高数 请问解法二划线部分怎么来的?
2个回答
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对称区间内对奇函数积分为零,添加的那项对整体积分没影响
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(1)
tany = 1/coty
=> arctany = arcot(1/y)
=> arctan(e^x) = arccot[ 1/e^(x) ] = arccot[e^(-x)]
(ii)
arctan(e^x) +arccot(e^x) =1
I
=5∫(-π/4-> π/4) 5cosx .arctane^x dx
=5∫(-π/4-> π/4) 5cosx .arccote^x dx
2I
=5∫(-π/4-> π/4) cosx .arctane^x dx +5∫(-π/4-> π/4) 5cosx .arccote^x dx
=5∫(-π/4-> π/4) cosx .(arctane^x +arccote^x )dx
=5∫(-π/4-> π/4) cosx dx
I = (5/2) ∫(-π/4-> π/4) cosx dx
tany = 1/coty
=> arctany = arcot(1/y)
=> arctan(e^x) = arccot[ 1/e^(x) ] = arccot[e^(-x)]
(ii)
arctan(e^x) +arccot(e^x) =1
I
=5∫(-π/4-> π/4) 5cosx .arctane^x dx
=5∫(-π/4-> π/4) 5cosx .arccote^x dx
2I
=5∫(-π/4-> π/4) cosx .arctane^x dx +5∫(-π/4-> π/4) 5cosx .arccote^x dx
=5∫(-π/4-> π/4) cosx .(arctane^x +arccote^x )dx
=5∫(-π/4-> π/4) cosx dx
I = (5/2) ∫(-π/4-> π/4) cosx dx
追问
arctan(e^x) +arccot(e^x) =1 请问这一步为什么等于1
追答
不好意思应该是 arctan(e^x) +arccot(e^x) =π/2
tan(A+B)= (tanA+tanB)/(1- tanA.tanB)
A+B
= arctan[(tanA+tanB)/(1- tanA.tanB) ]
A=arctan(e^x) , B=arctan(e^(-x) )
arctan(e^x) + B=arctan(e^(-x) )
=arctan[ e^x+e^(-x) )/(1- 1) ]
=arctan(∞)
=π/2
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