第1个回答 2016-11-16
令A=∫(0,π/2) sin^4x/(sin^4x+cos^4x)dx
B=∫(0,π/2) cos^4x/(sin^4x+cos^4x)dx
则A+B=∫(0,π/2)dx=π/2
A-B=∫(0,π/2) (sin^4x-cos^4x)/(sin^4x+cos^4x)dx
=∫(0,π/2) (sin^2x-cos^2x)/(1-2sin^2xcos^2x)dx
=∫(0,π/2) (2cos2x)/[sin^2(2x)-2]dx
=∫(0,π/2) d(sin2x)/[sin^2(2x)-2]
=(1/2√2)*ln|(sin2x-√2)/(sin2x+√2)||(0,π/2)
=0
所以A=B=π/4
即∫(0,π/2) sin^4x/(sin^4x+cos^4x)dx=π/4追问
B=∫(0,π/2) cos^4x/(sin^4x+cos^4x)dx
则A+B=∫(0,π/2)dx=π/2
A-B=∫(0,π/2) (sin^4x-cos^4x)/(sin^4x+cos^4x)dx
=∫(0,π/2) (sin^2x-cos^2x)/(1-2sin^2xcos^2x)dx
=∫(0,π/2) (2cos2x)/[sin^2(2x)-2]dx
=∫(0,π/2) d(sin2x)/[sin^2(2x)-2]
=(1/2√2)*ln|(sin2x-√2)/(sin2x+√2)||(0,π/2)
=0
所以A=B=π/4
即∫(0,π/2) sin^4x/(sin^4x+cos^4x)dx=π/4追问
这样的题只能用这个方法吗
追答不一定的，要具体问题具体分析
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