+0  
 
0
361
4
avatar

integral(sin^-1(x)dx)

Guest Dec 17, 2015

Best Answer 

 #1
avatar+90988 
+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
Sort: 

4+0 Answers

 #1
avatar+90988 
+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+90988 
0

Thanks Hayley :)

Melody  Dec 18, 2015
 #3
avatar+8621 
+5

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

Hayley1  Dec 18, 2015
 #4
avatar+90988 
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

13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details