求从0到π/2 (sinxcosx/1+cos^2x)dx

求从0到π/2 (sinxcosx/1+cos^2x)dx
求从0到π/2 (sinxcosx/1+cos^2x)dx

求从0到π/2 (sinxcosx/1+cos^2x)dx
1+cos²x=1+(1+cos2x)/2=(2cos2x+3)/2
所以原式=∫2sinxcosx/(2cos2x+3) dx
=∫sin2x/(2cos2x+3) dx
=-∫1/(2cos2x+3) d(cos2x)
=-1/2∫1/(2cos2x+3) d(2cos2x+3)
=-1/2*ln(2cos2x+3)
x=π/2,=0
x=0,=-1/2*ln5
所以原式=1/2*ln5