Looking to integrate arcsin(x)*dx from 0 to 1 by method of int by parts. You get an improper integral when doing this, but want to do it regardless.

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- Apr 19th 2010, 05:06 PMWartonMortonint by parts
Looking to integrate arcsin(x)*dx from 0 to 1 by method of int by parts. You get an improper integral when doing this, but want to do it regardless.

- Apr 19th 2010, 05:10 PMdwsmith
- Apr 19th 2010, 05:16 PMWartonMorton
so du = 1/sqrt(1-x^2)

v=1

Can't quite put it all together. - Apr 19th 2010, 05:20 PMdwsmith
$\displaystyle u=arcsinx$

$\displaystyle du=\frac{1}{\sqrt{-x^2}}dx$

$\displaystyle dv=dx$

$\displaystyle v=\int dx=x$

$\displaystyle uv-\int vdu$ - Apr 19th 2010, 05:21 PMharish21
$\displaystyle u = arcsinx \implies du = \frac{1}{\sqrt{1-x^2}} dx$

and,

$\displaystyle dv = dx \implies v = x$

using integration by parts, the expression is

$\displaystyle uv - \int v. \text{du}$

$\displaystyle = x \times arcsin(x) - \int \frac{x}{\sqrt{1-x^2}} dx$

Now for $\displaystyle \int \frac{x}{\sqrt{1-x^2}} dx$, use substitution:

$\displaystyle u = 1-x^2$

and integrate