Problem with Indefinite Integral $intfrac {cos^4x}{sin^3x} dx$












1












$begingroup$


I'm stuck with this integral



$intfrac {cos^4x}{sin^3x} dx$



which I rewrote as



$int csc^3x cos^4xdx$



then after using the half angle formula twice for $cos^4x$ I got this



$frac 14int csc^3x (1+cos(2x))(1+cos(2x))dx$



then after solving those products I got these integrals



$frac 14 {int csc^3xdx+2int csc^3x cos(2x)dx + int csc^3x cos^2(2x)dx}$



I do know how to solve the $int csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!






calculus indefinite-integrals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm stuck with this integral



    $intfrac {cos^4x}{sin^3x} dx$



    which I rewrote as



    $int csc^3x cos^4xdx$



    then after using the half angle formula twice for $cos^4x$ I got this



    $frac 14int csc^3x (1+cos(2x))(1+cos(2x))dx$



    then after solving those products I got these integrals



    $frac 14 {int csc^3xdx+2int csc^3x cos(2x)dx + int csc^3x cos^2(2x)dx}$



    I do know how to solve the $int csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!






    calculus indefinite-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm stuck with this integral



      $intfrac {cos^4x}{sin^3x} dx$



      which I rewrote as



      $int csc^3x cos^4xdx$



      then after using the half angle formula twice for $cos^4x$ I got this



      $frac 14int csc^3x (1+cos(2x))(1+cos(2x))dx$



      then after solving those products I got these integrals



      $frac 14 {int csc^3xdx+2int csc^3x cos(2x)dx + int csc^3x cos^2(2x)dx}$



      I do know how to solve the $int csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!






      calculus indefinite-integrals







      share|cite|improve this question











      $endgroup$




      I'm stuck with this integral



      $intfrac {cos^4x}{sin^3x} dx$



      which I rewrote as



      $int csc^3x cos^4xdx$



      then after using the half angle formula twice for $cos^4x$ I got this



      $frac 14int csc^3x (1+cos(2x))(1+cos(2x))dx$



      then after solving those products I got these integrals



      $frac 14 {int csc^3xdx+2int csc^3x cos(2x)dx + int csc^3x cos^2(2x)dx}$



      I do know how to solve the $int csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!






      calculus indefinite-integrals




      calculus indefinite-integrals






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 8 '16 at 14:43









      imranfat

      8,06141632




      8,06141632










      asked Feb 8 '16 at 14:37









      CryoCodexCryoCodex

      175




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          2 Answers
          2






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          3












          $begingroup$

          Let $t=cos x$, then
          $$int frac{cos^4 x}{sin^3 x}dx=-int frac{t^4}{(1-t^2)^2}dt.$$
          Can you proceed?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 14:47










          • $begingroup$
            Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
            $endgroup$
            – choco_addicted
            Feb 8 '16 at 14:58










          • $begingroup$
            Ohh i got it now! haha thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:03










          • $begingroup$
            To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
            $endgroup$
            – Travis
            Dec 29 '18 at 2:07



















          0












          $begingroup$

          Write $cos^4x=(1-sin^2x)(1-sin^2x)$ and expand. Then split the fraction. You'll then need to integrate $csc^3x$ , $csc x$ and $sin x$ all of which is standard






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'll try doing it this way too thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:12











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Let $t=cos x$, then
          $$int frac{cos^4 x}{sin^3 x}dx=-int frac{t^4}{(1-t^2)^2}dt.$$
          Can you proceed?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 14:47










          • $begingroup$
            Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
            $endgroup$
            – choco_addicted
            Feb 8 '16 at 14:58










          • $begingroup$
            Ohh i got it now! haha thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:03










          • $begingroup$
            To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
            $endgroup$
            – Travis
            Dec 29 '18 at 2:07
















          3












          $begingroup$

          Let $t=cos x$, then
          $$int frac{cos^4 x}{sin^3 x}dx=-int frac{t^4}{(1-t^2)^2}dt.$$
          Can you proceed?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 14:47










          • $begingroup$
            Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
            $endgroup$
            – choco_addicted
            Feb 8 '16 at 14:58










          • $begingroup$
            Ohh i got it now! haha thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:03










          • $begingroup$
            To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
            $endgroup$
            – Travis
            Dec 29 '18 at 2:07














          3












          3








          3





          $begingroup$

          Let $t=cos x$, then
          $$int frac{cos^4 x}{sin^3 x}dx=-int frac{t^4}{(1-t^2)^2}dt.$$
          Can you proceed?






          share|cite|improve this answer









          $endgroup$



          Let $t=cos x$, then
          $$int frac{cos^4 x}{sin^3 x}dx=-int frac{t^4}{(1-t^2)^2}dt.$$
          Can you proceed?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 8 '16 at 14:41









          choco_addictedchoco_addicted

          8,09261947




          8,09261947












          • $begingroup$
            Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 14:47










          • $begingroup$
            Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
            $endgroup$
            – choco_addicted
            Feb 8 '16 at 14:58










          • $begingroup$
            Ohh i got it now! haha thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:03










          • $begingroup$
            To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
            $endgroup$
            – Travis
            Dec 29 '18 at 2:07


















          • $begingroup$
            Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 14:47










          • $begingroup$
            Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
            $endgroup$
            – choco_addicted
            Feb 8 '16 at 14:58










          • $begingroup$
            Ohh i got it now! haha thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:03










          • $begingroup$
            To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
            $endgroup$
            – Travis
            Dec 29 '18 at 2:07
















          $begingroup$
          Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
          $endgroup$
          – CryoCodex
          Feb 8 '16 at 14:47




          $begingroup$
          Yeah I can proceed from there but I don't really understand what's going on with that substitution sorry if it is a silly question but what's going on behind it? I'm guessing you're using $sen^2 x = 1-cos^2 x$ on the denominator but i don't get how it got powered by 2, again sorry if it is a silly question or something!
          $endgroup$
          – CryoCodex
          Feb 8 '16 at 14:47












          $begingroup$
          Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
          $endgroup$
          – choco_addicted
          Feb 8 '16 at 14:58




          $begingroup$
          Your guessing is right. Since $dt=-sin x dx $, denominator except minus sign will be $sin^4 x $. Then apply $sin^2 x=1- cos^2 x$.
          $endgroup$
          – choco_addicted
          Feb 8 '16 at 14:58












          $begingroup$
          Ohh i got it now! haha thanks man!
          $endgroup$
          – CryoCodex
          Feb 8 '16 at 15:03




          $begingroup$
          Ohh i got it now! haha thanks man!
          $endgroup$
          – CryoCodex
          Feb 8 '16 at 15:03












          $begingroup$
          To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
          $endgroup$
          – Travis
          Dec 29 '18 at 2:07




          $begingroup$
          To see how you might come to this substitution, note that the integrand is a product of an even power of $cos x$ and an odd power of $sin x$.
          $endgroup$
          – Travis
          Dec 29 '18 at 2:07











          0












          $begingroup$

          Write $cos^4x=(1-sin^2x)(1-sin^2x)$ and expand. Then split the fraction. You'll then need to integrate $csc^3x$ , $csc x$ and $sin x$ all of which is standard






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'll try doing it this way too thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:12
















          0












          $begingroup$

          Write $cos^4x=(1-sin^2x)(1-sin^2x)$ and expand. Then split the fraction. You'll then need to integrate $csc^3x$ , $csc x$ and $sin x$ all of which is standard






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'll try doing it this way too thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:12














          0












          0








          0





          $begingroup$

          Write $cos^4x=(1-sin^2x)(1-sin^2x)$ and expand. Then split the fraction. You'll then need to integrate $csc^3x$ , $csc x$ and $sin x$ all of which is standard






          share|cite|improve this answer











          $endgroup$



          Write $cos^4x=(1-sin^2x)(1-sin^2x)$ and expand. Then split the fraction. You'll then need to integrate $csc^3x$ , $csc x$ and $sin x$ all of which is standard







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 1:53









          DavidG

          1




          1










          answered Feb 8 '16 at 14:41









          imranfatimranfat

          8,06141632




          8,06141632












          • $begingroup$
            I'll try doing it this way too thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:12


















          • $begingroup$
            I'll try doing it this way too thanks man!
            $endgroup$
            – CryoCodex
            Feb 8 '16 at 15:12
















          $begingroup$
          I'll try doing it this way too thanks man!
          $endgroup$
          – CryoCodex
          Feb 8 '16 at 15:12




          $begingroup$
          I'll try doing it this way too thanks man!
          $endgroup$
          – CryoCodex
          Feb 8 '16 at 15:12


















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