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

Revision as of 13:47, 28 October 2021

$\int {\frac {du}{\sqrt {a^{2}-u^{2}}}}=\arcsin \left({\frac {u}{a}}\right)+C$

$\int {\frac {du}{a^{2}+u^{2}}}={\dfrac {1}{a}}\arctan \left({\dfrac {u}{a}}\right)+C$

Failed to parse (syntax error): {\displaystyle \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\text{\arcsec} \left(\dfrac{|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> \int\frac{du}{\sqrt{a^2 - u^2}} = \arcsin \left(\frac{u}{a}\right) + C </math>
-\int\frac{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right) + C
+<math> \int\frac{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right) + C </math>
-\int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{1}{a}\text{\arcsec} \left(\dfrac{|u|}{a}\right) + C </math>
+<math> \int\frac{du}{u\sqrt{u^2 - a^2}} =\frac{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/MdsAvt9y5ds Integrating using Inverse Trigonometric Functions] by  patrickJMT