## Subject III. (10 points)

Calculate: $I=\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{(\arcsin x)^{2}}{\sqrt{1-x^{2}}+1+x} d x$

Prof. Ilie Diaconu, "Avram Iancu" Theoretical High School, Cluj-Napoca
## Subject III.

With the change of variable $\mathrm{x}=-\mathrm{t}$, we have:

$I=\int_{1 / 2}^{-1 / 2} \frac{(\arcsin (-t))^{2}}{\sqrt{1-t^{2}}+1-t}(-d t)=\int_{-1 / 2}^{1 / 2} \frac{(\arcsin t)^{2}}{\sqrt{1-t^{2}}+1-t} d t=\int_{-1 / 2}^{1 / 2} \frac{(\arcsin x)^{2}}{\sqrt{1-x^{2}}+1-x} d x$

By addition, we obtain:

$2 I=\int_{-1 / 2}^{1 / 2}(\arcsin x)^{2} \cdot\left(\frac{1}{\sqrt{1-x^{2}}+1+x}+\frac{1}{\sqrt{1-x^{2}}+1-x}\right) d x=$

$=\int_{-1 / 2}^{1 / 2}(\arcsin x)^{2} \cdot \frac{2\left(\sqrt{1-x^{2}}+1\right)}{\left(\sqrt{1-x^{2}}+1\right)^{2}-x^{2}} d x=\int_{-1 / 2}^{1 / 2}(\arcsin x)^{2} \cdot \frac{2\left(\sqrt{1-x^{2}}+1\right)}{2\left(1-x^{2}+\sqrt{1-x^{2}}\right)} d x=$

$=\int_{-1 / 2}^{1 / 2} \frac{(\arcsin x)^{2}}{\sqrt{1-x^{2}}} d x=2 \int_{0}^{1 / 2} \frac{(\arcsin x)^{2}}{\sqrt{1-x^{2}}} d x=\left.\frac{2}{3}(\arcsin x)^{3}\right|_{0} ^{1 / 2}=\frac{2}{3} \cdot\left(\frac{\pi}{6}\right)^{3} \Rightarrow I=\frac{\pi^{3}}{648}$

(10p)