MathematicsEasy210×since 2002Q3340∫0πxf(sinx)dx\int\limits_0^\pi {xf\left( {\sin x} \right)dx}0∫πxf(sinx)dx is equal toAπ∫0πf(cosx)dx\pi \int\limits_0^\pi {f\left( {\cos x} \right)dx}π0∫πf(cosx)dxB π∫0πf(sinx)dx\,\pi \int\limits_0^\pi {f\left( {sinx} \right)dx}π0∫πf(sinx)dxCπ2∫0π/2f(sinx)dx{\pi \over 2}\int\limits_0^{\pi /2} {f\left( {sinx} \right)dx}2π0∫π/2f(sinx)dxDπ∫0π/2f(cosx)dx\pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}π0∫π/2f(cosx)dxCheck answerSkip