JEE MainMathematicsDefinite IntegrationMCQ+4 / −1∫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)dxAttempt in chatSharePreviousNextMore from Definite Integration−23π∫−2π[(x+π)3+cos2(x+3π)]dx is equal to2006 · MCQLet f:R→R be a differentiable function having f(2)=6, f′(2)=(481). Then x→2lim6∫f(x)x−24t3dt…2005 · MCQThe value of −π∫π1+axcos2dx,a>0, is2005 · MCQIf I1=0∫12x2dx,I2=0∫12x3dx,I3=1∫22x2dx and I4=1∫22x3dx then2005 · MCQThe value of integral, 3∫69−x+xxdx is2005 · MCQThe value of I=0∫π/21+sin2x(sinx+cosx)2dx is2004 · MCQThe value of −2∫31−x2dx is2004 · MCQIf 0∫πxf(sinx)dx=A0∫π/2f(sinx)dx, then A is2004 · MCQ