integral(sin^-1(x)dx)

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

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

Thanks Hayley :)

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

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

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