被积函数为cosx分之1 dx

求不定积分∫dx/cosx
∫dx/cosx=∫d(x+π/2)/sin(x+π/2)=∫d(x+π/2)/[2sin[(x+π/2)/2]cos[(x+π/2)/2]
=∫d[(x+π/2)/2]/{tan[(x+π/2)/2]cos²[(x+π/2)/2]}=∫d[tan[(x+π/2)/2]}/tan[(x+π/2)/2]
=lntan[(x+π/2)/2]+C=ln[1-cos(x+π/2)]/sin(x+π/2)]+C
=ln[csc(x+π/2)-cot(x+π/2)]+C=ln(secx+tanx)+C

∫[0，pi/2]dx/(1+(cosx)^2) = ∫[0，pi/2]dx/((sinx)^2+2(cosx)^2) = ∫[0，pi/2]dx/(cosx)^2[(tanx)^2+2] = ∫[0，pi/2

∫1/cosx*dx
=∫secxdx
=∫secx*(secx+tanx)/(secx+tanx)dx
=∫(sec^2x+secxtanx)/(secx+tanx)dx
=∫1/(secx+tanx)*d(secx+tanx)
=ln|secx+tanx|+c