Задача 42420 Найти длину дуги кривой [m] y = \ln \sin...

barca121

2019-12-09 17:37:27

Найти длину дуги кривой [m] y = \ln \sin x [/m] от [m] x = \frac{\pi}{3} [/m] до [m] x = \frac{\pi}{2} [/m]

sova

2019-12-09 19:26:22

f(x)=lnsinx
f`(x)=(1/sinx)*(sinx)`=cosx/sinx=ctgx



L= ∫ ^(π/2)_(π/3)sqrt(1+(ctgx)^2) dx= ∫ ^(π/2)_(π/3)sqrt(1/sin^2x) dx=∫ ^(π/2)_(π/3)(1/sinx) dx=ln|tg(x/2)|)||^(π/2)_(π/3)=

=ln|tg(π/4)|-ln|tg(π/6)|=ln1-ln(1/sqrt(3))=0-ln3^(-1/2))=(1/2)ln3


О т в е т. (1/2)ln3