3
$\begingroup$

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

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?

$\endgroup$
6
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ $\LaTeX$ 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
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

For alternative way:

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.