定积分,求详细的计算过程
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∫(0->π/2) (acosx+bsinx)^2 dx
=∫(0->π/2) [a^2.(cosx)^2+2absinx.cosx + b^2.(sinx)^2 ] dx
=(1/2)∫(0->π/2) [a^2.(1+cos2x)+4absinx.cosx + b^2.(1-cos2x) ] dx
=(1/2)∫(0->π/2) [a^2.(1+cos2x)+2absin2x + b^2.(1-cos2x) ] dx
=(1/2) [a^2.(x+(1/2)sin2x)-abcos2x + b^2.(x-(1/2)sin2x) ]| (0->π/2)
=(1/4)π(a^2 +b^2) + ab
=∫(0->π/2) [a^2.(cosx)^2+2absinx.cosx + b^2.(sinx)^2 ] dx
=(1/2)∫(0->π/2) [a^2.(1+cos2x)+4absinx.cosx + b^2.(1-cos2x) ] dx
=(1/2)∫(0->π/2) [a^2.(1+cos2x)+2absin2x + b^2.(1-cos2x) ] dx
=(1/2) [a^2.(x+(1/2)sin2x)-abcos2x + b^2.(x-(1/2)sin2x) ]| (0->π/2)
=(1/4)π(a^2 +b^2) + ab
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