计算定积分I=∫(0→π)f(sinx)/[f(sinx)+f(cosx)]*dx,其中f(x)为连续函数,且f(sin
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计算定积分I=∫(0→π)f(sinx)/[f(sinx)+f(cosx)]*dx,其中f(x)为连续函数,且f(sinx)+f(cosx)不等于0
修改一下,上下限是:(0→π/2)
修改一下,上下限是:(0→π/2)
![计算定积分I=∫(0→π)f(sinx)/[f(sinx)+f(cosx)]*dx,其中f(x)为连续函数,且f(sin](/uploads/image/z/168480-0-0.jpg?t=%E8%AE%A1%E7%AE%97%E5%AE%9A%E7%A7%AF%E5%88%86I%3D%E2%88%AB%280%E2%86%92%CF%80%29f%28sinx%29%2F%5Bf%28sinx%29%2Bf%28cosx%29%5D%2Adx%2C%E5%85%B6%E4%B8%ADf%28x%29%E4%B8%BA%E8%BF%9E%E7%BB%AD%E5%87%BD%E6%95%B0%2C%E4%B8%94f%28sin)
令u=π/2 -x
则x=π/2 -u
原积分=∫(π/2→0) f( sin(π/2 -u) ) / [f( sin(π/2 -u) ) + f( cos(π/2 -u) )] d(π/2 -u)
= - ∫(π/2→0) f(cosu) / [f(cosu) + f(sinu)] du
= ∫(0→π/2) f(cosu) / [f(cosu) + f(sinu)] du
积分变量的字符对积分的结果没有影响,因此
∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx = ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
则
2 ∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx + ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2) [f(sinx)+f(cosx)] / [f(sinx)+f(cosx)] dx
=∫(0→π/2) 1 dx
= x|(0→π/2)
= π/2 - 0
= π/2
则原积分 = π/4
则x=π/2 -u
原积分=∫(π/2→0) f( sin(π/2 -u) ) / [f( sin(π/2 -u) ) + f( cos(π/2 -u) )] d(π/2 -u)
= - ∫(π/2→0) f(cosu) / [f(cosu) + f(sinu)] du
= ∫(0→π/2) f(cosu) / [f(cosu) + f(sinu)] du
积分变量的字符对积分的结果没有影响,因此
∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx = ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
则
2 ∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx + ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2) [f(sinx)+f(cosx)] / [f(sinx)+f(cosx)] dx
=∫(0→π/2) 1 dx
= x|(0→π/2)
= π/2 - 0
= π/2
则原积分 = π/4
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