高等数学中的不定积分

Question

用换元法做

Answer 1

如下

Answer 2

∫dx/[(sinxcosx^4)
=∫ -dcosx/[(sinx^2cosx^4]
=∫-dcosx/[cosx^4(1-cosx^2)]
cosx=u
=∫-du/[u^4(1-u^2)]
=∫[(u^2-1)-u^2]du/[u^4(1-u^2)]
=∫-du/u^4-∫du/u^2(1-u^2)
=(1/3)(1/u^3) -[∫du/(1-u^2)-∫du/u^2]
=(1/3)(1/u^3)+∫du/u^2 -∫du/(u-1)(u+1)
=(1/3)(1/u^3)-(1/u)-(1/2)ln|(u+1)/(u-1)| +C
=(1/3)(1/cosx^3) -1/cosx -(1/2) |(cosx+1)/(cosx-1)|+C

Answer 3

原积分=∫ sinx / [(sinx)^2*(cosx)^4]dx
=∫ -1 / [ (1-(cosx)^2 ) * (cosx)^4 ] dcosx
=∫ 1 / [ (1-(cosx)^2 ) * (cosx)^2 ] d(1/cosx) 记1/cosx=t
=∫ t^4 / [ t^2 -1] dt
=∫ t^2+1+1/(2t -2) - 1/(2t+2) dt
= t^3/3+t+1/2 *ln|t -1| +1/2 *ln|t +1| +c
= [1/cosx]^3 /3 + [1/cosx] +1/2 *ln| [1/cosx] -1| + 1/2 *ln| [1/cosx] +1| +c