7.1 Integration By Parts/10: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
 
(3 intermediate revisions by the same user not shown)
Line 1: Line 1:
<math> f'(x)= \int_{}^{}\sin^{-1}(x)\cdot dx </math> <br><br>
<math> f'(x)= \int_{}^{}\sin^{-1}(x)\cdot dx </math> <br><br>
<math>\int_{}^{}\sin^{-1}(x)dx</math> = <math>x\sin^{-1}(x)-\int_{}^{}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}\frac{1}{\sqrt{u}}du</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}u^{-\frac{1}{2}}du</math> = <math>x\sin^{-1}(x)+\frac{1}{2}(2u^{frac{1}{2}}</math> = <math>x\sin^{-1}(x)+\sqrt{u}</math> = <math>x\sin^{-1}(x)+\sqrt{1-x^2}</math>
<math>\int_{}^{}\sin^{-1}(x)dx</math> = <math>x\sin^{-1}(x)-\int_{}^{}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}\frac{1}{\sqrt{u}}du</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}u^{-\frac{1}{2}}du</math> = <math>x\sin^{-1}(x)+\frac{1}{2}(2u^{\frac{1}{2}})</math>  
= <math>x\sin^{-1}(x)+\sqrt{u}</math> = <math>x\sin^{-1}(x)+\sqrt{1-x^2}+C</math>
 
 
<math>{u}</math> = <math>{1-x^2}</math>
 
<math>{du}</math> = <math>{-2x}</math>
 
<math>{-\frac{1}{2}du}</math> = <math>{x}dx</math>
 
 
<math>{u}</math> = <math>{\sin^{-1}(x)}</math> ,  <math>{dv}</math> = <math>dx</math>
 
<math>{du}</math> = <math>{\frac{1}{\sqrt{1-x^2}}dx}</math> ,  <math>{v}</math> = <math>{x}</math>

Latest revision as of 01:49, 27 November 2022



= = = = = =


=

=

=


= , =

= , =

Retrieved from "https://wiki.dvaezazizi.com/index.php?title=7.1_Integration_By_Parts/10&oldid=5379"