Trigonometric functions what is: $\int_{\pi/4}^{\pi/3}f(\cot\theta)\,d\theta$?

Question

Given that: $$\int_{\pi/6}^{\pi/4}f(\tan\theta)\,d\theta=2017$$ What is the value of the following integral: $$\int_{\pi/4}^{\pi/3}f(\cot\theta)\,d\theta$$

Answer 1

Using the change of variable
$x= \pi/2 -\theta$ $$ 2017 =\int_{\pi/6}^{\pi/4}f(\tan x)\,dx=\int_{\pi/6}^{\pi/4}f\left(\frac{\sin x}{\cos x}\right)dx=\int_{\pi/3}^{\pi/4}f\left(\frac{\sin\left(\frac\pi2-\theta\right)}{\cos\left(\frac\pi2-\theta\right)}\right)d\theta \\ \hspace{-4.45cm}=-\int_{\pi/3}^{\pi/4}f\left(\frac{\cos\theta}{\sin\theta}\right)d\theta=\int^{\pi/3}_{\pi/4}f(\cot\theta)\,d\theta $$

That is

$$ 2017 =\int_{\pi/6}^{\pi/4}f(\tan\theta)\,d\theta =\int^{\pi/3}_{\pi/4}f(\cot\theta)\,d\theta $$