- #1
agro
- 46
- 0
I find it really hard to solve
[tex]\int\arcsin{x}\;dx[/tex]
I tried using integration by part with these values:
[tex]
\begin{array}{rl}
u=\arcsin{x},&dv=dx\\
du=\frac{1}{\sqrt{1-x^2}}\;dx,&v=x\\
\end{array}
[/tex]
Which yields
[tex]
\begin{equationarray}
\int\arcsin{x}\;dx&=&x\arcsin{x}-\int\frac{x}{\sqrt{1-x^2}}\;dx
\end{equationarray}
[/tex]
After that I tried various substitutions/integration by parts but didn't get any simpler form... Can anyone help my desperate self :) (maybe hints...)
Thanks a lot beforehand...
[tex]\int\arcsin{x}\;dx[/tex]
I tried using integration by part with these values:
[tex]
\begin{array}{rl}
u=\arcsin{x},&dv=dx\\
du=\frac{1}{\sqrt{1-x^2}}\;dx,&v=x\\
\end{array}
[/tex]
Which yields
[tex]
\begin{equationarray}
\int\arcsin{x}\;dx&=&x\arcsin{x}-\int\frac{x}{\sqrt{1-x^2}}\;dx
\end{equationarray}
[/tex]
After that I tried various substitutions/integration by parts but didn't get any simpler form... Can anyone help my desperate self :) (maybe hints...)
Thanks a lot beforehand...