Difference between revisions of "Integrals Resulting in Inverse Trigonometric Functions"
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<p><math>\int \frac{4}{\sqrt{16-x^2}}\text{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}}\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>\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}}\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><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> |
Revision as of 16:37, 15 January 2022
Contents
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 and . 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
Licensing
Content obtained and/or adapted from:
- Integrals Resulting in Inverse Trigonometric Functions, LibreTexts: Mathematics under a CC BY-SA-NC license