Given,
f(x)=2+xcosx2−xcosx, g(x)=logex
⇒g(f(x))=loge(2+xcosx2−xcosx)
⇒g(f(−x))=loge(2+(−x)cos(−x)2−(−x)cos(−x))
⇒g(f(−x))=loge(2−xcosx2+xcosx)
⇒g(f(−x))=−log(2+xcosx2−xcosx)
⇒g(f(−x))=–g(f(x))
Hence, g(f(x)) is an odd function.
By using the property of definite integration, \displaystyle {\int }_{-a}^{a}f(x)dx={\begin{matrix}2{\int }_{0}^{a} & f(x)dx,f(-x)=f(x) \\ 0, & f(-x)=-f(x)\end{matrix}, we can write
∫−4π4πg(f(x))dx=0=loge1