对xcos^2xdx求定积分,上限2派=360✀下限0

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2024-11-19 03:29:20
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回答1:

∫(0到2π) x*cos�0�5x dx=∫(0到2π) x*1/2*(1+cos2x) dx=1/2*∫(0到2π) x dx+1/2*∫(0到2π) x*cos2x dx=1/2*1/2*x�0�5(0到2π)+1/2*∫(0到2π) x*cos2x dx 对于∫(0到2π) x*cos2x dx=∫(0到2π) x*d(1/2*sin2x)=1/2*x*sin2x(0到2π)-1/2*∫(0到2π) sin2x dx=0+1/4*cos2x(0到2π)=0∴原式=1/2*1/2*x�0�5(0到2π)+0=π�0�5