1个回答
展开全部
dt/dx = -cost
dt/cost = -dx
sect dt = - dx
∫sectdt
=∫dt/cost
=∫costdt/cos²t
=∫dsint/cos²t
=∫dsint/(1-sin²t)
=(1/2)[∫dsint/(sint+1)-∫dsint/(sint-1)]
=(1/2)(ln|sint+1|-ln|sint-1|)+C
=(1/2)ln|(sint+1)/(sint-1)|+C (对数里银衫橘分塌核子分母都乘以sint+1)
=(1/2)ln|(sint+1)²/cos²t|+C
=ln|(sint+1)/cost|+C
=ln|tant+sect|+C
所以锋团
ln|tant+sect|=-x+C1
tant+sect=Ce^(-x)
所以tan(y-x)+sec(y-x)=Ce^(-x)
dt/cost = -dx
sect dt = - dx
∫sectdt
=∫dt/cost
=∫costdt/cos²t
=∫dsint/cos²t
=∫dsint/(1-sin²t)
=(1/2)[∫dsint/(sint+1)-∫dsint/(sint-1)]
=(1/2)(ln|sint+1|-ln|sint-1|)+C
=(1/2)ln|(sint+1)/(sint-1)|+C (对数里银衫橘分塌核子分母都乘以sint+1)
=(1/2)ln|(sint+1)²/cos²t|+C
=ln|(sint+1)/cost|+C
=ln|tant+sect|+C
所以锋团
ln|tant+sect|=-x+C1
tant+sect=Ce^(-x)
所以tan(y-x)+sec(y-x)=Ce^(-x)
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询