Let I1=∫aπ−a xf(sinx)dx,I2=∫aπ−a f(sinx)dx, then I2 is equal to
π2I1
πI1
2πI1
2I1
GivenI1=∫aπ−a xf(sinx)dxand I2=∫aπ−a f(sinx)dx
Now, I1=∫aπ−a xf(sinx)dx=∫aπ−a (π−x)f[sin(π−x)]dx=∫aπ−a (π−x)f(sinx)dx=∫aπ−a πf(sinx)dx−I1⇒ 2I1=πI2⇒ I2=2πI1