Question

Вычислить длины дуг кривых

Answer 1

$\displaystyle\\L=\int\limits^{\dfrac{\pi}{2} }_{\dfrac{\pi}{3} } {\sqrt{1+((\ln(\sin(x))')^2}} \, dx= \int\limits^{\dfrac{\pi}{2} }_{\dfrac{\pi}{3} } \sqrt{1+\bigg(\frac{1}{\sin(x)}*\cos(x)\bigg)^2 }\ dx=\\\\\\=\int\limits^{\dfrac{\pi}{2} }_{\dfrac{\pi}{3} }\sqrt{1+\cot^2(x)}\ dx=\int\limits^{\dfrac{\pi}{2} }_{\dfrac{\pi}{3} } \csc(x)\ dx=-\ln(\mid\csc(x)+\cot(x) \mid)\mid^{\frac{\pi}{2} }_{\frac{\pi}{3}} =$

$\displaystyle\\=-\ln(\mid\csc(\frac{\pi}{2} )+\cot(\frac{\pi}{2} ) \mid)-(-\ln(\mid\csc(\frac{\pi}{3} )+\cot(\frac{\pi}{3} ) \mid))=-\ln(\mid 1+0 \mid)+\\\\\\+\ln(\mid \frac{2\sqrt{3}}{3}+\frac{\sqrt{3}}{3}\mid)=-\ln(1)+\ln(\sqrt{3})=\ln\sqrt{3}=\frac{\ln(3)}{2}$