Question

一道高数题积分题求助？

Answer 1

Answer 2

令t=x-π/2，则x=t+π/2，dx=dt
原式=∫(-π/2,π/2) (t+π/2)sin^2(t+π/2)dt
=∫(-π/2,π/2) tcos^2tdt+(π/2)*∫(-π/2,π/2) cos^2tdt
因为tcos^2t是奇函数，所以∫(-π/2,π/2) tcos^2tdt=0
因为cos^2t是偶函数，所以∫(-π/2,π/2) cos^2tdt=2∫(0,π/2) cos^2tdt
原式=(π/2)*2∫(0,π/2) cos^2tdt
=(π/2)*∫(0,π/2) (1+cos2t)dt
=(π/2)*[t+(1/2)*sin2t]|(0,π/2)
=(π^2)/4

Answer 3

Answer 4

方法如下图所示，
请认真查看，
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追答