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? $\frac{\pi}{12}$?

Let $\frac{\pi}{12}\cdot y=t\implies \frac{\pi}{12}dy=dt\iff dy=\frac{12}{\pi}dt$
We have $$\int \tan^2\left(\frac{\pi}{12}\cdot y\right)dy=\int \tan^2 (t) \frac{12}{\pi}dt$$ $$=\frac{12}{\pi}\int \tan^2(t)dt$$
$$=\frac{12}{\pi}\int (\sec^2(t)-1)dt$$
$$=\frac{12}{\pi}\int\sec^2(t)dt-\frac{12}{\pi}\int dt$$
$$=\frac{12}{\pi}\tan(t)-\frac{12}{\pi}\cdot t+C$$
Setting $t=\frac{\pi}{12}\cdot y$
$$=\frac{12}{\pi}\tan\left(\frac{\pi}{12}\cdot y\right)-\frac{12}{\pi}\cdot \frac{\pi}{12}\cdot y+C$$
$$=\color{red}{\frac{12}{\pi}\tan\left(\frac{\pi}{12}\cdot y\right)-y+C}$$
