[SOLVED] Integration by Parts Help

**GreySweatshirt306** · February 12th 2009, 02:02 PM

Does anyone know how to do the integral of "4x arcsin(x)dx" ? Help is much appreciated.

**Danny** · February 12th 2009, 03:17 PM

Originally Posted by GreySweatshirt306

Does anyone know how to do the integral of "4x arcsin(x)dx" ? Help is much appreciated.

Use integration by parts $u = \sin^{-1}x,\;\;\;dv = 4x\,dx$

**Nacho** · February 12th 2009, 04:04 PM

$ \begin{gathered} I = \int {4x\arcsin (x)dx} \hfill \\ u = \arcsin (x) \Rightarrow du = \frac{1} {{\sqrt {1 - x^2 } }} \wedge dv = x \Rightarrow v = x^2 \hfill \\ \end{gathered} $

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

Now

$ \int {\frac{{ - x^2 }} {{\sqrt {1 - x^2 } }}dx} = \int {\frac{{1 - x^2 + 1}} {{\sqrt {1 - x^2 } }}} dx = \int {\sqrt {1 - x^2 } } dx + \int {\frac{1} {{\sqrt {1 - x^2 } }}dx} $

Can you continue?

**Scott H** · February 12th 2009, 04:11 PM

Correction:

$\int\frac{-x^2}{\sqrt{1-x^2}}\,dx=\int\frac{1-x^2-1}{\sqrt{1-x^2}}\,dx=\int\sqrt{1-x^2}\,dx-\int\frac{1}{\sqrt{1-x^2}}\,dx.$

**Nacho** · February 13th 2009, 03:56 PM

Originally Posted by Scott H

Correction:

$\int\frac{-x^2}{\sqrt{1-x^2}}\,dx=\int\frac{1-x^2-1}{\sqrt{1-x^2}}\,dx=\int\sqrt{1-x^2}\,dx-\int\frac{1}{\sqrt{1-x^2}}\,dx.$

Always I happened me that

thank you

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