How to prove Integral [ f(tan(x)) , from 0 to Pi/2]dx = Integral[ f(cot(x)) , from 0 to Pi/2]dx

While knowing Integral [ f(sin(x)) , from 0 to Pi/2]dx = Integral[ f(cos(x)) , from 0 to Pi/2]dx

Let z = π/2 - x <==> x = π/2 - z.
So, dx = -dz.
Bounds: x = 0 ==> z = π/2, and x = π/2 ==> z = 0.

Moreover, tan(π/2 - z) = sin(π/2 - z)/cos(π/2 - z) = cos z/sin z = cot z.

Therefore,
∫(x = 0 to π/2) f(tan x) dx
= ∫(z = π/2 to 0) f(cot z) * -dz
= ∫(z = 0 to π/2) f(cot z) dz
= ∫(x = 0 to π/2) f(cot x) dx, dummy variable change.

I hope this helps!