定积分。

如题所述

第1个回答  2015-03-03

第2个回答  2015-03-03
∫(0->π)√(1+sin2x)dx
=∫(0->π)√(sin^2x+2sinxcosx+cos^2x)dx
=∫(0->π)|sinx+cosx|dx
=∫(0->3π/4) (sinx+cosx)dx +∫(3π/4->π) -(sinx+cosx)dx
=∫(0->3π/4)sinxdx+∫(0->3π/4)cosxdx-∫(3π/4->π)sinxdx-∫(3π/4->π)cosxdx
=-cosx|(0->3π/4)+sinx|(0->3π/4)+cosx|(3π/4->π)-sinx|(3π/4->π)
=-(-√2/2-1)+(√2/2-0) +(-1-(-√2/2))-(0-√2/2)
=√2/2+1+√2/2 -1+√2/2+√2/2
=2√2本回答被提问者采纳
第3个回答  2015-03-03
我会追答

∫(0->π)√(1+sin2x)dx
=∫(0->π)√(sin^2x+2sinxcosx+cos^2x)dx
=∫(0->π)|sinx+cosx|dx
=∫(0->3π/4) (sinx+cosx)dx +∫(3π/4->π) -(sinx+cosx)dx
=∫(0->3π/4)sinxdx+∫(0->3π/4)cosxdx-∫(3π/4->π)sinxdx-∫(3π/4->π)cosxdx
=-cosx|(0->3π/4)+sinx|(0->3π/4)+cosx|(3π/4->π)-sinx|(3π/4->π)
=-(-√2/2-1)+(√2/2-0) +(-1-(-√2/2))-(0-√2/2)
=√2/2+1+√2/2 -1+√2/2+√2/2
=2√2

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