Range of a function involving integration

Question

Find the range of the function for $\alpha \in \mathbb{R}$, $$f(\alpha) = \int_{\tan^{-1}(\alpha)}^{\cot^{-1}(\alpha)}{\frac{\tan(x)}{\tan(x) + \cot(x)}}dx$$

My attempt: Please verify if my way of solution is correct.

$$f(\alpha) = \int_{\tan^{-1}(\alpha)}^{\cot^{-1}(\alpha)}{\frac{\tan(x)}{\tan(x) + \cot(x)}}dx$$ $$\text{or } f(\alpha) = \int_{\tan^{-1}(\alpha)}^{\cot^{-1}(\alpha)}{2\sin^2(x)}dx$$ $$\text{or }\, f(\alpha) = \frac{1}{2}\int_{\tan^{-1}(\alpha)}^{\cot^{-1}(\alpha)}\left({1-\cos(2x)}\right)dx$$ $$\text{or }\, f(\alpha) = \frac{1}{2}\left({x-0.5\sin(2x)}\right)|_{\tan^{-1}(\alpha)}^{\cot^{-1}(\alpha)}$$ $$\text{or }\, f(\alpha) = \frac{\pi}{4} - \tan^{-1}{\alpha}$$

So range is $(\frac{-\pi}{4},\frac{3\pi}{4})$

Answer 1

You have a mistake in your first to second step, since you have an excess $2$, since it is :

$$\frac{\tan(x)}{\tan(x) + \cot(x)} = \sin^2(x)$$

The next identity used is correct. The mistake having the $2$ there makes you have a mistaken $\frac{1}{2}$ as a coefficient in :

$$"\text{or }\, f(\alpha) = \frac{1}{2}\left({x-0.5\sin(2x)}\right)|_{\tan^{-1}(\alpha)}^{\cot^{-1}(\alpha)}"$$

Also, as a tip, please use \sin, \cos, \tan, \cot etc between dollars for coding and typesetting mathematical functions in $LaTeX$.

Answer 2

Hint:

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

$$I+I=?$$