Question

求函数f(x)=cos2x/cosx+sinx在闭区间[0,派/2]上的定极分

Answer 1

∫(0→π/2)cos2x/cosx+sinx dx
=∫(0→π/2)(cosx+sinx)(cosx-sinx)/(cosx+sinx) dx
=∫(0→π/2) (cosx-sinx) dx
=(sinx+cosx) | (0→π/2)
=(sin(π/2)+cos(π/2))-(sin(0)+cos(0))
=1-1
=0