Question

求解积分，要详细过程

Answer 1

∫costx(2+x)/4dx
=1/2∫costxdx+1/4∫xcostxdx
=1/(2t)∫dsintx+1/(4t)∫xdsintx
=sintx/2t+(xsintx-∫sintxdx)/4t +C
=sintx/2t+(xsintx+1/t ∫dcostx)/4t+C
=sintx/2t +xsintx/4t+costx/(4t^2) +C

∫(-2→0) costx(2+x)/4dx
=1/(4t^2)+sin2t/2t-2sin2t/4t-cos2t/(4t^2)
=(1-cos2t)/4t^2
=(sint)^2/2t^2

∫sintx(2-x)/4dx
=1/2∫sintxdx+1/4∫x(-sintx)dx
=-1/(2t)∫dcostx+1/(4t)∫xdcostx
=-costx/2t+(xcostx-∫costxdx)/4t +C
=-costx/2t+(xcostx-1/t ∫dsintx)/4t+C
=-costx/2t +xcostx/4t-sintx/(4t^2) +C

i∫(0→2) sintx(2-x)/4dx
=i[-cos2t/2t +2cos2t/4t-sin2t/(4t^2)+1/2t]
=i/2t(1-sin2t/2t)

所以原式=(sint)^2/2t^2 +i/2t(1-sin2t/2t)