Why do different substitutions give different results for $\int\cot(x)\csc(x)^2\,dx?$ I am having some issues when trying to integrate this function. First of all I have to decide to make a $U$ substitution either for $\csc(x)$ or for $\cot(x)$, both of them are acceptable substitutions, I believe. Now, when I solve the integral after doing a $U$ substitution for $\csc(x)$ I get a different answer than when I do a $U$ substitution for $\cot(x)$. After observing that the answers were not matching I decided to rely on technology. I used my TI-89 calculator that gave me a different answer than when I do it on wolfram-alpha. 
Meaning that I am back to the beginning. My question is which is the right answer? Why are these two reliable calculators giving different answers? 
$\frac{-\csc(x)^2}{2}$  and  $\frac{-\cot(x)^2}{2}$
are the two answers I got. 
 A: The reason why you are getting different answers is the importance of the $+C$. Read this: https://brilliant.org/discussions/thread/the-importance-of-c/
To answer your answer, both are right. 
Also note that:
$$\cot^2(x)=\csc^2(x)-1$$
and as $1$ is just constant, it is accounted for in the $+C$, merely the $C$ in the two forms are different.
Edit:
$$\frac{-\cot^2(x)}{2}=\frac{-\csc^2(x)+1}{2}=\frac{-\csc^2(x)}{2}+\frac{1}{2}$$
Now what is $\frac{1}{2}+C$. It can be made into another $C$.
And you are left with $$-\frac{\csc^2(x)}{2}+C$$
A: Here is another integral you can preform different substitutions and get different looking answers but still are in the same family of functions: $I=\int 2 \sin(x) \cos(x) dx \text{ well first we can say this:} \int 2 \sin(x) \cos(x) dx=\int \sin(2x) dx=\frac{-1}{2}\cos(2x)+C \text{ there is also we can use the subsitution } u=\sin(x) \text{ then so } du=cos(x) dx \text{ so we would have } \int 2 \sin(x) \cos(x) dx= \int 2udu =2 \frac{u^2}{2}+C=u^2+C=(\sin(x))^2+C=\sin^2(x)+C \text{ there is also the other substitution we could do } v=\cos(x)  \text{ which will give } I= \frac{-\cos^2(x)}{2}+C $ We can show these answers are the same using trig identities. And keeping in mind... that a known constant+an unknown constant is still an unknown constant
A: This doesn't answer the question directly but maybe it helps to illustrate some of the underlying confusion.
Take $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$
We integrate both sides to get:
$-\cos(x)=\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}-...$
Oops! But we're missing the first term of negative cosine!?!
$-\cos(x)=-1+\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}-...$
So what happened to the extra $-1$ term?!?
As you can C (pun intended!), there is no canonical choice of the $+C$.
And hence there is no canonical choice of the $+C$ because the $+C$ is just a product of how you write and solve the integral. Of course this doesn't mess up any applications, because when you take definite integrals, the $+C$ just falls back out;
$(f(x))|_{x=a}^{x=b}$ equals $(f(x)+C)|_{x=a}^{x=b}$ for every number that you can think of, $C$, since the $C$'s in the difference $(f(x)+C)|_{x=a}^{x=b}$ will always just cancel with each other (that is, regardless of what value is being used). If this seems weird that's ok. But think of it pictorially: if every point on a graph is shifted upwards, then the difference between two graphs don't change, right?
You will see that evaluating $\left(\frac{-\csc^2(x)}{2}\right)|_1^2=\left(\frac{-\cot^2(x)}{2}\right)|_1^2$. Think in terms of definite rather than indefinite integration. What is the point of indefinite integration then? To find a function whose derivative is the original function. And you can see that you've done that perfectly, which is all you needed to do - so actually you're done. Indeed, you can check that differentiating both of them with respect to $x$ gives you back your original function, that is, cotangent of $x$ times cosecant squared of $x$, but that's not the point. If you take the integrals properly, then you shouldn't have to worry about such things. Math always works itself out in the end. Just go with it and you'll see you get the right answer, but now you know why.
Hope this helps. If not I'm honestly open to talk more, and to modify my answer as is deemed appropriate.
