Задача 45348 ...

найти длину дуги

r = 1/φ от φ = 3/4 до φ = 4/3

По формуле:

r=(1/ φ )
r`( φ )=–1/ φ ²

√r²+(r`)²=√(1/φ)²+(–1/ φ ²)²d φ=(1/ φ )√ φ ²+1

L= ∫ ^4/3 _3/4 (√ φ ²+1/ φ )d φ =


замена переменной:
φ =tgt ⇒ d φ =dt/cos²t

φ ²+1=tg²+1=1/cos²t

√φ ²+1=1/cost


=∫ ^{arctg( 4/3)} _arctg(3/4) (1/cost)/ tgt )d t/cos²t =

=∫ ^{arctg( 4/3)} _arctg(3/4) dt/(sint·cos²t)


1=cos²t+sin²t


=∫ ^{arctg( 4/3)} _arctg(3/4) (sin²t+cos²t)dt/(sint·cos²t)

=∫ ^{arctg( 4/3)} _arctg(3/4) (sint/cos²t)dt + ∫ ^{arctg( 4/3)} _arctg(3/4) dt/sint=

=((1/cost)+ln |tg(t/2)|)|^(arctg(4/3)_(arctg(3/4)=

=1/cos(arctg(4/3))–1/cos(arctg(3/4))+ln|tg(arctg(4/3))/2|–ln|tg(arctg(3/4))/2|

Теперь упрощаем

arctg(4/3)= α ⇒ tg α =(4/3) ⇒ cos α =? tg( α /2)=?

cos α =√1/(1+tg² α )=√9/25=3/5 ⇒ sin α =4/5


tg( α /2)=sin α /(1+cos α )=1/2



arctg(3/4)= β ⇒ tg α =(3/4) ⇒ cos α =? tg( α /2)=?

cos α =√1/(1+tg² α )=√16/25=4/5 ⇒ sin α =3/5


tg( α /2)=sin α /(1+cos α )=1/3

О т в е т. =(5/3)–(5/4)+ln(1/2|–ln|1/3|=(5/12)+ln(3/2)