Evaluating $int cot x csc^2x ,mathrmdx$ with $u=cot x$

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$newcommanddmathrmd$




Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$




Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$



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






calculus integration trigonometry indefinite-integrals







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  • You have not completely turned the integrand to a function of $u$. Try do more.
    – xbh
    Aug 30 at 17:42











  • $csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
    – xbh
    Aug 30 at 17:44







  • 2




    Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
    – xbh
    Aug 30 at 17:51










  • I had confused it with another formula, but I got it now. Thank you, @xbh.
    – Mauricio Mendes
    Aug 30 at 17:58






  • 2




    $LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
    – Mohammad Zuhair Khan
    Aug 30 at 17:59















up vote
3
down vote

favorite












$newcommanddmathrmd$




Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$




Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$



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






calculus integration trigonometry indefinite-integrals







share|cite|improve this question























  • You have not completely turned the integrand to a function of $u$. Try do more.
    – xbh
    Aug 30 at 17:42











  • $csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
    – xbh
    Aug 30 at 17:44







  • 2




    Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
    – xbh
    Aug 30 at 17:51










  • I had confused it with another formula, but I got it now. Thank you, @xbh.
    – Mauricio Mendes
    Aug 30 at 17:58






  • 2




    $LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
    – Mohammad Zuhair Khan
    Aug 30 at 17:59













up vote
3
down vote

favorite









up vote
3
down vote

favorite











$newcommanddmathrmd$




Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$




Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$



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






calculus integration trigonometry indefinite-integrals







share|cite|improve this question















$newcommanddmathrmd$




Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$




Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$



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






calculus integration trigonometry indefinite-integrals




calculus integration trigonometry indefinite-integrals






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edited Sep 6 at 9:23









Nathanael Skrepek

1,5071515




1,5071515










asked Aug 30 at 17:39









Mauricio Mendes

968




968











  • You have not completely turned the integrand to a function of $u$. Try do more.
    – xbh
    Aug 30 at 17:42











  • $csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
    – xbh
    Aug 30 at 17:44







  • 2




    Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
    – xbh
    Aug 30 at 17:51










  • I had confused it with another formula, but I got it now. Thank you, @xbh.
    – Mauricio Mendes
    Aug 30 at 17:58






  • 2




    $LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
    – Mohammad Zuhair Khan
    Aug 30 at 17:59

















  • You have not completely turned the integrand to a function of $u$. Try do more.
    – xbh
    Aug 30 at 17:42











  • $csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
    – xbh
    Aug 30 at 17:44







  • 2




    Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
    – xbh
    Aug 30 at 17:51










  • I had confused it with another formula, but I got it now. Thank you, @xbh.
    – Mauricio Mendes
    Aug 30 at 17:58






  • 2




    $LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
    – Mohammad Zuhair Khan
    Aug 30 at 17:59
















You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42





You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42













$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44





$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44





2




2




Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51




Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51












I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58




I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58




2




2




$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59





$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59











2 Answers
2






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up vote
3
down vote



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You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
$$
intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
$$
On the other hand, rewriting the integral as
$$
intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
$$
is much easier.






share|cite|improve this answer



























    up vote
    1
    down vote













    For alternative way:



    $$int cot x csc^2 x dx$$
    $$=int fraccos x dxsin^3 x$$



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






    share|cite|improve this answer






















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
      $$
      intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
      $$
      On the other hand, rewriting the integral as
      $$
      intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
      $$
      is much easier.






      share|cite|improve this answer
























        up vote
        3
        down vote



        accepted










        You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
        $$
        intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
        $$
        On the other hand, rewriting the integral as
        $$
        intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
        $$
        is much easier.






        share|cite|improve this answer






















          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
          $$
          intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
          $$
          On the other hand, rewriting the integral as
          $$
          intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
          $$
          is much easier.






          share|cite|improve this answer












          You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
          $$
          intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
          $$
          On the other hand, rewriting the integral as
          $$
          intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
          $$
          is much easier.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 30 at 20:11









          egreg

          167k1281189




          167k1281189




















              up vote
              1
              down vote













              For alternative way:



              $$int cot x csc^2 x dx$$
              $$=int fraccos x dxsin^3 x$$



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






              share|cite|improve this answer


























                up vote
                1
                down vote













                For alternative way:



                $$int cot x csc^2 x dx$$
                $$=int fraccos x dxsin^3 x$$



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






                share|cite|improve this answer
























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  For alternative way:



                  $$int cot x csc^2 x dx$$
                  $$=int fraccos x dxsin^3 x$$



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






                  share|cite|improve this answer














                  For alternative way:



                  $$int cot x csc^2 x dx$$
                  $$=int fraccos x dxsin^3 x$$



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







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 30 at 20:36









                  amWhy

                  190k27221433




                  190k27221433










                  answered Aug 30 at 18:02









                  Entrepreneur

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