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

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<math> \int\frac{du}{\sqrt{a^2−u^2}} = \arcsin \left(\frac{u}{a}\right)+C
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<math> \int\frac{du}{\sqrt{a^2 - u^2}} = \arcsin \left(\frac{u}{a}\right) + C </math>
  
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\int\frac{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right)+C
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<math> \int\frac{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right) + C </math>
  
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\int\frac{du}{u\sqrt{u^2−a^2}} =\dfrac{1}{a}\text{arcsec} \left(\dfrac{|u|}{a}\right)+C </math>
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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>
  
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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
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===Example 1===
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<p>Evaluate the integral</p>
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<p class="mt-align-center"><math> \int\dfrac{dx}{\sqrt{4 - 9x^2}}.</math></p>
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<p><strong>Solution</strong></p>
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<p>Substitute <math> u=3x</math>. Then <math> du=3dx</math> and we have</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>
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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>
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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>
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==Resources==
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*[https://youtu.be/AE-0gXXx_j0 Integration into Inverse trigonometric functions using Substitution] by The Organic Chemistry Tutor
 +
 
 +
*[https://youtu.be/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT
 +
 
 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{du}{\sqrt{a^2 - u^2}} = \arcsin \left(\frac{u}{a}\right) + C }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right) + C }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\arcsec \left(\dfrac{|u|}{a}\right) + C }


Example 1

Evaluate the integral

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\dfrac{dx}{\sqrt{4 - 9x^2}}.}

Solution

Substitute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=3x} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=3dx} and we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.}

Applying the formula with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=2, } we obtain

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.}

Example 2

Evaluate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\text{dx} } .

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 16-x^2} . We handle each separately.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{4}{\sqrt{16-x^2}}\text{dx} = 4\arcsin\frac{x}{4} + C.}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{x}{\sqrt{16-x^2}}\text{dx} } : Set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 16-x^2} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{du} = -2x\text{dx} } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\text{dx} = -\text{du} /2} . We have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Combining these together, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\text{dx} = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.}

Resources

Licensing

Content obtained and/or adapted from:

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