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Integral of $sin^5xcdotcos^14x$



The Next CEO of Stack OverflowEvaluate $int frac1sin xcos x dx $How to solve $int cos(2x)cos(3x) dx$?Integral $int limits _0 ^pi |sin x + cos x|; dx$Integrate $int fracsin(3x)cos x~dx$Finding $int ^pi_-pisin(nx)cos(mx)dx$Going from integral of $dx$ to integral of $dtheta$?Fourier Series for $f(x)=sin(x)+cos(2x)$Evaluate the definite integral: $intlimits_0^pi/2fracsin x-cos xsqrt1-sin 2x, dx$Improper integral of $exp[-(a+bi)x]log x$Integral of $sin^5(x)cos(x)$










5












$begingroup$


I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.



$$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$



My work:



$$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$



$$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$



substitute $cos(x)$ for $u$.
$du=sin(x)dx$



$$intlimits_0^fracpi2(1-u^2)^2u^14du$$



$$intlimits_0^fracpi2(1-2u+u^4)u^14du$$



$$intlimits_0^fracpi2u^14-2u^16+u^18du$$



$$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$



$$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$



$$0-0+0-(1-1+1)$$



$$-1+1-1=-1$$



The correct answer is $frac84845$



Where did I go wrong?










share|cite|improve this question











$endgroup$
















    5












    $begingroup$


    I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.



    $$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$



    My work:



    $$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$



    $$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$



    substitute $cos(x)$ for $u$.
    $du=sin(x)dx$



    $$intlimits_0^fracpi2(1-u^2)^2u^14du$$



    $$intlimits_0^fracpi2(1-2u+u^4)u^14du$$



    $$intlimits_0^fracpi2u^14-2u^16+u^18du$$



    $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$



    $$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$



    $$0-0+0-(1-1+1)$$



    $$-1+1-1=-1$$



    The correct answer is $frac84845$



    Where did I go wrong?










    share|cite|improve this question











    $endgroup$














      5












      5








      5





      $begingroup$


      I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.



      $$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$



      My work:



      $$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$



      $$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$



      substitute $cos(x)$ for $u$.
      $du=sin(x)dx$



      $$intlimits_0^fracpi2(1-u^2)^2u^14du$$



      $$intlimits_0^fracpi2(1-2u+u^4)u^14du$$



      $$intlimits_0^fracpi2u^14-2u^16+u^18du$$



      $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$



      $$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$



      $$0-0+0-(1-1+1)$$



      $$-1+1-1=-1$$



      The correct answer is $frac84845$



      Where did I go wrong?










      share|cite|improve this question











      $endgroup$




      I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.



      $$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$



      My work:



      $$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$



      $$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$



      substitute $cos(x)$ for $u$.
      $du=sin(x)dx$



      $$intlimits_0^fracpi2(1-u^2)^2u^14du$$



      $$intlimits_0^fracpi2(1-2u+u^4)u^14du$$



      $$intlimits_0^fracpi2u^14-2u^16+u^18du$$



      $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$



      $$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$



      $$0-0+0-(1-1+1)$$



      $$-1+1-1=-1$$



      The correct answer is $frac84845$



      Where did I go wrong?







      integration definite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 18 at 11:30









      MarianD

      1,7391617




      1,7391617










      asked Mar 18 at 10:56









      HarukuHaruku

      353




      353




















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
            $endgroup$
            – Haruku
            Mar 18 at 11:09











          • $begingroup$
            Your welcome. Good luck!!
            $endgroup$
            – Mostafa Ayaz
            Mar 18 at 11:10


















          1












          $begingroup$

          $$u = cos x$$
          $$du = -sin x ,dx$$



          (and not $du = sin x ,dx$, as you wrote).






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
            $$u=cos(x)$$
            then the limit of integration change to



            $$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)_colorredcos0^colorredcosfracpi2=F(cos(x))_colorred0^colorredfracpi2$$






            share|cite|improve this answer











            $endgroup$













              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
                $endgroup$
                – Haruku
                Mar 18 at 11:09











              • $begingroup$
                Your welcome. Good luck!!
                $endgroup$
                – Mostafa Ayaz
                Mar 18 at 11:10















              4












              $begingroup$

              It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
                $endgroup$
                – Haruku
                Mar 18 at 11:09











              • $begingroup$
                Your welcome. Good luck!!
                $endgroup$
                – Mostafa Ayaz
                Mar 18 at 11:10













              4












              4








              4





              $begingroup$

              It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...






              share|cite|improve this answer









              $endgroup$



              It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 18 at 11:02









              Mostafa AyazMostafa Ayaz

              18.1k31040




              18.1k31040











              • $begingroup$
                And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
                $endgroup$
                – Haruku
                Mar 18 at 11:09











              • $begingroup$
                Your welcome. Good luck!!
                $endgroup$
                – Mostafa Ayaz
                Mar 18 at 11:10
















              • $begingroup$
                And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
                $endgroup$
                – Haruku
                Mar 18 at 11:09











              • $begingroup$
                Your welcome. Good luck!!
                $endgroup$
                – Mostafa Ayaz
                Mar 18 at 11:10















              $begingroup$
              And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
              $endgroup$
              – Haruku
              Mar 18 at 11:09





              $begingroup$
              And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
              $endgroup$
              – Haruku
              Mar 18 at 11:09













              $begingroup$
              Your welcome. Good luck!!
              $endgroup$
              – Mostafa Ayaz
              Mar 18 at 11:10




              $begingroup$
              Your welcome. Good luck!!
              $endgroup$
              – Mostafa Ayaz
              Mar 18 at 11:10











              1












              $begingroup$

              $$u = cos x$$
              $$du = -sin x ,dx$$



              (and not $du = sin x ,dx$, as you wrote).






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                $$u = cos x$$
                $$du = -sin x ,dx$$



                (and not $du = sin x ,dx$, as you wrote).






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  $$u = cos x$$
                  $$du = -sin x ,dx$$



                  (and not $du = sin x ,dx$, as you wrote).






                  share|cite|improve this answer









                  $endgroup$



                  $$u = cos x$$
                  $$du = -sin x ,dx$$



                  (and not $du = sin x ,dx$, as you wrote).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 18 at 11:05









                  MarianDMarianD

                  1,7391617




                  1,7391617





















                      0












                      $begingroup$

                      There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
                      $$u=cos(x)$$
                      then the limit of integration change to



                      $$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)_colorredcos0^colorredcosfracpi2=F(cos(x))_colorred0^colorredfracpi2$$






                      share|cite|improve this answer











                      $endgroup$

















                        0












                        $begingroup$

                        There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
                        $$u=cos(x)$$
                        then the limit of integration change to



                        $$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)_colorredcos0^colorredcosfracpi2=F(cos(x))_colorred0^colorredfracpi2$$






                        share|cite|improve this answer











                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
                          $$u=cos(x)$$
                          then the limit of integration change to



                          $$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)_colorredcos0^colorredcosfracpi2=F(cos(x))_colorred0^colorredfracpi2$$






                          share|cite|improve this answer











                          $endgroup$



                          There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
                          $$u=cos(x)$$
                          then the limit of integration change to



                          $$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)_colorredcos0^colorredcosfracpi2=F(cos(x))_colorred0^colorredfracpi2$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Mar 20 at 11:44

























                          answered Mar 18 at 11:26









                          miracle173miracle173

                          7,38022247




                          7,38022247



























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