Could someone please help me evaluate this integral? The integral is
$$
\int_{0}^{1/\sqrt{\vphantom{\large A}2\,}}
{\arccos\left(x\right) \over \sqrt{\vphantom{\large A}1 -x^2\,}\,}\;{\rm d}x 
$$
I was just wondering if I could use substitution to solve this problem or if I had to solve it a different way.

We have to evaluate the definite integral:
$$\int_0^{1/\sqrt{2}}\frac{\arccos x}{\sqrt{1-x^2}} \ dx$$
Just use u-substitution. Let:
$$u=\arccos x \implies du=-dx$$
$$\int_0^{1/\sqrt{2}}u \ -du=-\int_0^{1/\sqrt{2}}u \ du=\left.-\frac{u^2}{2}\right|_0^{1/\sqrt{2}}$$
Substituting $\arccos x$ back for $u$:
$$\int_0^{1/\sqrt{2}}\frac{\arccos x}{\sqrt{1-x^2}} \ dx=\left.-\frac{1}{2}(\arccos x)^2+C \ \right|_0^{1/\sqrt{2}}$$
Now we just evaluate the definite integral.
$$\left.-\frac{1}{2}(\arccos x)^2+C \ \right|_0^{1/\sqrt{2}}=\left[-\frac{1}{2}\arccos \left(\frac{1}{\sqrt{2}}\right)^2+C\right]-\left[-\frac{1}{2}(\arccos 0 )^2+ C\right]$$
You should know that $\arccos \left(\frac{1}{\sqrt{2}}\right)=\dfrac{\pi}{4}$ and $\arccos 0=\dfrac{\pi}{2}$.
$$-\frac{1}{2}\cdot\left(\frac{\pi}{4}\right)^2+C+\frac{1}{2}\cdot\left(\frac{\pi}{2}\right)^2-C$$
$$=-\frac{1}{2}\cdot\frac{\pi^2}{16}+\frac{1}{2}\cdot\frac{\pi^2}{4}$$
$$=\frac{\pi^2}{8}-\frac{\pi^2}{32}$$
$$=\frac{4\pi^2}{32}-\frac{\pi^2}{32}$$
$$=\frac{3\pi^2}{32}$$
$$\displaystyle \color{green}{\therefore \int_0^{1/\sqrt{2}}\dfrac{\arccos x}{\sqrt{1-x^2}} \ dx=\dfrac{3\pi^2}{32}}$$
Hope I helped!
