# Thread: [SOLVED] Integration by Parts Help

1. ## [SOLVED] Integration by Parts Help

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

2. 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 $u = \sin^{-1}x,\;\;\;dv = 4x\,dx$

3. $
\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?

4. Correction:

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

5. 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