1. ## Integrate the Following:

Hello, and thanks for the help in advance...I have a final in this tomorrow and need to get this stuff down pat.

I'm not sure how to go about integrating the following:

x*arcsin(x)dx

I know that substitution is always worth trying, but neither substitution nor integration by parts seems to work well here. Would anyone have any suggestions?

2. Originally Posted by NBrunk
Hello, and thanks for the help in advance...I have a final in this tomorrow and need to get this stuff down pat.

I'm not sure how to go about integrating the following:

x*arcsin(x)dx

I know that substitution is always worth trying, but neither substitution nor integration by parts seems to work well here. Would anyone have any suggestions?
Use Integration by parts...

let $u = arcsin(x)$ and $dv = x dx$

then $du = \frac{1}{\sqrt{1-x^2}}$ and $v = \frac{x^2}{2}$

Then perform: $\int x \times arcsin(x)dx = uv - \int v .\mbox{du}$

You can check your work here

3. Originally Posted by NBrunk
Hello, and thanks for the help in advance...I have a final in this tomorrow and need to get this stuff down pat.

I'm not sure how to go about integrating the following:

x*arcsin(x)dx

I know that substitution is always worth trying, but neither substitution nor integration by parts seems to work well here. Would anyone have any suggestions?
using parts, I sense that you're probably having a problem integrating

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

if so, use a trig sub ... $x = \sin{t}$

$dx = \cos{t} \, dt$

$\frac{1}{2} \int \frac{\sin^2{t}}{\sqrt{1-\sin^2{t}}} \cos{t} \, dt$

$\frac{1}{2} \int \frac{\sin^2{t}}{\cos{t}} \cos{t} \, dt$

$\frac{1}{2} \int \sin^2{t} \, dt$

$\frac{1}{4} \int 1 - \cos(2t) \, dt$

$\frac{t}{4} - \frac{\sin(2t)}{8}$

$\frac{t}{4} - \frac{\sin{t}\cos{t}}{4}
$

back substitute ...

$\frac{\arcsin{x}}{4} - \frac{x \sqrt{1-x^2}}{4} + C
$

so, you final solution should be ...

$\frac{x^2}{2}\arcsin{x} - \frac{\arcsin{x}}{4} + \frac{x \sqrt{1-x^2}}{4} + C$