Question

请教各位，这两道高等数学不定积分题该怎么做？

多谢多谢

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Answer 1

I. 原式=∫(1-sinx)/[cosx(1-sinx)(1+sinx)]dx
=∫(1-sinx)/(cosx)^3
=∫(secx)^3-∫sinx/(cosx)^3dx
∫sec³xdx=
∫secxdtanx=
secxtanx-∫tanxdsecx
=secxtanx-∫secxtan²xdx
=secxtanx-∫secx(sec²x-1)dx
=secxtanx-∫sec³xdx+∫secxdx
=secxtanx-∫sec³xdx+ln|secx+tanx|
则∫sec³xdx=1/2secxtanx+1/2ln|secx+tanx|+C
∫sinx/(cosx)^3dx= -∫1/(cosx)^3dxd(cosx)=1/[3(cosx)^2]+C
所以，原式=1/2secxtanx+1/2ln|secx+tanx|+1/[3(cosx)^2]+C
II. 分子分母同除以(sinx)^2
原式=∫(1/(sinx)^2)/(1+cotanx) dx
=-∫1/(1+cotanx) d(cotanx)
=-ln|1+cotanx|+C

Answer 2

15. I = ∫cosxdx/[(1+sinx)(cosx)^2] = ∫dsinx/[(1+sinx)^2(1-sinx)]
= (1/4)∫[2/(1+sinx)^2 + 1/(1+sinx) + 1/(1-sinx)]dsinx
= (-1/2)/(1+sinx) + (1/4)ln|(1+sinx)/(1-sinx)| + C
= (-1/2)/(1+sinx) + (1/2)ln|(1+sinx)/cosx| + C
16. I = ∫dx/[(sinx)^2(1+cotx)] = -∫dcotx/(1+cotx) = - ln|1+cotx| + C

Answer 3

公式公式，套公式就是了

Answer 4

Answer 5