Evaluating $\int \cot x \csc^2x \,\mathrm{d}x$ with $u=\cot x$

Question

$\newcommand{\d}{\mathrm{d}}$

Evaluate the integral using the indicated substituion. $$\int \cot x \csc^2x \,\d{x}, \qquad u= \cot x .$$

Differentiating both sides of $u$, then making the substitution: $$ \begin{align} u &= \phantom{-}\cot x, \\ \d u &= -\cot x\csc x \,\d{x}, \\ \d x &= -\frac{\d u}{u \csc x}. \end{align}$$ $$\int -\frac{u\csc^2 x \,\d{u}}{u\csc x} = \int -\csc x \,\d{u}. $$

Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?

You have not completely turned the integrand to a function of $u$. Try do more. — xbh, Aug 30 '18 at 17:42
Also, $\mathrm d\cot(x) = -\csc^2(x) \mathrm dx$, not $-\cot(x) \csc (x)\mathrm dx$. — xbh, Aug 30 '18 at 17:51
I had confused it with another formula, but I got it now. Thank you, @xbh. — Mauricio Mendes, Aug 30 '18 at 17:58
$\LaTeX$ Tip: Try using \cot x, \csc x and \mathrm dx to get $\cot x$, $\csc x$ and $\mathrm dx$ respectively. — For the love of maths, Aug 30 '18 at 17:59
@MauricioMendes You are welcome. That one seems to be $\mathrm d \csc(x)$. — xbh, Aug 30 '18 at 17:59
Does this answer your question? Getting different answers when integrating using different techniques — Anne Bauval, Oct 25 '23 at 19:12

score 3 · Accepted Answer · answered Aug 30 '18 at 20:11

You have $du=-\csc^2x\,dx$, rather than your wrong differentiation. This implies the integral is $$ \int\cot x\csc^2x\,dx=\int-u\,du=-\frac{1}{2}u^2+c=-\frac{1}{2}\cot^2x+c $$ On the other hand, rewriting the integral as $$ \int\frac{\cos x}{\sin^3x}\,dx=\int(\sin x)^{-3}d(\sin x)=-\frac{1}{2}\frac{1}{\sin^2x}+c $$ is much easier.

score 1 · Answer 2 · edited Aug 30 '18 at 20:36

1

For alternative way:

$$\int \cot x \csc^2 x dx$$ $$=\int \frac{\cos x dx}{\sin^3 x}$$

Now you can advance taking $\sin x = z$ .

edited Aug 30 '18 at 20:36

amWhy

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answered Aug 30 '18 at 18:02

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Evaluating $\int \cot x \csc^2x \,\mathrm{d}x$ with $u=\cot x$

2 Answers2