Question

Даю 80 баллов. Доделать задание "вычислить длину дуги кривой, заданной уравнением"

Answer 1

$y=ln\, cosx\; \; ,\; \; \frac{\pi }{3}\leq x\leq \frac{2\pi }{3}\\\\l= \int\limits^{b}_{a}\sqrt{1+(y')^2}\, dx\\\\y'=\frac{1}{cosx}\cdot (-sinx)=-tgx\\\\\sqrt{1+(y')^2}=\sqrt{1+tg^2x} =\sqrt{\frac{1}{cos^2x}}=\frac{1}{|cosx|}=\frac{1}{cosx}\; ,\\\\t.k.\; cosx\ \textgreater \ 0\; \; pri\; \; \; \frac{\pi }{3}\leq x\leq \frac{2\pi }{3}\\\\l=\int\limits^{\frac{2\pi}{3}}_{\frac{\pi}{3}} \frac{1}{cosx}\, dx=ln|tg(\frac{x}{2}+\frac{\pi}{4})|\Big |_{\frac{\pi}{3}}^{\frac{2\pi }{3}}=\\\\=ln|tg(\frac{\pi }{3}+\frac{\pi}{4})|-ln|tg(\frac{\pi}{6}+\frac{\pi}{4})|=ln|tg\frac{7\pi }{12}|-ln|tg\frac{5\pi }{12}|$

$P.S.\; \; \int \frac{dx}{cosx}=\int \frac{dx}{sin(\frac{\pi }{2}+x)}=\int \frac{dx}{sin(2\cdot (\frac{\pi}{4}+\frac{x}{2}))}=\\\\=\int \frac{dx}{2\cdot sin(\frac{\pi}{4}+\frac{x}{2})\cdot cos(\frac{\pi}{4}+\frac{x}{2})}=\int \frac{dx}{2\cdot \frac{sin(\frac{\pi}{4}+\frac{x}{2})}{cos(\frac{\pi}{4}+\frac{x}{2})}\cdot cos^2(\frac{\pi}{4}+\frac{x}{2})}=\\\\=\int \frac{dx}{2\cdot tg(\frac{\pi}{4}+\frac{x}{2})\cdot cos^2(\frac{\pi}{4}+\frac{x}{2})}=\int \frac{d(tg(\frac{\pi}{4}+\frac{x}{2}))}{tg(\frac{\pi}{4}+\frac{x}{2})}=ln|tg(\frac{x}{2}+\frac{\pi}{4})|+C$