11.09.2008, 23:53
общий
это ответ
Здравствуйте, Ольга!
Пусть x=tg t, тогда dx=dt/〖cos〗^2 t, x=1 t=π/4 ; x=3 t= π/3
∫_1^3▒√(x^2+1)/x^2 dx=∫_(π/4)^(π/3)▒〖√(〖tg〗^2 t+1)/(〖tg〗^2 t) dt/(〖cos〗^2 t)〗=∫_(π/2)^(π/3)▒dt/(〖sin〗^2 t*cost)=∫_(π/2)^(π/3)▒(〖cos〗^2 t+〖sin〗^2 t)/(〖sin〗^2 t*cost) dt=∫_(π/4)^(π/3)▒dt/cost+∫_(π/4)^(π/3)▒cost/(〖sin〗^2 t) dt=├ [ln|tgt+1/cost|-1/sint]┤| (π/3)¦(π/4)=ln|tgπ/3+1/(cosπ/3)|-1/(sinπ/3)-ln|tgπ/4+1/(cosπ/4)|+1/(sinπ/4)
Приложение:
Пусть x=tg t, тогда dx=dt/〖cos〗^2 t, x=1 t=π/4 ; x=3 t= π/3
∫_1^3▒√(x^2+1)/x^2 dx=∫_(π/4)^(π/3)▒〖√(〖tg〗^2 t+1)/(〖tg〗^2 t) dt/(〖cos〗^2 t)〗=∫_(π/2)^(π/3)▒dt/(〖sin〗^2 t*cost)=∫_(π/2)^(π/3)▒(〖cos〗^2 t+〖sin〗^2 t)/(〖sin〗^2 t*cost) dt=∫_(π/4)^(π/3)▒dt/cost+∫_(π/4)^(π/3)▒cost/(〖sin〗^2 t) dt=├ [ln|tgt+1/cost|-1/sint]┤| (π/3)¦(π/4)=ln|tgπ/3+1/(cosπ/3)|-1/(sinπ/3)-ln|tgπ/4+1/(cosπ/4)|+1/(sinπ/4)