Question

计算这个积分

Answer 1

∫[-π/4,π/4]x/(1+sinx)dx=∫[-π/4,π/4]x(1-sinx)/cos^2xdx
=∫[-π/4,π/4]xd(tanx-1/cosx)
=x(tanx-1/cosx)|[-π/4,π/4]-∫[-π/4,π/4](tanx-1/cosx)dx
=-π/√2+∫[-π/4,π/4](1-sinx)/cosxdx
=-π/√2+∫[-π/4,π/4]cosx/(1+sinx)dx
=-π/√2+ln(1+sinx)|[-π/4,π/4]
=-π/√2+2ln(√2+1)

Answer 2

∫x/(1+sinx)dx=∫xd[∫1/(1+sinx)]=x∫1/(1+sinx)-∫[∫1/(1+sinx)dx]dx

∫1/(1+sinx)dx=∫(1-sinx)/(cosx)^2dx=∫(secx)^2dx+∫dcosx/(cosx)^2
=tanx-1/cosx=tanx-secx

∫[∫1/(1+sinx)dx]dx=∫(tanx-secx)dx=-ln|cosx|-ln|tanx+secx|=-ln(1+sinx)

所以∫x/(1+sinx)dx=x(tanx-secx)+ln(1+sinx)+C
(-π/4,π/4)∫x/(1+sinx)dx=-√2π/2+ln(3+√2)