第1个回答 2019-03-17
I = ∫dx/[(sinx)^2 cosx] = ∫cosxdx/[(sinx)^2 (cosx)^2]
= ∫dsinx/{(sinx)^2 [1-(sinx)^2]} (u = sinx)
= ∫du/[u^2(1-u^2)] = ∫[1/u^2+1/(1-u^2)]du
= ∫du/u^2 + (1/2)∫[1/(1-u)+1/(1+u)]du
= -1/u + (1/2)[-ln|1-u| + ln|1+u|] + C
= -1/sinx + (1/2)ln|(1+sinx)/(1-sinx)| + C
另题仿做即可。本回答被提问者采纳