Integration of $\int_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx$ ?? I was solving the integration of inverse trigonometric function and faced a question which i find it hard to understand. I need to find the definite integration of this function.
$$\int_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx$$
I tried to use the substitutional method by 
$$u= \sin^{-1}(x)$$
and getting $\dfrac{du}{dx}= \dfrac{1}{\sqrt{1-x^2}}$
and $dx= du(\sqrt{1-x^2})$
but when i substitute that into the function, it does not make any sense.
This is where i got stuck 
(not even sure if i did in the right way or not..)
am i doing it right? Should I use another method to approach to the answer?
(Sorry if this question is duplicating, i could not find an appropriate answer..)
 A: $$
\int_0^{1/2} \underbrace{(\sin^{-1} x)}_{u} \underbrace{\left( \frac{dx}{\sqrt{1-x^2}} \right)}_{du} = \int u\,du
$$
When $x=0$ then $u=0$ and when $x=1/2$ then $u=\pi/6$, so you actually get
$$
\int_0^{\pi/6} u\,du.
$$
A: Let $y=\arcsin x$, then $x=\sin y\;\Rightarrow\;dx=\cos y\ dy=\sqrt{1-x^2}\ dy$, then
$$
\begin{align}
\require{cancel}\int_{x=0}^{\Large\frac12}\frac{\arcsin x}{\sqrt{1-x^2}}\ dx&=\int_{x=0}^{\Large\frac12}\frac{y}{\cancel{\sqrt{1-x^2}}}\cdot\cancel{\sqrt{1-x^2}}\ dy\\
&=\int_{x=0}^{\Large\frac12} y\ dy\\
&=\left.\frac12y^2\right|_{x=0}^{\Large\frac12}\\
&=\frac12\arcsin^2\left(\frac12\right)-\frac12\arcsin^2\left(0\right)\\
&=\large\color{blue}{\frac{\pi^2}{72}}.
\end{align}
$$
A: Notice the derivative of $\mathrm{Arcsin} \,x$ is $\frac{1}{\sqrt{1-x^2}}$, hence your integrand is of the form $u u'$ with $u=\mathrm{Arcsin}\,x$, and a primitive is $u^2/2$. Thus
$$\int_{0}^{\frac{1}{2}}\frac{\mathrm{Arcsin}\, x}{\sqrt{1-x^2}} dx= \frac12\left[\mathrm{Arcsin}^2\,x\right]_0^{1/2}=\frac12\mathrm{Arcsin}^2\,\frac12=\frac{\pi^2}{72}$$
