Question

求一道定积分

Answer 1

解：
∵∫<-π/4,π/4>[e^x•cos²x/(e^x+1)]dx=∫<-π/4,π/4>[cos²x/(e^(-x)+1)]dx (分子分母同除e^x)
=∫<-π/4,π/4>[cos²x/(e^x+1)]dx (用-x代换x，并化简)
∴∫<-π/4,π/4>[e^x•cos²x/(e^x+1)]dx=(1/2)[2∫<-π/4,π/4>[e^x•cos²x/(e^x+1)]dx]
=(1/2)[∫<-π/4,π/4>[e^x•cos²x/(e^x+1)]dx+∫<-π/4,π/4>[e^x•cos²x/(e^x+1)]dx]
=(1/2)[∫<-π/4,π/4>[e^x•cos²x/(e^x+1)]dx+∫<-π/4,π/4>[cos²x/(e^x+1)]dx]
=(1/2)∫<-π/4,π/4>[e^x•cos²x/(e^x+1)+cos²x/(e^x+1)]dx
=(1/2)∫<-π/4,π/4>[(e^x+1)cos²x/(e^x+1)]dx
=(1/2)∫<-π/4,π/4>cos²xdx
=∫<0,π/4>cos²xdx
=(1/2)∫<0,π/4>[1+cos(2x)]dx (应用倍角公式)
=(1/2)(π/4+1/2)
=(π+2)/8。