∫[0,π/4] (tanx)^3dx=∫[0,π/4][(secx)^2-1]tanxdx=∫[0,π/4](secx)^2tanxdx -∫[0,π/4]tanxdx=∫[0,π/4]tanxdtanx +∫[0,π/4]dcosx/cosx=(1/2)tanx|[0,π/4] +ln|cosx| |[0,π/4]=1/2+lncos(π/4)最后答案是1/2(1-ln2)啊···你得解答最后几步看不懂·····1/2+lncos(π/4) =1/2+ln(√2/2)=1/2-ln√2=1/2-(1/2)ln2=(1/2)(1-ln2)=∫[0,π/4](secx)^2tanxdx -∫[0,π/4]tanxdx(tanx)'=(secx)^2 sinxdx/cosx=-dcosx/cosx=dlncosx=∫[0,π/4]tanxdtanx +∫[0,π/4]dcosx/cosx