If \[\int\limits_0^\pi {xf\left( {\sin x} \right)} dx = A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\], then what is the value of \[A\]?
A. \[2\pi \]
B. \[\pi \]
C. \[\dfrac{\pi }{4}\]
D. 0

Hint: Here, an equation of the definite integral is given. First, simplify the left-hand side integral by applying the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]. Then, add the left-hand side and this simplified integral and solve it. After that, apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\] and solve the integral. In the end, compare the integral with the right-hand side of the originally given integral and get the required answer.



Formula Used:Integration rules:
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]



Complete step by step solution:The given equation of the definite integral is \[\int\limits_0^\pi {xf\left( {\sin x} \right)} dx = A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\].
Let consider,
\[I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin \left( {\pi - x} \right)} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)} dx\]
Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx + \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi f\left( {\sin x} \right)} dx\]
\[ \Rightarrow 2I = \pi \int\limits_0^\pi {f\left( {\sin x} \right)} dx\]
Here, \[f\left( {\sin x} \right) = f\left( {\sin \left( {\pi - x} \right)} \right)\] because \[\sin \left( {\pi - x} \right) = \sin x\] .
So, apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\].
We get,
\[ \Rightarrow 2I = 2\pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
\[ \Rightarrow I = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]

Now compare the integral with the right-hand side of the originally given integral.
We get,
\[A\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
\[ \Rightarrow A = \pi \]



Option ‘A’ is correct



Note:
Below are some rules to find the definite integral of a function by splitting it into parts:
\[\int\limits_0^{2a} {f\left( x \right)} dx = \int\limits_0^a {f\left( x \right)} dx + \int\limits_0^a {f\left( {2a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = 0\] if \[f\left( {2a - x} \right) = - f\left( x \right)\]