If ∫dxsinxcosx=log|f(x)|+C then
f(x)=sinx+cosx
f(x)=tanx
f(x)=sec2x
none of these
∫dxsinxcosx=∫sin2x+cos2xsinxcosxdx=∫(tanx+cotx)dx=−log|cosx|+log|sinx|+C=log|tanx|+C. Thus f(x)=tanx.