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

From Department of Mathematics at UTSA
Jump to navigation Jump to search
 
Line 28: Line 28:
 
<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


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

Licensing

Content obtained and/or adapted from: