If `int_0^pi x f(sinx) dx=A int_0^(pi/2) f(sinx)dx`, then `A` is (A) `pi/2` (B) `pi` (C) `0` (D) `2pi`

Explanation

Given:`int_0^pi x f(sinx) dx=A int_0^(pi/2) f(sinx)dx`

As `int_0^(2a) f(x)dx=2 int_0^a f(x)dx`

So,` A int_0^(pi/2) f(sinx)dx=pi/2 int_0^pi f(sinx)dx=int_0^pi x f(sinx) dx`

`implies A int_0^(pi/2) f(sinx)dx=pi/2xx2 int_0^(pi/2) f(sinx)dx`

`implies A int_0^(pi/2) f(sinx)dx=pi int_0^(pi/2) f(sinx)dx`

`implies A=pi`

Hence correct option is (b).