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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?

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  • $\begingroup$ You have not completely turned the integrand to a function of $u$. Try do more. $\endgroup$ – xbh Aug 30 '18 at 17:42
  • $\begingroup$ $\csc^2(x) = 1 +\cot^2(x) = 1 + u^2$. $\endgroup$ – xbh Aug 30 '18 at 17:44
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    $\begingroup$ Also, $\mathrm d\cot(x) = -\csc^2(x) \mathrm dx$, not $-\cot(x) \csc (x)\mathrm dx$. $\endgroup$ – xbh Aug 30 '18 at 17:51
  • $\begingroup$ I had confused it with another formula, but I got it now. Thank you, @xbh. $\endgroup$ – Mauricio Mendes Aug 30 '18 at 17:58
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    $\begingroup$ $\LaTeX$ Tip: Try using \cot x, \csc x and \mathrm dx to get $\cot x$, $\csc x$ and $\mathrm dx$ respectively. $\endgroup$ – Mohammad Zuhair Khan Aug 30 '18 at 17:59
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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.

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For alternative way:

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

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

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