Integration of $int_0^frac{pi}{6} cos^{-3}2x sin2x , dx $?












0












$begingroup$


I tried substituting $x=frac{cos t}{2}$



but I didn't got anywhere.



Thanks!






calculus integration definite-integrals







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    0












    $begingroup$


    I tried substituting $x=frac{cos t}{2}$



    but I didn't got anywhere.



    Thanks!






    calculus integration definite-integrals







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I tried substituting $x=frac{cos t}{2}$



      but I didn't got anywhere.



      Thanks!






      calculus integration definite-integrals







      share|cite|improve this question









      $endgroup$




      I tried substituting $x=frac{cos t}{2}$



      but I didn't got anywhere.



      Thanks!






      calculus integration definite-integrals




      calculus integration definite-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 16:38









      RaghavRaghav

      557




      557






















          2 Answers
          2






          active

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          4












          $begingroup$

          HINT: Rewrite the integrand as
          $$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:43












          • $begingroup$
            Do you mean the integrand is $arccos^3 2x ,sin 2x$?
            $endgroup$
            – Bernard
            Dec 11 '18 at 16:46










          • $begingroup$
            It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
            $endgroup$
            – Shubham Johri
            Dec 11 '18 at 16:47












          • $begingroup$
            @Bernard Yeah I meant that.
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:47










          • $begingroup$
            @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:48



















          0












          $begingroup$

          Hint:



          For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$



          choose $cos(2x)=y$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            HINT: Rewrite the integrand as
            $$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:43












            • $begingroup$
              Do you mean the integrand is $arccos^3 2x ,sin 2x$?
              $endgroup$
              – Bernard
              Dec 11 '18 at 16:46










            • $begingroup$
              It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
              $endgroup$
              – Shubham Johri
              Dec 11 '18 at 16:47












            • $begingroup$
              @Bernard Yeah I meant that.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:47










            • $begingroup$
              @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:48
















            4












            $begingroup$

            HINT: Rewrite the integrand as
            $$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:43












            • $begingroup$
              Do you mean the integrand is $arccos^3 2x ,sin 2x$?
              $endgroup$
              – Bernard
              Dec 11 '18 at 16:46










            • $begingroup$
              It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
              $endgroup$
              – Shubham Johri
              Dec 11 '18 at 16:47












            • $begingroup$
              @Bernard Yeah I meant that.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:47










            • $begingroup$
              @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:48














            4












            4








            4





            $begingroup$

            HINT: Rewrite the integrand as
            $$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$






            share|cite|improve this answer









            $endgroup$



            HINT: Rewrite the integrand as
            $$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 11 '18 at 16:40









            FrpzzdFrpzzd

            22.9k841109




            22.9k841109












            • $begingroup$
              wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:43












            • $begingroup$
              Do you mean the integrand is $arccos^3 2x ,sin 2x$?
              $endgroup$
              – Bernard
              Dec 11 '18 at 16:46










            • $begingroup$
              It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
              $endgroup$
              – Shubham Johri
              Dec 11 '18 at 16:47












            • $begingroup$
              @Bernard Yeah I meant that.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:47










            • $begingroup$
              @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:48


















            • $begingroup$
              wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:43












            • $begingroup$
              Do you mean the integrand is $arccos^3 2x ,sin 2x$?
              $endgroup$
              – Bernard
              Dec 11 '18 at 16:46










            • $begingroup$
              It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
              $endgroup$
              – Shubham Johri
              Dec 11 '18 at 16:47












            • $begingroup$
              @Bernard Yeah I meant that.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:47










            • $begingroup$
              @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
              $endgroup$
              – Raghav
              Dec 11 '18 at 16:48
















            $begingroup$
            wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:43






            $begingroup$
            wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:43














            $begingroup$
            Do you mean the integrand is $arccos^3 2x ,sin 2x$?
            $endgroup$
            – Bernard
            Dec 11 '18 at 16:46




            $begingroup$
            Do you mean the integrand is $arccos^3 2x ,sin 2x$?
            $endgroup$
            – Bernard
            Dec 11 '18 at 16:46












            $begingroup$
            It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
            $endgroup$
            – Shubham Johri
            Dec 11 '18 at 16:47






            $begingroup$
            It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
            $endgroup$
            – Shubham Johri
            Dec 11 '18 at 16:47














            $begingroup$
            @Bernard Yeah I meant that.
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:47




            $begingroup$
            @Bernard Yeah I meant that.
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:47












            $begingroup$
            @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:48




            $begingroup$
            @ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
            $endgroup$
            – Raghav
            Dec 11 '18 at 16:48











            0












            $begingroup$

            Hint:



            For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$



            choose $cos(2x)=y$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Hint:



              For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$



              choose $cos(2x)=y$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint:



                For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$



                choose $cos(2x)=y$






                share|cite|improve this answer









                $endgroup$



                Hint:



                For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$



                choose $cos(2x)=y$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 17:33









                lab bhattacharjeelab bhattacharjee

                225k15157275




                225k15157275






























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