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integral(sin^-1(x)dx)

Guest Dec 17, 2015

Best Answer 

 #1
avatar+93289 
+25

This one looks like a challenge for me.  :)

 

integral(sin^-1(x)dx)

 

I am going to use integration by parts.

 

\(\boxed{\int\;u\;\frac{dv}{dx}\;dx=uv-\int\;v\;\frac{du}{dx}\;dx}\\~\\ Let\;\; u=asin(x)\qquad and \qquad \frac{dv}{dx}=1\\ then \;\;\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\qquad and \qquad v=x\\ so\\ \int\;asin(x)\;*1\;dx=asin(x)*x-\int\;x\;\frac{1}{\sqrt{1-x^2}}\;dx\\ \int\;asin(x)\;dx=x*asin(x)-\int\;\frac{x}{\sqrt{1-x^2}}\;dx\\ \)

 

 

Now I need to work out what this is

 

\(\int\;\frac{x}{\sqrt{1-x^2}}\;dx\\~\\ let\; g=1-x^2\\ \frac{dg}{dx}=-2x\\ dx=\frac{dg}{-2x}\\~\\ \int\;\frac{x}{\sqrt{1-x^2}}\;dx\\~\\ =\int\;\frac{x}{\sqrt{g}}\;dx\\~\\ =\int\;\frac{x}{\sqrt{g}}*\;\frac{dg}{-2x}\\~\\ =\;-0.5\int\;g^{-0.5}\;dg\\~\\ =-0.5*\frac{g^{0.5}}{0.5}\\~\\ =-\;g^{0.5}\\~\\ =-\;(1-x^2)^{0.5}\\~\\ =-\;\sqrt{1-x^2}\\~\\ \)

so

 

\(\int\;asin(x)\;dx=x*asin(x)-\int\;\frac{x}{\sqrt{1-x^2}}\;dx\\     =\int\;asin(x)\;dx=x*asin(x)- -\;\sqrt{1-x^2}+c\\ =\int\;asin(x)\;dx=x*asin(x)+\;\sqrt{1-x^2}+c\\ or\\ =\int\;sin^{-1}(x)\;dx=x\;sin^{-1}(x)+\sqrt{1-x^2}+c\\\)

Melody  Dec 18, 2015
 #1
avatar+93289 
+25
Best Answer

This one looks like a challenge for me.  :)

 

integral(sin^-1(x)dx)

 

I am going to use integration by parts.

 

\(\boxed{\int\;u\;\frac{dv}{dx}\;dx=uv-\int\;v\;\frac{du}{dx}\;dx}\\~\\ Let\;\; u=asin(x)\qquad and \qquad \frac{dv}{dx}=1\\ then \;\;\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\qquad and \qquad v=x\\ so\\ \int\;asin(x)\;*1\;dx=asin(x)*x-\int\;x\;\frac{1}{\sqrt{1-x^2}}\;dx\\ \int\;asin(x)\;dx=x*asin(x)-\int\;\frac{x}{\sqrt{1-x^2}}\;dx\\ \)

 

 

Now I need to work out what this is

 

\(\int\;\frac{x}{\sqrt{1-x^2}}\;dx\\~\\ let\; g=1-x^2\\ \frac{dg}{dx}=-2x\\ dx=\frac{dg}{-2x}\\~\\ \int\;\frac{x}{\sqrt{1-x^2}}\;dx\\~\\ =\int\;\frac{x}{\sqrt{g}}\;dx\\~\\ =\int\;\frac{x}{\sqrt{g}}*\;\frac{dg}{-2x}\\~\\ =\;-0.5\int\;g^{-0.5}\;dg\\~\\ =-0.5*\frac{g^{0.5}}{0.5}\\~\\ =-\;g^{0.5}\\~\\ =-\;(1-x^2)^{0.5}\\~\\ =-\;\sqrt{1-x^2}\\~\\ \)

so

 

\(\int\;asin(x)\;dx=x*asin(x)-\int\;\frac{x}{\sqrt{1-x^2}}\;dx\\     =\int\;asin(x)\;dx=x*asin(x)- -\;\sqrt{1-x^2}+c\\ =\int\;asin(x)\;dx=x*asin(x)+\;\sqrt{1-x^2}+c\\ or\\ =\int\;sin^{-1}(x)\;dx=x\;sin^{-1}(x)+\sqrt{1-x^2}+c\\\)

Melody  Dec 18, 2015
 #2
avatar+93289 
0

Thanks Hayley :)

Melody  Dec 18, 2015
 #3
avatar+8613 
+5

Who said it was me...? ^.-

Hayley1  Dec 18, 2015
 #4
avatar+93289 
0

Ah Hayley, thanks also be due to Sir CPhill and Serf Guest,

 

The Queen hath special magical powers Mistress Hayley, She can see all with her chrystal sphere.

Morgan La Faye loaned The Queen the orb so She could better govern Camelot!

 

By order of:

Queen Guinevere.

Sovereign of the Magical Kingdom of Camelot.

Melody  Dec 19, 2015
edited by Melody  Dec 19, 2015

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