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

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<math> \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\arcsec \left(\dfrac{|u|}{a}\right) + C </math>
 
<math> \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\arcsec \left(\dfrac{|u|}{a}\right) + C </math>
  
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===Example 1===
 
<p>Evaluate the integral</p>
 
<p>Evaluate the integral</p>
  
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<p style="text-align: center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.</math></p>
 
<p style="text-align: center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.</math></p>
  
<p>Applying the formula with <math>\( a=2,\)</math> we obtain</p>
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<p>Applying the formula with <math> a=2, </math> we obtain</p>
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<p class="mt-indent-3" style="text-align:center;"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.</math></p>
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===Example 2===
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<p>Evaluate <math> \int \frac{4-x}{\sqrt{16-x^2}}\text{dx} </math>.</p>
  
<p class="mt-indent-3" style="text-align:center;"><math>\[ \int;\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.\]</math></p>
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<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>
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<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>
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<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>
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<p><math>\int \frac{4}{\sqrt{16-x^2}}\text{dx}  = 4\arcsin\frac{x}{4} + C.</math></p>
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<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>
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<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>
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<p>Combining these together, we have</p>
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<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==
[https://youtu.be/AE-0gXXx_j0 Integration into Inverse trigonometric functions using Substitution] by The Organic Chemistry Tutor
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*[https://youtu.be/AE-0gXXx_j0 Integration into Inverse trigonometric functions using Substitution] by The Organic Chemistry Tutor
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*[https://youtu.be/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT
  
[https://youtu.be/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT
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==Licensing==
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Content obtained and/or adapted from:
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* [https://math.libretexts.org/Courses/Monroe_Community_College/MTH_211_Calculus_II/Chapter_5%3A_Integration/5.7%3A_Integrals_Resulting_in_Inverse_Trigonometric_Functions_and_Related_Integration_Techniques Integrals Resulting in Inverse Trigonometric Functions, LibreTexts: Mathematics] under a CC BY-SA-NC license

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: