计算定积分∫(-π,π)sin^2mxdx

题目

计算定积分∫(-π,π)sin^2mxdx

答案

∫(-π,π)sin^2mxdx=∫(-π,π){(1-cos2mx)/2}dx
=π-[∫(-π,π)cos2mxd2mx]/4m
=π-[sin(2mπ)-sin(-2mπ)]/4m
=π-sin(2mπ)/2m