# Limit of $\int_1^a \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}},a\rightarrow1^+$

I would like to find the limit of $$\int_1^a \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}}$$

when $$a\rightarrow1^+$$ It seems that $$\int_1^a \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}}\sim_{a\rightarrow 1^+} \pi$$

What bothers me is that $a$ is in the integrand and I cannot find an equivalent of $$\frac{1}{\sqrt{t(t-1)(a-t)}}$$ when $a\rightarrow1^+$

Moreover the integral $$\int \frac{\mathrm dt}{\sqrt{t(t-1)(a-t)}}$$ "cannot be computed", is not simple.

Do you have any idea?

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Don't say "cannot be computed". It is an elliptic integral: $$U(a):=\int_{1}^{a} \frac{dt}{\sqrt{t (t - 1) (a - t)}} = \frac{2 K \Bigl(\sqrt{\frac{a - 1}{a}}\Bigr)}{\sqrt{a}}$$ and of course $2K(0) = \pi$.
Now, of course, the question is to show $K(0)=\pi/2$ directly, without using knowledge of elliptic integrals. For that I chose a change of variables: $u=(t-1)/(a-1)$ to make this an integral from $0$ to $1$ $$U(a) = \int_0^1\frac{du}{\sqrt{u(1-u)(1+(a-1)u)}}$$ Now the limit at $a=1$ is clear: $$U(1) = \int_0^1 \frac{du}{\sqrt{u(1-u)}} = \pi$$