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


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

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