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Integrating x*arcsin(x)


Function

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Hi everyone

 

Another one for tomorrow:

 

[math]I=\int{x\cdot\arcsin{x}\text{ d}x}[/math]

 

[math]I=\int{\arcsin{x}\text{ d}\left(\frac{x^2}{2}\right)}[/math]

 

[math]I=\frac{1}{2}\left(x^2\cdot\arcsin{x}-\int{x^2\text{ d}(\arcsin{x})}\right)[/math]

 

[math]I=\frac{1}{2}(a(x)- b(x))[/math]

 

[math]b(x)=\int{\frac{x^2}{\sqrt{1-x^2}}\text{ d}x}[/math]

 

Let [math]\sqrt{1-x^2}=u[/math]

 

[math]\text{d}u=\frac{-x\text{ d}x}{\sqrt{1-x^2}}[/math]

 

[math]\text{d}x=\frac{-\sqrt{1-x^2}\text{ d}u}{x}[/math]

 

[math]b(x)=\int{\frac{x^2\sqrt{1-x^2}\text{ d}u}{-x\sqrt{1-x^2}}}[/math]

 

[math]b(x)=-\int{x\text{ d}u}[/math]

 

[math]u=\sqrt{1-x^2}\Leftrightarrow x=\pm\sqrt{1-u^2}[/math]

 

[math]b(x)=-\int{\left(1-u^2\right)^{\frac{1}{2}}\text{ d}u}[/math]

 

(Or the opposite, but also when I fill in that one in the total result, my calculator won't give a constant term (see later))

 

[math]b(x)=\cdots = -\frac{1}{2}\left(\frac{2}{3}(1-u^2)^{\frac{3}{2}}\right)+C=\cdots =-\frac{1}{3}x^3+C[/math]

 

[math]I=\frac{1}{2}\left(x^2\arcsin{x}+\frac{1}{3}x^3\right)+C[/math]

 

Now, my book gives the following answer:

 

[math]\frac{1}{4}\left((2x^2-1)\arcsin{x}+x\sqrt{1-x^2}\right)+C[/math]

 

I used my TI-84 Plus to see if they do differ only from one constant, but they don't. My answer should thus be incorrect.

 

Can someone help me?

 

Thanks!


I found a video online showing the solution, but I don't really see where my way of working it out went wrong..?

Edited by Function
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sunitswn91, please note the Homework Help forum rules: "A simple reminder to all: this is the "Homework Help" forum, not the "Homework Answers" forum. We will not do your work for you, only point you in the right direction. Posts that do give the answers may be removed."

 

Please do not offer to do all the work for people, or have them contact you off SFN.

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