If f:R→R, g:R→R are continuous functions then the value of the integral
∫−π/2π/2 [(f(x)+f(−x))(g(x) – g(–x))] dx is
1
0
-1
π
Let F(x) =(f(x)+f(-x))(g(x)-g(-x)) so F(−x)=(f(−x)+f(x))(g(−x)−g(x))=−F(x).
Hence ∫−π/2π/2 f(x)dx=0. (since F(x) is odd function)