题目

求 (sinx)^4/(cosx)^3不定积分

提问时间：2020-12-27

答案

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