <math>u = 16-x^2</math>, so <math>du = - | + | <p><math>\int\frac{x}{\sqrt{16-x^2}}\text{dx} </math>: Set <math>u = 16-x^2</math>, so <math>\text{du} = -2x\text{dx} <math> and <math>x\text{dx} = -\text{du} /2</math>. We have</p> |
− | <p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx | + | <p><math>\begin{align} \int\frac{x}{\sqrt{16-x^2}}\text{dx} = \int\frac{-\text{du} /2}{\sqrt{u}}\\ = -\frac12\int \frac{1}{\sqrt{u}}\text{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><math> \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.</math></p> | + | <p><math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx} = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.</math></p> |
==Resources== | ==Resources== |
Revision as of 14:33, 28 October 2021
Example 1
Evaluate the integral
Solution
Substitute . Then and we have
Applying the formula with we obtain
Example 2
Evaluate .
Solution
This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:
/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