Given, $$g(a)=\lim _\limits{h \rightarrow 0^{+}} \int_h^{1-h} t^{-a}(1-t)^{a-1} d t$$ is differentiable for $$a \in(0,1)$$.

$$\begin{aligned} & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \int_h^{1-h} \frac{1}{\sqrt{t-t^2}} d t \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \int_h^{1-h} \frac{d t}{\sqrt{\left(\frac{1}{2}\right)^2-\left[t^2-2 \cdot \frac{1}{2} t+\left(\frac{1}{2}\right)^2\right]}} \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}}^{1-h} \frac{d t}{\sqrt{\left(\frac{1}{2}\right)^2-\left(t-\frac{1}{2}\right)^2}} \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}}\left[\sin ^{-1}\left(\frac{\left(t-\frac{1}{2}\right)}{\frac{1}{2}}\right)\right]_h^{1-h} \\ & \Rightarrow g\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}}[\sin (1-2 h)-\sin (2 h-1)] \\ & \Rightarrow g\left(\frac{1}{2}\right)=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi \end{aligned}$$

Hint:

(i) For the integral of $$\int \frac{d x}{\sqrt{a x^2+b x+c}}$$, Convert $$a x^2+b x+c$$ into perfect square like $$a x^2+b x+c=a\left[\left(x+\frac{b}{2 a}\right)^2 \pm \lambda^2\right]$$.

(ii) $$\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$$