If f :R→R and g:R→R are one to one, real valued functions, then the value of the integral ∫−ππ [f(x)+f(−x)][g(x)−g(−x)]dx is
0
π
1
None of these
Let,ϕ(x)=[f(x)+f(−x)][g(x)−g(−x)]
then ϕ(−x)=[f(−x)+f(x)][g(−x)−g(x)]ϕ(−x)=−ϕ(x)
⇒it odd function.
∴ ∫−ππ ϕ(x)dx=0⇒ ∫−ππ [f(x)+f(−x)][g(x)−g(−x)]dx=0