Integrate $sqrt{(1-cos(x))^3}$












0












$begingroup$


I need help solving the following integration problem:
$$int sqrt{(1-cos(x))^3} dx$$



I know I am supposed to use a trig identity so that I can do a u-sub, but don't know which one or how.






integration







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    0












    $begingroup$


    I need help solving the following integration problem:
    $$int sqrt{(1-cos(x))^3} dx$$



    I know I am supposed to use a trig identity so that I can do a u-sub, but don't know which one or how.






    integration







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      I need help solving the following integration problem:
      $$int sqrt{(1-cos(x))^3} dx$$



      I know I am supposed to use a trig identity so that I can do a u-sub, but don't know which one or how.






      integration







      share|cite|improve this question











      $endgroup$




      I need help solving the following integration problem:
      $$int sqrt{(1-cos(x))^3} dx$$



      I know I am supposed to use a trig identity so that I can do a u-sub, but don't know which one or how.






      integration




      integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 30 '18 at 11:50









      KM101

      5,9251524




      5,9251524










      asked Nov 30 '18 at 11:45









      SamSam

      1887




      1887






















          2 Answers
          2






          active

          oldest

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          5












          $begingroup$

          Hint:



          Since $cos(2x) = 1 - 2 sin^2(x)$, $1-cos x = 2 sin^2(frac{x}{2})$:



          $$int 2 sqrt{2} bigg(sin^2left(frac{x}{2}right)bigg)^{3/2} dx$$
          $$= 2sqrt{2} int left| sin^3left(frac{x}{2}right) right| dx$$



          Now from $sin(3x) = 3 sin x - 4 sin^3 x$, $4 sin^3(x) = 3 sin(x) - sin(3x)$, so you can use the identity $sin^3(x) = frac{3}{4}sin(x) - frac{1}{4}sin(3x)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
            $endgroup$
            – Sam
            Nov 30 '18 at 11:53








          • 3




            $begingroup$
            Because $2cdotfrac{3}{2} = 3$.
            $endgroup$
            – KM101
            Nov 30 '18 at 11:55










          • $begingroup$
            Can you explain how you got from $cos(2theta)$ to $1-costheta$?
            $endgroup$
            – Sam
            Nov 30 '18 at 11:59








          • 1




            $begingroup$
            @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
            $endgroup$
            – Toby Mak
            Nov 30 '18 at 12:01










          • $begingroup$
            It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
            $endgroup$
            – Bernard
            Nov 30 '18 at 12:03





















          1












          $begingroup$

          $$int sqrt{left(1-cos left(xright)right)^3} = 4sqrt{2}int :frac{y^3}{left(1+y^2right)^{frac{5}{2}}}dy=2sqrt{2}int :frac{1}{z^{frac{3}{2}}}-frac{1}{z^{frac{5}{2}}}dz$$
          You will get this result by applying these substitutions in sequence
          $$y=tanleft(frac{x}{2}right)qquadqquad z= 1+y^2$$
          $$int sqrt{left(1-cos left(xright)right)^3}=color{red}{4sqrt{2}left(frac{1}{3sec ^3left(frac{x}{2}right)}-frac{1}{sec left(frac{x}{2}right)}right)+C}$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

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            active

            oldest

            votes









            5












            $begingroup$

            Hint:



            Since $cos(2x) = 1 - 2 sin^2(x)$, $1-cos x = 2 sin^2(frac{x}{2})$:



            $$int 2 sqrt{2} bigg(sin^2left(frac{x}{2}right)bigg)^{3/2} dx$$
            $$= 2sqrt{2} int left| sin^3left(frac{x}{2}right) right| dx$$



            Now from $sin(3x) = 3 sin x - 4 sin^3 x$, $4 sin^3(x) = 3 sin(x) - sin(3x)$, so you can use the identity $sin^3(x) = frac{3}{4}sin(x) - frac{1}{4}sin(3x)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:53








            • 3




              $begingroup$
              Because $2cdotfrac{3}{2} = 3$.
              $endgroup$
              – KM101
              Nov 30 '18 at 11:55










            • $begingroup$
              Can you explain how you got from $cos(2theta)$ to $1-costheta$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:59








            • 1




              $begingroup$
              @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
              $endgroup$
              – Toby Mak
              Nov 30 '18 at 12:01










            • $begingroup$
              It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
              $endgroup$
              – Bernard
              Nov 30 '18 at 12:03


















            5












            $begingroup$

            Hint:



            Since $cos(2x) = 1 - 2 sin^2(x)$, $1-cos x = 2 sin^2(frac{x}{2})$:



            $$int 2 sqrt{2} bigg(sin^2left(frac{x}{2}right)bigg)^{3/2} dx$$
            $$= 2sqrt{2} int left| sin^3left(frac{x}{2}right) right| dx$$



            Now from $sin(3x) = 3 sin x - 4 sin^3 x$, $4 sin^3(x) = 3 sin(x) - sin(3x)$, so you can use the identity $sin^3(x) = frac{3}{4}sin(x) - frac{1}{4}sin(3x)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:53








            • 3




              $begingroup$
              Because $2cdotfrac{3}{2} = 3$.
              $endgroup$
              – KM101
              Nov 30 '18 at 11:55










            • $begingroup$
              Can you explain how you got from $cos(2theta)$ to $1-costheta$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:59








            • 1




              $begingroup$
              @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
              $endgroup$
              – Toby Mak
              Nov 30 '18 at 12:01










            • $begingroup$
              It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
              $endgroup$
              – Bernard
              Nov 30 '18 at 12:03
















            5












            5








            5





            $begingroup$

            Hint:



            Since $cos(2x) = 1 - 2 sin^2(x)$, $1-cos x = 2 sin^2(frac{x}{2})$:



            $$int 2 sqrt{2} bigg(sin^2left(frac{x}{2}right)bigg)^{3/2} dx$$
            $$= 2sqrt{2} int left| sin^3left(frac{x}{2}right) right| dx$$



            Now from $sin(3x) = 3 sin x - 4 sin^3 x$, $4 sin^3(x) = 3 sin(x) - sin(3x)$, so you can use the identity $sin^3(x) = frac{3}{4}sin(x) - frac{1}{4}sin(3x)$.






            share|cite|improve this answer











            $endgroup$



            Hint:



            Since $cos(2x) = 1 - 2 sin^2(x)$, $1-cos x = 2 sin^2(frac{x}{2})$:



            $$int 2 sqrt{2} bigg(sin^2left(frac{x}{2}right)bigg)^{3/2} dx$$
            $$= 2sqrt{2} int left| sin^3left(frac{x}{2}right) right| dx$$



            Now from $sin(3x) = 3 sin x - 4 sin^3 x$, $4 sin^3(x) = 3 sin(x) - sin(3x)$, so you can use the identity $sin^3(x) = frac{3}{4}sin(x) - frac{1}{4}sin(3x)$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 30 '18 at 12:04

























            answered Nov 30 '18 at 11:49









            Toby MakToby Mak

            3,49811128




            3,49811128












            • $begingroup$
              Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:53








            • 3




              $begingroup$
              Because $2cdotfrac{3}{2} = 3$.
              $endgroup$
              – KM101
              Nov 30 '18 at 11:55










            • $begingroup$
              Can you explain how you got from $cos(2theta)$ to $1-costheta$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:59








            • 1




              $begingroup$
              @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
              $endgroup$
              – Toby Mak
              Nov 30 '18 at 12:01










            • $begingroup$
              It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
              $endgroup$
              – Bernard
              Nov 30 '18 at 12:03




















            • $begingroup$
              Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:53








            • 3




              $begingroup$
              Because $2cdotfrac{3}{2} = 3$.
              $endgroup$
              – KM101
              Nov 30 '18 at 11:55










            • $begingroup$
              Can you explain how you got from $cos(2theta)$ to $1-costheta$?
              $endgroup$
              – Sam
              Nov 30 '18 at 11:59








            • 1




              $begingroup$
              @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
              $endgroup$
              – Toby Mak
              Nov 30 '18 at 12:01










            • $begingroup$
              It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
              $endgroup$
              – Bernard
              Nov 30 '18 at 12:03


















            $begingroup$
            Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
            $endgroup$
            – Sam
            Nov 30 '18 at 11:53






            $begingroup$
            Why is the it not $int sin^frac{3}{2}frac{x}{2} dx$?
            $endgroup$
            – Sam
            Nov 30 '18 at 11:53






            3




            3




            $begingroup$
            Because $2cdotfrac{3}{2} = 3$.
            $endgroup$
            – KM101
            Nov 30 '18 at 11:55




            $begingroup$
            Because $2cdotfrac{3}{2} = 3$.
            $endgroup$
            – KM101
            Nov 30 '18 at 11:55












            $begingroup$
            Can you explain how you got from $cos(2theta)$ to $1-costheta$?
            $endgroup$
            – Sam
            Nov 30 '18 at 11:59






            $begingroup$
            Can you explain how you got from $cos(2theta)$ to $1-costheta$?
            $endgroup$
            – Sam
            Nov 30 '18 at 11:59






            1




            1




            $begingroup$
            @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
            $endgroup$
            – Toby Mak
            Nov 30 '18 at 12:01




            $begingroup$
            @Sam Take the negative sign on both sides: $-cos(2x) = -1 + 2 sin^2(x)$. Then add $1$ to both sides; $1 - cos(2x) = 2 sin^2(x)$. Then if $x = frac{u}{2}$, then $1 - cos(u) = 2 sin^2(frac{u}{2})$.
            $endgroup$
            – Toby Mak
            Nov 30 '18 at 12:01












            $begingroup$
            It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
            $endgroup$
            – Bernard
            Nov 30 '18 at 12:03






            $begingroup$
            It should be $;{displaystyleint}Bigl|sin^3bigl(frac x2bigr)Bigr|,mathrm dx$.
            $endgroup$
            – Bernard
            Nov 30 '18 at 12:03













            1












            $begingroup$

            $$int sqrt{left(1-cos left(xright)right)^3} = 4sqrt{2}int :frac{y^3}{left(1+y^2right)^{frac{5}{2}}}dy=2sqrt{2}int :frac{1}{z^{frac{3}{2}}}-frac{1}{z^{frac{5}{2}}}dz$$
            You will get this result by applying these substitutions in sequence
            $$y=tanleft(frac{x}{2}right)qquadqquad z= 1+y^2$$
            $$int sqrt{left(1-cos left(xright)right)^3}=color{red}{4sqrt{2}left(frac{1}{3sec ^3left(frac{x}{2}right)}-frac{1}{sec left(frac{x}{2}right)}right)+C}$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $$int sqrt{left(1-cos left(xright)right)^3} = 4sqrt{2}int :frac{y^3}{left(1+y^2right)^{frac{5}{2}}}dy=2sqrt{2}int :frac{1}{z^{frac{3}{2}}}-frac{1}{z^{frac{5}{2}}}dz$$
              You will get this result by applying these substitutions in sequence
              $$y=tanleft(frac{x}{2}right)qquadqquad z= 1+y^2$$
              $$int sqrt{left(1-cos left(xright)right)^3}=color{red}{4sqrt{2}left(frac{1}{3sec ^3left(frac{x}{2}right)}-frac{1}{sec left(frac{x}{2}right)}right)+C}$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $$int sqrt{left(1-cos left(xright)right)^3} = 4sqrt{2}int :frac{y^3}{left(1+y^2right)^{frac{5}{2}}}dy=2sqrt{2}int :frac{1}{z^{frac{3}{2}}}-frac{1}{z^{frac{5}{2}}}dz$$
                You will get this result by applying these substitutions in sequence
                $$y=tanleft(frac{x}{2}right)qquadqquad z= 1+y^2$$
                $$int sqrt{left(1-cos left(xright)right)^3}=color{red}{4sqrt{2}left(frac{1}{3sec ^3left(frac{x}{2}right)}-frac{1}{sec left(frac{x}{2}right)}right)+C}$$






                share|cite|improve this answer









                $endgroup$



                $$int sqrt{left(1-cos left(xright)right)^3} = 4sqrt{2}int :frac{y^3}{left(1+y^2right)^{frac{5}{2}}}dy=2sqrt{2}int :frac{1}{z^{frac{3}{2}}}-frac{1}{z^{frac{5}{2}}}dz$$
                You will get this result by applying these substitutions in sequence
                $$y=tanleft(frac{x}{2}right)qquadqquad z= 1+y^2$$
                $$int sqrt{left(1-cos left(xright)right)^3}=color{red}{4sqrt{2}left(frac{1}{3sec ^3left(frac{x}{2}right)}-frac{1}{sec left(frac{x}{2}right)}right)+C}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 '18 at 11:57









                AmarildoAmarildo

                1,769816




                1,769816






























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