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Find $\int-\csc^2x\cot xdx$

Using $\int f'f^{n}dx=\frac{f^{n+1}}{n+1}$ rule, I am getting two answers for the above question. Considering that the derivatives of $\csc x$ and $\cot x$ are $-\csc x\cot x$ and $-\csc^2x$ respectively, depending on which one of $\cot x$ or $\csc x$ i take as my $f$ I'm getting the answers as $(\cot^2x)/2 + c$ or $(\csc^2x)/2+c$ . Which one of my above answers is incorrect and why?

amWhy
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user1241091
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    Please use MathJax to format your equations. Both formulas are correct, since they differ by a constant. – Andrei Oct 25 '23 at 14:21
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    A single function has multiple anti-derivatives separated by a constant. $\csc^2$ and $\cot ^2$ differ by a constant. So both answers are correct and are the same, apart from the constant – TurlocTheRed Oct 25 '23 at 14:31

3 Answers3

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Computing the difference of the 2 results :
$\dfrac{\csc^2 x} {2} - \dfrac{\cot^2x}{2} = \dfrac{1 - \cos^2 x} {2 \sin^2 x} = \dfrac{\sin ^2 x} {2 \sin^2 x} = \dfrac12$

Thus, as noted by TurlocTheRed,since the difference is a constant, there is no contradiction and the 2 answers are correct.

M.B
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$$\csc^2x=\frac{1}{\sin^2x}=\frac{\sin^2x+\cos^2x}{\sin^2x}=1+\cot ^2x$$

Andrei
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$$ \begin{aligned}\int-\csc ^2 x \cot x d x = & \int \csc x d(\csc x) \\ = & \frac{\csc ^2 x}{2}+C_1 \\ = & \frac{\csc ^2 x}{2}-\frac{1}{2}+C_1-\frac{1}{2} \\ = & \frac{\csc ^2 x-1}{2}+C_2 \\ = & \frac{\cot ^2 x}{2}+C_2 \end{aligned} $$

Lai
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