The following result is from a paper. The author says it is not hard to show that:



> 
> $$\lim\_{t\to 1}\dfrac{1-t}{\sqrt{1+pt}}\int\_{0}^{t}\dfrac{a(1+pa)}{(1-a)^2}\left(4a\left[1-\left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt}\right]\right)^{-1/2}da=\dfrac{\pi}{4}\sqrt{p+1}$$ 
> 
> 
> 


But I try use Taylor formula $$\left(4a\left[1-\left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt}\right]\right)^{-1/2}=\dfrac{1}{2\sqrt{a}}\left(1-\dfrac{1}{2}\left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt}+o( \left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt})\right)$$
still can't Solve this problem. So how to solve this problem?
Thanks 



How to prove this integral equality with limits
I assume that $p>-1$. We change the variables, at first to $1-a=x$, $1-t=\varepsilon\rightarrow +0$, we need to check that
$$
\varepsilon\int\_{\varepsilon}^1 \frac{1+p(1-x)}{x^2}(1-x)^{-1/2}\left[1-\left(\frac{\varepsilon}{x}\right)^2\frac{1+p(1-x)}{1+p(1-\varepsilon)}\right]^{-1/2}dx\to \frac{\pi}2 (p+1).
$$
Note that the integral from $1/2$ to 1 is bounded (the singularity of $(1-x)^{-1/2}$ is summable), so we may replace upper limit to $1/2$ (this is to make the multiple $(1-x)^{-1/2}$ bounded.) Next, we denote $x=\varepsilon\cdot \tau$, the task becomes
$$
\int\_1^{1/(2\varepsilon)}\frac{1+p-p\varepsilon \tau}{\tau^2}(1-\varepsilon \tau)^{-1/2}\left(1-\tau^{-2}\cdot \frac{1+p-p\varepsilon\tau}{1+p-p\varepsilon}\right)^{-1/2}d\tau\rightarrow \frac{\pi}2 (p+1).
$$
If we replace $\varepsilon$ everywhere to 0 and upper limit to infinity, we get $$
\int\_1^{\infty} \frac{(1+p)d\tau}{\tau^2\sqrt{\tau^2-1}}=\frac{\pi}2(1+p)
$$
as desired (for evaluating integral denote $\tau=1/\cos \theta$, for example). Thus it suffices to justify this replacement of $\varepsilon$ to 0. This is routine, the most delicate part is with 
$$
1-\tau^{-2}\cdot \frac{1+p-p\varepsilon\tau}{1+p-p\varepsilon}=1-\tau^{-2}+\tau^{-2}\frac{p\varepsilon (\tau-1)}{1+p-p\varepsilon}=(1-\tau^{-2})(1+O(\varepsilon \tau)),
$$
which suffices. 


