So I seem to be reaching 2 seemingly distinct answers when evaluating the following integral with two approaches:
$$ \int{\cot(10z)\csc^4(10z)}\;dz $$
In the first approach, I break $\csc^4(10z)$ up into $\csc^2(10z)\csc^2(10z)$ and substitute in $\csc^2(10z) = \cot^2(10z) + 1$ for one of them, like so:
$$ \int{\cot(10z)(\cot^2(10z) + 1)\csc^2(10z)}\;dz $$
Then, I u-substitute in $ u = \cot(10z) $ and $ du = -10\csc^2(10z)\;dz $ to get
$$ {-\frac{1}{10}\int{u(u^2 + 1)}\;du} $$ $$ = {-\frac{1}{10}\int{u^3 + u}\;du} $$ $$ = -\frac{u^4}{40} - \frac{u^2}{20} + C $$ $$ = \bbox[yellow,5px,border:2px solid red]{-\frac{\cot^4(10z)}{40} - \frac{\cot^2(10z)}{20} + C} \tag{1} $$
It seems like a completely legitimate approach to me. However, if I use another technique where I break $ \csc^4(10z) $ up into $ \csc^3(10z)\csc(10z) $ instead, and then u-substitute $ u = \csc(10z) $ with $ -\frac{du}{10} = \csc(10z)\cot(10z)\;dz $, I get
$$ {-\frac{1}{10}\int{u^3}\;du} $$ $$ = \bbox[yellow,5px,border:2px solid red]{-\frac{\csc^4(10z)}{40} + C} \tag{2} $$
Now, these are apparently not equal as plugging in $ z = 1 $ results in about $ -0.2604 $ for $ (1) $ and $ -0.2854 $ for $ (2) $. What is going on here? Why am I reaching this contradiction?
Taking your first answer,
$\frac{-\cot^2(10z)}{40} \left[\cot^2(10z)+2 \right] + c$
$\frac{-\cot^2(10z)}{40} \left[\csc^2(10z)-1+2 \right] + c$
$\frac{-\cot^2(10z)}{40} \left[\cot^2(10z)+1 \right] + c$
$\frac{1-\csc^2(10z)}{40} \left[1+\cot^2(10z) \right] + c$
$\frac{1 - \csc^4(10z)}{40} + c$
$\frac{1}{40} - \frac {\csc^4(10z)}{40} + c$
$\frac{- \csc^4(10z)}{40} + c'$