Integrate $\tan^2(\frac{\pi}{12} \cdot y)$ Integrate $\tan^2(\frac{\pi}{12} \cdot y)$
Wolfram gives the answer:
$$\frac{12 \tan(\frac{\pi y}{12})}{\pi} -y + \text{constant}$$
But why is the value of $\tan^2$ not getting differentiated?
According to this rule, the answer should be:
$$\frac{\tan(\frac{\pi}{12} \cdot y)}{\frac{\pi}{12} \cdot (2-1)}$$
($\dfrac{\pi}{12}$ is a constant so it's not differentiated)
Also, what is meant by aX? In my case, what's my "aX"? $\frac{\pi}{12} \cdot y$? So does the "a" simply means what's before the variable y? $\frac{\pi}{12}$?

Recall that $\int \tan^2 x\ dx = \int {\sin^2 x \over 1 - \sin^2 x} dx = -x + \tan x$, we find:
$\int \tan^2 \left( {\pi y \over 12} \right) d y = \frac{12 \tan \left(\frac{\pi  y}{12}\right)}{\pi }-y$
