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As the title may suggest, I come today with a question regarding a (seemingly) simple trigonometric integral, that being $$\int_{}^{} \csc^2(x)\cot(x) \,dx$$ I encountered this question in a Calculus textbook, on the chapter involving u-substitution: the following is how I initially solved it.

Set $u = \csc(x)$. Then, $du = -\cot(x)\csc(x)dx$, and so:

$$\int_{}^{} \csc^2(x)\cot(x) \,dx = - \int_{}^{} -\csc^2(x)\cot(x) \,dx = - \int_{}^{} u \,du = - \frac{1}{2} u^2 + C = - \frac{1}{2} \csc^2(x) + C $$

That's all fine and dandy, but upon checking the answer key, I noticed another similar solution.

This time, we set $u = \cot(x)$. In this scenario, $du = -\csc^2(x)$, and hence:

$$\int_{}^{} \csc^2(x)\cot(x) \,dx = - \int_{}^{} u \,du = - \frac{1}{2} u^2 + C = - \frac{1}{2} \cot^2(x) + C $$

As far as I can tell, both of these methods are completely valid, but doesn't that then lead us to the conclusion that:

$$- \frac{1}{2} \cot^2(x) = - \frac{1}{2} \csc^2(x)$$ $$ \cot^2(x) = \csc^2(x)$$

Evidently, this can't be right, but I'm unsure of what went wrong here. Could somebody enlighten me on if there's something I'm missing, or if one of these methods is somehow incorrect?

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    $\begingroup$ Why do you have the same C both times? $\endgroup$ Apr 9 at 21:27
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    $\begingroup$ It's due to a trig identity and the faulty assumption that the C's are the same in both cases. My team lost first place in a statewide tournament because my solution was rejected because it didn't match the key. The next day I was able to prove they were the same with a trig identity, but it was too late. $\endgroup$ Apr 10 at 2:40

2 Answers 2

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They differ by a constant:

\begin{align*} \cot^{2}(x) = \frac{\cos^{2}(x)}{\sin^{2}(x)} = \frac{\cos^{2}(x) + \sin^{2}(x)}{\sin^{2}(x)} - 1 = \frac{1}{\sin^{2}(x)} - 1 = \csc^{2}(x) - 1 \end{align*}

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This is because of the pesky $+C$ integration constant you must add at the end of every indefinite integral. There's a trig identity similar to $\sin^2\theta+\cos^2\theta=1$ that relates $\csc x$ to $\cot x$, namely

$$1+\cot^2 x=\csc^2 x$$

Substituting that into the first integral result gives

\begin{align*} \int\csc^2 x\cot x\,\mathrm dx & =-\frac 12\csc^2 x+C\\ & =-\frac 12\left(1+\cot^2 x\right)+C\\ & =\color{blue}{-\frac 12\cot^2 x}+C-\frac 12 \end{align*}

The term highlighted in blue is the second integral result from substituting $u=\cot x$.

Since subtracting $\tfrac 12$ from any constant gives another constant, we can define another constant, say $C'$, as the last two terms: $C'\equiv C-\frac 12$ and get the result derived from the second substitution.

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