what is the antiderivative of ∫arcsinxdx

Question

anonymous · Answer

Integrate by parts:
$$\int\arcsin x~dx$$
Let $u=\arcsin x$ and $dv=dx$, so that $du=\dfrac{dx}{\sqrt{1-x^2}}$ and $v=x$:
$$uv-\int v~du=x\arcsin x-\int\frac{x}{\sqrt{1-x^2}}~dx$$

anonymous · Answer

A substitution will clear up the remaining integral, $t=1-x^2\Rightarrow-\dfrac{1}{2}dt=x~dx$.

mkmkasim · Answer

what happens to the square root though. Would the substitution just be 1/u?

turingtest · Answer

the square root stays, simply treat it as a negative exponent

mkmkasim · Answer

so it would be 1/sqrt(u)?

turingtest · Answer

nevermind, this is why i say watch the differential...
what is dt= in @SithsAndGiggles sustitution?

mkmkasim · Answer

1-x^2

turingtest · Answer

@mkmkasim ,no, that is t, what is dt?

mkmkasim · Answer

should be -2xdx, then you would bring -1/2 out in front of the integral.

turingtest · Answer

right, *then* you have dt/sqrt(t) as the integrand

mkmkasim · Answer

oaky then you sub in 1-x^2 for t, so the final answer should be 

xarcsinx + 1/2(1/sqrt(1-x^2))

turingtest · Answer

you didn't do the last integral though...

anonymous · Answer

@TuringTest