Question

求一个简单的不定积分,∫dx/sinx

Answer 1

∫(cscx)dx=∫(1/sinx)dx

=∫(sinx/sin²x)dx

=∫d(-cosx)/(1-cos²x),令u=cosx,只是用u代替cosx而不是换元积分法

=-∫du/(1-u²)

=-(1/2)∫[1/(1-u)+1/(1+u)]du

=-(1/2)∫du/(1-u)-(1/2)∫du/(1+u)

=(1/2)[∫d(1-u)/(1-u)-∫d(1+u)/(1+u)]

=(1/2)[ln|1-u|-ln|1+u|]+C

=(1/2)ln|(1-cosx)/(1+cosx)|+C

=ln|√[(1-cosx)/(1+cosx)]|+C

=ln|√[(1-cosx)/(1+cosx)*(1-cosx)/(1-cosx)]|+C

=ln|√[(1-cosx)²/(1-cos²x)]|+C

=ln|(1-cosx)/sinx|+C

=ln|cscx-cotx|+C

这个方法容易推导,但是比你那个简单吗?