If $ \int\limits_0^\pi {xf(\sin x)dx = A} \int\limits_0^{\pi /2} {f(\sin (x))dx} $ , then $ A $ is
1. $ 0 $
2. $ \pi $
3. $ \dfrac{\pi }{4} $
4. $ 2\pi $

Hint: Use the properties of definite integral to simplify LHS in terms of RHS. Use the properties in accordance with the requirements of the integral and finally compare the given integral to obtain value to find the value of A.

Complete step-by-step answer:
 $ \int\limits_0^\pi {xf(\sin x)dx = A} \int\limits_0^{\pi /2} {f(\sin (x))dx} $ . . . (1)
Let $ {I_1} = \int\limits_0^\pi {xf(\sin x)dx} $ . . . (2)
We have a property,
 $ \int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } $
By using this property, we can write equation (2) as
 $ {I_1} = \int\limits_0^\pi {(\pi - x)f(\sin (\pi - x))dx} $
 $ {I_1} = \int\limits_0^\pi {(\pi - x)f(\sin x)dx} $ $ \left( {\because \sin (\pi - x) = \sin x} \right) $ . . . (3)
Adding (2) and (3)
 $ \Rightarrow 2{I_1} = \int\limits_0^\pi {[xf(\sin x) + (\pi - x)f(\sin x)]dx} $
Taking $ f(\sin x) $ common, we get
 $ 2{I_1} = \int\limits_0^\pi {[(x + \pi - x)f(\sin x)]dx} $
 $ \Rightarrow 2{I_1} = \int\limits_0^\pi {\pi f(\sin x)dx} $
 $ \Rightarrow 2{I_1} = \pi \int\limits_0^\pi {f(\sin x)dx} $
We have a property,
 $ \int\limits_0^{2a} {f(x)dx = 2} \int\limits_0^a {f(2a - x)dx,} $ if $ f(2a - x) = f(x) $
 $ \because \sin (\pi - x) = \sin x, $ we can use above property to get
 $ 2{I_1} = 2\pi \int\limits_0^{\pi /2} {f(\sin x)dx.} $
 $ \Rightarrow {I_1} = \pi \int\limits_0^{\pi /2} f (\sin x)dx $
Substituting this value of $ {I_1} $ in equation (1) we get
 $ \pi \int\limits_0^{\pi /2} f (\sin x)dx = A\int\limits_0^{\pi /2} {f(\sin (x))dx} $
Cancelling the common terms, we get
 $ A = \pi $
Therefore, from the above explanation, the correct answer is, option (2) $ \pi $

So, the correct answer is “Option 2”.

Note: You won’t be able to solve this question without using the properties of definite integral. This question is an ideal example to let you know that, using properties of definite integral to solve the question of definite integrals is useful.