Question

Решите пожалуйста, заранее спасибо

Answer 1

$\int\limits^1_0 {\frac{\sqrt{arctg x}}{1+x^2}} \, dx; \\ \\ t=arctgx; \ \ dt=\frac{dx}{1+x^2} \\ \\ \int\limits^1_0 {\frac{\sqrt{t}}{1+x^2}}\cdot (1+x^2) \, dt; = \int\limits^1_0 {\sqrt{t} \, dt;=\int\limits^1_0 {t^\frac{1}{2} \, dt;=\frac{t^\frac{3}{2}}{\frac{3}{2}}|^1_0 = \frac{2}{3}\cdot \sqrt{t^3}|^1_0=\frac{2}{3}\cdot \sqrt{arctg^3x}|^1_0=$

$=\frac{2}{3}\cdot ( \sqrt{arctg^31}-\sqrt{arctg^30})=\frac{2}{3}\cdot (\sqrt{(\frac{\pi}{4})^3}-\sqrt{0^3})=\frac{2}{3}\cdot \sqrt{(\frac{\pi}{4})^3}=\frac{2}{3}\cdot \sqrt{\frac{\pi^3}{64}}= \\ \\ = \frac{2}{3}\cdot \frac{\sqrt{\pi^3}}{8}=\frac{\sqrt{\pi^3}}{12}$