Proof definite integral $\int_α^{π/2-α}f(\tan{x})dx=\frac{\pi}{2}-α $

Question

Given:

$$f(x) +f\left(\frac{1}{x}\right) =1$$
I am trying to prove that
$$\int_α^{π/2-α}f(\tan{x})dx=\frac{\pi}{2}-α $$

Answer 1

Using $$\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx,$$

$$I=\int_\alpha^{\frac\pi2-\alpha}f(\tan x)dx=\int_\alpha^{\frac\pi2-\alpha}f(\cot x)dx$$

$$I+I=\int_\alpha^{\frac\pi2-\alpha}[f(\tan x)+f(\cot x)]dx$$

Now set $x=\tan y$ in $f(x)+f\left(\dfrac1x\right)=1$