tanx的开方+cosx的开方在0到π/2上的定积分
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tanx的开方+cosx的开方在0到π/2上的定积分
tanx的开方+cosx的开方在0到π/2上的定积分
tanx的开方+cosx的开方在0到π/2上的定积分
先考虑A=∫(tanx)^(1/2)dx 令t=(tanx)^(1/2) 则t∈[0,∞] 2tdt=[(tanx)^2+1]dx dx=2tdt/(t^4+1)
A=∫2t^2dt/(1+t^4) =∫(t^2-1)dt/(t^4+1)+∫(t^2+1)dt/(t^4+1)
= ∫d(t+1/t)/[(t+1/t)^2-2] +∫d(t-1/t)/[(t-1/t)^2+2]
=2^(1/2)*1/4*In{[x^2-2^(1/2)*x+1]/[x^2+2^(1/2)*x+1]} + 2^(1/2)*1/2*arctan[(x^2-1)/(2^(1/2)*x)] +C
(代入上下限)=1/2*2^(1/2)*π
再考虑B= ∫(cosx)^(1/2)dx =∫[1-2*(sin(x/2))^2]^(1/2)dx =2*∫[1-2*(sinx)^2]^(1/2)dx x∈[0,π/4]
令sint=2^(1/2)*sinx则t∈[0,π/2] costdt=2^(1/2)cosxdx
dx=2^(1/2)*1/2*costdt/[1-1/2*(sint)^2]^(1/2)
原积分=2^(1/2)*∫(cost)^2/dt/[1-1/2*(sint)^2]^(1/2) 有积分区间[0,π/2]
=2*2^(1/2)*∫[1-1/2*(sint)^2]^(1/2)dt - 2^(1/2)*∫dt/[1-1/2*(sint)^2]^(1/2)
=-2^(1/2)*K(1/2*2^(1/2))+2*2^(1/2)*E(1/2*2^(1/2))
注:K(k),E(k)为第一,二类完全椭圆积分,为未定无理数