Help with the Integral of $x \arcsin x\,dx$ So I've started the integration using integration by parts, but I'm stuck on the second part of it.
$$ \int_0^\frac{1}{2} x \sin^{-1}x dx $$


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*Step 1. Set up my variables
       $$ u = \arcsin x $$ 
       $$du = \frac{1}{\sqrt{1-x^{2}}} dx $$ 
       $$ dv = x dx $$
       $$ v = \frac{x^{2}}{2} $$

*Step 2. $uv - \int v\,du$
      $$ \frac{x^2}{2} \arcsin x   - \frac{1}{2} \int x^{2}\frac{1}{\sqrt{1-x^{2}}}$$

*Step 3. Now I need to integrate $ \frac{x^2}{\sqrt{1-x^2}} $
This is where I'm stuck. Do I do a trig sub? Let $x = \sin\theta$ which $1-\sin^2\theta$ would then equal $\sqrt{\cos^2\theta}$ turning the second part into $\int \sec \theta$?
Edit. Attempting to continue the problem with the suggestions provided.


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*Step 3 (continued). $$ \frac{-1}{2} \int \frac{x^2}{\sqrt{1-x^2}}dx $$
$$ \text{Let }x = \sin \theta $$
$$ -\frac{1}{2} \int \frac{\sin^2\theta \cos\theta}{\sqrt{1-\sin^2\theta}} $$
$$ -\frac{1}{2} \int \frac{\sin^2\theta \cos\theta}{\sqrt{\cos^2\theta}} $$
$$ -\frac{1}{2} \int \frac{\sin^2\theta \cos\theta}{\cos\theta} $$
$$ -\frac{1}{2} \int \sin^2\theta  $$
$$ -\frac{1}{2} \int \frac{1 - \cos2\theta}{2} $$
$$ -\frac{1}{4} [\theta-\sin\theta \cos\theta] $$

*Step 4. Find what $\theta$ is equal to in terms of x.
Since we let $x = \sin\theta$, that would mean we have $$x^2 + b^2 = 1^2$$
So, does that mean I substitute $$ 1 - x$$ in for $\theta$?


If you want to continue with step (3), then you can use integration by parts again with $u=x$
$$ \int \frac{x^2}{\sqrt{1-x^2}} dx = -x\sqrt {-{x}^{2}+1}+\int \!\sqrt {1-{x}^{2}}{dx}.$$
For the last integral, you can use the trig. subs $x=\sin(\theta)$.
