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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?

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  • $\begingroup$ The first sanity check is to set $C=0$ in both and plug in a few independent values of $z$ to see if they both differ by the same constant every time $\endgroup$ – user14972 Apr 17 '17 at 3:35
  • $\begingroup$ Try to use $k$ as constant in the first and use the identity $csc^2x=1+\cot^2x$ $\endgroup$ – Juniven Apr 17 '17 at 3:37
  • $\begingroup$ Just checking one value isn't enough to establish that they are different. The solution isn't 1 function, but a set of functions. $\endgroup$ – DanielV Apr 17 '17 at 5:32
  • $\begingroup$ @DanielV Here you can clearly see that $(1)$ and $(2)$ differ by only a constant $(0.025)$, and as such evaluating the indefinite integrals with any two limits, the same ones for both, you can see that $\int{(1)(x)} = \int{(2)(x)}$. Interesting. $\endgroup$ – R. Kap Apr 17 '17 at 6:09
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Taking your first answer,

$\frac{-\cot^2(10z)}{40} \left[\cot^2(10z)+2 \right] + c$

$\frac{-\cot^2(10z)}{40} \left[\csc^2(10z)-1+2 \right] + c$

$\frac{-\cot^2(10z)}{40} \left[\cot^2(10z)+1 \right] + c$

$\frac{1-\csc^2(10z)}{40} \left[1+\cot^2(10z) \right] + c$

$\frac{1 - \csc^4(10z)}{40} + c$

$\frac{1}{40} - \frac {\csc^4(10z)}{40} + c$

$\frac{- \csc^4(10z)}{40} + c'$

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