Let
I=∫0πf(x)sinxdx...(1)
Now using the property ∫abf(x)dx=∫abf(a+b−x)dx we get,
⇒I=∫0πf(π−x)sin(π−x)dx
⇒I=∫0πf(π−x)sinxdx...(2)
Adding (1)&(2), we get
2I=∫0π[f(x)+f(π−x)]sinxdx
⇒2I=π2∫0πsinxdx {as given f(x)+f(π−x)=π2}
⇒2I=π2[−cosx]0π
⇒2I=2π2
⇒I=π2
Hence this is the correct option.