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?
Best Answer
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*}