Find limit of $\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt}$ Given that for each $a\in (0, 1)$,  $\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt}$ exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$. Find the value of $g(\frac{1}{2})$
My approach
$g(a)=\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^a}{{\left( {1 - t} \right)}^{ - 1}}dt} $
$g(a)=\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{{\left( {\frac{{1 - t}}{t}} \right)}^a}{{\left( {1 - t} \right)}^{ - 1}}dt}$
$g(a)=\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\frac{1}{{1 - t}}{{\left( {\frac{{1 - t}}{t}} \right)}^a}dt} $
From here onward not able to approach

$$g\left(\frac{1}{2}\right) = \int_0^1 \frac{dx}{\sqrt{x(1-x)}}$$
Now, if $u = \sqrt{x} \implies du = \frac{dx}{2\sqrt{x}}$
$$\implies g\left(\frac{1}{2}\right) = 2\int_0^1\frac{du}{\sqrt{1-u^2}} = 2[\arcsin(u)]_0^1 = \pi$$
