Calculating improper integral $\int\limits_0^1 \frac{\arcsin(x)}{\sqrt{1-x^2}}\,dx$

Question

Calculate improper integral $\displaystyle \int_0^1\dfrac{\arcsin(x)}{\sqrt{1-x^2}}\,dx$

We had the following equation to calculate improper integrals (2nd style):
Let f in$\left(a,b\right]$ unbounded, but $\forall \varepsilon >0$ in every subinterval $\left[a+\varepsilon,b\right]$ is bounded, we define:

$\displaystyle \int_a^b f(x)\,dx:=\lim_{\varepsilon\to0^+}\displaystyle \int_{a+\varepsilon}^b f(x)\, dx$

However, I only came up with this solution:
\begin{align} &\displaystyle \int_0^1 \dfrac{\arcsin(x)}{\sqrt{1-x^2}}\,dx &&\ \mid \ u=\arcsin x \to dx=\sqrt{1-x^2}\, du\quad u=\arcsin0=0 \quad u=\arcsin1=\pi/2\\ &=\displaystyle \int_0^{\frac\pi2}u\,du\\ &= \left[\frac{u^2}{2}\right]^{\frac\pi2}_0\\ &=\frac{\pi^2}{8} \end{align} My question is now, which is more accurate (and is this even correct)?

Answer 1

As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider

$$\int_0^b\frac{\arcsin(x)}{\sqrt{1-x^2}}\,dx,$$ then take the limit as $b\to 1^-.$ This will lead to the answer you found.

Answer 2

In fact you show only one solution, because the first approach remains theoretical. The only way that I know to handle this integral is by the arc sine change of variable, which makes the "improperness" disappear.

Answer 3

The problem isn't at $x=0$, but instead at $x=1$. However, your substitution seems to have eliminated this issue, so yes, your answer is correct.

Answer 4

$$I\;=\;\lim_{b \to 1} \int_{0}^{b} \frac{\arcsin x}{\sqrt{1-x^2}}\,dx$$ Let $$u\;=\;\arcsin x$$ Then $$du\;=\; \frac{1}{\sqrt{1-x^2}}\,dx$$ We have $$I\;=\;\lim_{b \to 1} \int_{0}^{b} (\arcsin x) \,d(\arcsin x)$$ $$=\;\lim_{b \to 1}\frac{1}{2}\arcsin^2x \Big\lvert_{0}^b\;=\;\frac{\pi^2}{8}$$