Difference between revisions of "Integrals Resulting in Inverse Trigonometric Functions"
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| − | <math> \begin{align} ∫\dfrac{du}{\sqrt{a^2−u^2}}&=\arcsin \left(\dfrac{u}{a}\right)+C \\ | + | <math> \begin{align} ∫\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 \end{align} </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 13:44, 28 October 2021
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} ∫\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 \end{align} } Integration into Inverse trigonometric functions using Substitution by The Organic Chemistry Tutor
Integrating using Inverse Trigonometric Functions by patrickJMT
