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

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

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
Commented Aug 30, 2018 at 17:42
• $\csc^2(x) = 1 +\cot^2(x) = 1 + u^2$.
– xbh
Commented Aug 30, 2018 at 17:44
• Also, $\mathrm d\cot(x) = -\csc^2(x) \mathrm dx$, not $-\cot(x) \csc (x)\mathrm dx$.
– xbh
Commented Aug 30, 2018 at 17:51
• I had confused it with another formula, but I got it now. Thank you, @xbh. Commented Aug 30, 2018 at 17:58
• $\LaTeX$ Tip: Try using \cot x, \csc x and \mathrm dx to get $\cot x$, $\csc x$ and $\mathrm dx$ respectively. Commented Aug 30, 2018 at 17:59

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.
$$\int \cot x \csc^2 x dx$$ $$=\int \frac{\cos x dx}{\sin^3 x}$$
Now you can advance taking $\sin x = z$ .