Монгол Бодлогын Сан

Заавар:

Бодолт: $t=\arctg x$ гэвэл $x=\tg t$ ба $t\in[-\pi/2;\pi/2]$ байна. $$\begin{align*} \dfrac{x}{\sqrt{1+x^2}}&=\dfrac{\tg t}{\sqrt{1+\tg^2t}}=\dfrac{\tg t}{\sqrt{1/\cos^2t}}\\ &=\dfrac{\tg t}{|1/\cos t|}=\dfrac{\sin t/\cos t}{1/\cos t}\\ &=\sin t=\sin(\arctg x) & &\color{red} {\leftarrow\cos t\ge 0\Rightarrow |1/\cos t|=1/\cos t} \end{align*}$$