Difference between revisions of "Integrals Resulting in Inverse Trigonometric Functions"
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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}}\text{dx} = \int \frac{4}{\sqrt{16-x^2}}\text{dx} - \int \frac{x}{\sqrt{16-x^2}}\text{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> |
Latest revision as of 16:38, 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:
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