A=∫√tan(x)dx,B=∫√cot(x)dx

A+B=∫tan(x)^(1/2)+cot(x)^(1/2)dx
=∫[sin(x)/cos(x)]^(1/2)+[cos(x)/sin(x)]^(1/2)dx
=∫[sin(x)+cos(x)]/[sin(x)cos(x)]^(1/2)dx

令x=y+π/4
=∫[sin(y+π/4)+cos(y+π/4)]/[sin(y+π/4)cos(y+π/2)]^(1/2)dx
而
sin(y+π/4)=[cos(y)-sin(y)]/2^(1/2)
cos(y+π/4)=[cos(y)+sin(y)]/2^(1/2)
所以
sin(y+π/4)+cos(y+π/4)=2^(1/2)cos(y)
sin(y+π/4)cos(y+π/4)=[cos(y)^2-sin(y)^2]/2
即∫[sin(y+π/4)+cos(y+π/4)]/[sin(y+π/4)cos(y+π/4)]^(1/2)dx
=∫2^(1/2)cos(y)/[[cos(y)^2-sin(y)^2]/2]^(1/2)dx
=2∫dsin(y)/[1-2sin(y)^2]^(1/2)
考虑
∫dt/√(1-2t^2)
=1/√2∫d(√2t)/√(1-2t^2)
=Arcsin(√2t)/√2
所以
2∫dsin(y)/[1-2sin(y)^2]^(1/2)
=√2Arcsin(√2sin(y))

继续计算A-B
=∫[sin(x)-cos(x)]/[sin(x)cos(x)]^(1/2)dx
而sin(y+π/4)-cos(y+π/4)=-2^(1/2)sin(y)
所以∫[sin(x)-cos(x)]/[sin(x)cos(x)]^(1/2)dx
=-2∫dcos(y)/[2cos(y)^2-1]^(1/2)
=√2Arccosh(√2cos(y))
因此
A=[(A+B)+(A-B)]/2=Arcsin(√2sin(y))/√2-Arccosh(√2cos(y))/√2

∵x=y+π/4
∴x∈[0,π/2]，y∈[-π/4,π/4]
当y=π/4时，
Arcsin(√2sin(π/4))/√2-Arccosh(√2cos(π/4))/√2
=Arcsin(1)/√2-Arccosh(1)/√2
=π/(2√2)
当y=-π/4时
Arcsin(√2sin(-π/4))/√2-Arccosh(√2cos(-π/4))/√2
=Arcsin(-1)/√2-Arccosh(1)/√2
=-π/(2√2)
于是∫[0,π/2]√tan(x)dx=π/√2

对于-2∫dcos(y)/[2cos(y)^2-1]^(1/2)
令cos(y)=sec(z)/√2
=-√2∫dsec(z)/tan(z)
=-√2∫sec(z)dz
=-√2ln|tan(z)+sec(z)|
=-√2ln|[2cos(y)^2-1]^(1/2)+√2cos(y)|
这时在[-π/4,π/4]上有定义

∫tan(x)^(1/3)dx
使t=tan(x)^(2/3)
=∫√tdArctan(t√t)
=3/2∫t/(t^3+1)dt
=1/2∫-1/(t+1)+(t+1)/(t^2-t+1)dt
=-ln(t+1)/2+1/4∫1/[(t-1/2)^2+3/4]d(t-1/2)^2+3/4∫1/[(t-1/2)^2+3/4]dt
=-ln(t+1)/2+ln(t^2-t+1)/4+3Arctan[2(t-1/2)/√3]/(2√3)
=-ln(tan(x)^(2/3)+1)/2+ln(tan(x)^(4/3)-tan(x)^(2/3)+1)/4+3Arctan[(2tan(x)^(2/3)-1)/√3]/(2√3)
求在0到π/2的定积分,x∈[0,π/2],t∈[0,+∞]
求-ln(t+1)/2+ln(t^2-t+1)/4在0和+∞的极限。
在0的极限，直接代入，得到0
在+∞的极限，
lim(x→+∞)-ln(t+1)/2+ln(t^2-t+1)/4
=lim(x→+∞)-ln(t^2+2t+1)/4+ln(t^2-t+1)/4
=lim(x→+∞)ln((t^2-t+1)/(t^2+2t+1))/4
=lim(x→+∞)ln(1)/4
=0
所以只需要考虑3Arctan[(2t-1)/√3]/(2√3)在0和+∞的极限。
在t→+∞时，Arctan[(2t-1)/√3]→π/2
因此在t→+∞时，原式→3π/(4√3)
在t→0时，3Arctan[(2t-1)/√3]/(2√3)=3Arctan(-1/√3)/(2√3)=-π/(4√3)
相减得π/√3