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> | ||
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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> | ||
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 }
Contents
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:
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} = \int \frac{4}{\sqrt{16-x^2}}\text{dx} - \int \frac{x}{\sqrt{16-x^2}}\text{dx} }
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
- 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
