indefinite integral question! I tried using integration by parts, $\int u\,{\rm d}v= uv- \int v\,{\rm d}u$, on the following integral: 
$$\int x\arcsin(8x)\,{\rm d}x$$ 
This is what I have so far, I don't know what to do after.
$$u= x$$
$$dv= \arcsin(8x)dx$$
$$\frac{x^2}{2}\arcsin(8x)- \int \frac{4x^2}{\sqrt{1-64x^2}}dx$$ 
 A: I think you can use a trig substitution on your last integral:
$$\int \frac{4x^2}{\sqrt{1-64x^2}}\, dx = \int \frac{4x^2}{\sqrt{1-16(4x^2)}}\, dx$$
$$\text{let $a=2x$}$$
$$\int \frac{a^2}{\sqrt{1-16a^2}}\, da$$
I think this is equal to an inverse trig function plus some constants (correct me if I'm wrong) so you'll get a pretty simple answer to the integral that you'll have to add to the other half of your integration by parts.
A: $u=8x  \ du=8dx \rightarrow x=\frac{u}{8}$
$=\int \frac{u}{8} arcsin (u) \frac{1}{8} du$
$=\frac{1}{8} \int \frac{u}{8} arcsin (u) du$
$=\frac{1}{8} \frac{1}{8} \int u arcsin (u) du$
Now apply integration by parts and you should get your answer
A: While integration by parts certainly works, it leads to a lot of complicated algebra. You can, if you wish to see an alternative way, avoid "complicated" integration by parts, here is how:
First u-sub $8x=t$ with $dx=\frac{1}{8}dt$ Integral becomes $\frac{1}{64}\int tarcsint\,{\rm d}t $ Now another u-sub: $arcsint=v$ with $dt=cosvdv$ With $t=sinv$ the integral now becomes:  $\frac{1}{64}\int vsinvcosv\,{\rm d}t $ Now using the double angle formula for sine we arrive at: $\frac{1}{128}\int vsin2v\,{\rm d}t $ Doing integration by parts here is lots simpler since the anti derivative of sine is just a cosine. You just need to back sub which is not too hard
