Question

求积分，要详细的过程

Answer 1

1/π∫[-π,π]xsin2xdx
=1/2π∫[-π,π]xsin2xd2x
=-1/2π∫[-π,π]xdcos2x
=-1/2π xcos2x[-π,π]+1/2π ∫[-π,π]cos2xdx
=-1/2π [πcos2π-(-π)cos(-2π)]+1/4π ∫[-π,π]cos2xd2x
=-1/2π[2π]+1/4π sin2x[-π,π]
=-1/2 + 1/4π[sin2π-sin(-2π)]
=-1/2 + 0
=-1/2

Answer 2

1/π∫(-π,π)xsin2xdx
=-1/(2π)∫(-π,π)xdcos2x
=-1/(2π)(xcos2x)+1/(2π)∫(-π,π)cos2xdx
=-1/(2π)(xcos2x)+1/(4π)∫(-π,π)dsin2x
=-1/(2π)(xcos2x)+1/(4π)sin2x |(-π,π)
=-1