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

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<math> \int\dfrac{du}{\sqrt{a^2−u^2}} & = \arcsin \left(\dfrac{u}{a}\right)+C \\ \int\dfrac{du}{a^2+u^2}&=\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right)+C \\ \int\dfrac{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}{\sqrt{a^2−u^2}} = \arcsin \left(\frac{u}{a}\right)+C
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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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\int\frac{du}{u\sqrt{u^2−a^2}} =\dfrac{1}{a}\text{arcsec} \left(\dfrac{|u|}{a}\right)+C </math>
  
 
[https://youtu.be/AE-0gXXx_j0 Integration into Inverse trigonometric functions using Substitution] by The Organic Chemistry Tutor
 
[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
 
[https://youtu.be/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT

Revision as of 13:45, 28 October 2021

Failed to parse (syntax error): {\displaystyle \int\frac{du}{\sqrt{a^2−u^2}} = \arcsin \left(\frac{u}{a}\right)+C \int\frac{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right)+C \int\frac{du}{u\sqrt{u^2−a^2}} =\dfrac{1}{a}\text{arcsec} \left(\dfrac{|u|}{a}\right)+C }

Integration into Inverse trigonometric functions using Substitution by The Organic Chemistry Tutor

Integrating using Inverse Trigonometric Functions by patrickJMT

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