Why can't I make the substitution $ u = sin (ax + b) $ to evaluate $ int sin (ax + b) cos (ax + b) dx$?












2












$begingroup$


Evaluate $ int sin (ax + b) cos (ax + b) dx$?



To do this, I started of by substituting $ u = sin (ax + b) $. That made $ du = cos (ax + b) cdot a $ and wrote the integral as $ frac 1a int u du $ to get the final answer as: $$ frac 1{2a} sin^2 (ax + b) $$
This answer however, is wrong. My textbook uses a different method and arrives at a different answer. I understand how to arrive at the (right) answer but I want to know why I can't get the same answer by substitution here.



My textbook starts off by rewriting the expression as $ frac {sin 2 (ax + b)}{2} $ and then substitutes $ 2 (ax + b) = u $ to get this answer: $$ - frac { cos 2 (ax + b) } {4a } $$






indefinite-integrals







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$endgroup$








  • 1




    $begingroup$
    What you found is indeed an antiderivative of $sin(ax+b)cos(ax+b)$.
    $endgroup$
    – drhab
    Jan 21 at 9:12






  • 1




    $begingroup$
    They used the identity $sin(2u)=2sin(u)cos(u)$
    $endgroup$
    – coreyman317
    Jan 21 at 9:13








  • 4




    $begingroup$
    The answers differ by a constant (the "C" that you omitted).
    $endgroup$
    – David Mitra
    Jan 21 at 9:13






  • 2




    $begingroup$
    $sin^2(ax+b)=dfrac{1-cos(2(ax+b))}2$; the $1/2$ gets absorbed in the arbitrary constant
    $endgroup$
    – Shubham Johri
    Jan 21 at 9:16








  • 3




    $begingroup$
    Possible duplicate of Integral of $sin x cos x$ using two methods differs by a constant?
    $endgroup$
    – Hans Lundmark
    Jan 21 at 9:18
















2












$begingroup$


Evaluate $ int sin (ax + b) cos (ax + b) dx$?



To do this, I started of by substituting $ u = sin (ax + b) $. That made $ du = cos (ax + b) cdot a $ and wrote the integral as $ frac 1a int u du $ to get the final answer as: $$ frac 1{2a} sin^2 (ax + b) $$
This answer however, is wrong. My textbook uses a different method and arrives at a different answer. I understand how to arrive at the (right) answer but I want to know why I can't get the same answer by substitution here.



My textbook starts off by rewriting the expression as $ frac {sin 2 (ax + b)}{2} $ and then substitutes $ 2 (ax + b) = u $ to get this answer: $$ - frac { cos 2 (ax + b) } {4a } $$






indefinite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What you found is indeed an antiderivative of $sin(ax+b)cos(ax+b)$.
    $endgroup$
    – drhab
    Jan 21 at 9:12






  • 1




    $begingroup$
    They used the identity $sin(2u)=2sin(u)cos(u)$
    $endgroup$
    – coreyman317
    Jan 21 at 9:13








  • 4




    $begingroup$
    The answers differ by a constant (the "C" that you omitted).
    $endgroup$
    – David Mitra
    Jan 21 at 9:13






  • 2




    $begingroup$
    $sin^2(ax+b)=dfrac{1-cos(2(ax+b))}2$; the $1/2$ gets absorbed in the arbitrary constant
    $endgroup$
    – Shubham Johri
    Jan 21 at 9:16








  • 3




    $begingroup$
    Possible duplicate of Integral of $sin x cos x$ using two methods differs by a constant?
    $endgroup$
    – Hans Lundmark
    Jan 21 at 9:18














2












2








2





$begingroup$


Evaluate $ int sin (ax + b) cos (ax + b) dx$?



To do this, I started of by substituting $ u = sin (ax + b) $. That made $ du = cos (ax + b) cdot a $ and wrote the integral as $ frac 1a int u du $ to get the final answer as: $$ frac 1{2a} sin^2 (ax + b) $$
This answer however, is wrong. My textbook uses a different method and arrives at a different answer. I understand how to arrive at the (right) answer but I want to know why I can't get the same answer by substitution here.



My textbook starts off by rewriting the expression as $ frac {sin 2 (ax + b)}{2} $ and then substitutes $ 2 (ax + b) = u $ to get this answer: $$ - frac { cos 2 (ax + b) } {4a } $$






indefinite-integrals







share|cite|improve this question











$endgroup$




Evaluate $ int sin (ax + b) cos (ax + b) dx$?



To do this, I started of by substituting $ u = sin (ax + b) $. That made $ du = cos (ax + b) cdot a $ and wrote the integral as $ frac 1a int u du $ to get the final answer as: $$ frac 1{2a} sin^2 (ax + b) $$
This answer however, is wrong. My textbook uses a different method and arrives at a different answer. I understand how to arrive at the (right) answer but I want to know why I can't get the same answer by substitution here.



My textbook starts off by rewriting the expression as $ frac {sin 2 (ax + b)}{2} $ and then substitutes $ 2 (ax + b) = u $ to get this answer: $$ - frac { cos 2 (ax + b) } {4a } $$






indefinite-integrals




indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 9:12







WorldGov

















asked Jan 21 at 9:06









WorldGovWorldGov

324111




324111








  • 1




    $begingroup$
    What you found is indeed an antiderivative of $sin(ax+b)cos(ax+b)$.
    $endgroup$
    – drhab
    Jan 21 at 9:12






  • 1




    $begingroup$
    They used the identity $sin(2u)=2sin(u)cos(u)$
    $endgroup$
    – coreyman317
    Jan 21 at 9:13








  • 4




    $begingroup$
    The answers differ by a constant (the "C" that you omitted).
    $endgroup$
    – David Mitra
    Jan 21 at 9:13






  • 2




    $begingroup$
    $sin^2(ax+b)=dfrac{1-cos(2(ax+b))}2$; the $1/2$ gets absorbed in the arbitrary constant
    $endgroup$
    – Shubham Johri
    Jan 21 at 9:16








  • 3




    $begingroup$
    Possible duplicate of Integral of $sin x cos x$ using two methods differs by a constant?
    $endgroup$
    – Hans Lundmark
    Jan 21 at 9:18














  • 1




    $begingroup$
    What you found is indeed an antiderivative of $sin(ax+b)cos(ax+b)$.
    $endgroup$
    – drhab
    Jan 21 at 9:12






  • 1




    $begingroup$
    They used the identity $sin(2u)=2sin(u)cos(u)$
    $endgroup$
    – coreyman317
    Jan 21 at 9:13








  • 4




    $begingroup$
    The answers differ by a constant (the "C" that you omitted).
    $endgroup$
    – David Mitra
    Jan 21 at 9:13






  • 2




    $begingroup$
    $sin^2(ax+b)=dfrac{1-cos(2(ax+b))}2$; the $1/2$ gets absorbed in the arbitrary constant
    $endgroup$
    – Shubham Johri
    Jan 21 at 9:16








  • 3




    $begingroup$
    Possible duplicate of Integral of $sin x cos x$ using two methods differs by a constant?
    $endgroup$
    – Hans Lundmark
    Jan 21 at 9:18








1




1




$begingroup$
What you found is indeed an antiderivative of $sin(ax+b)cos(ax+b)$.
$endgroup$
– drhab
Jan 21 at 9:12




$begingroup$
What you found is indeed an antiderivative of $sin(ax+b)cos(ax+b)$.
$endgroup$
– drhab
Jan 21 at 9:12




1




1




$begingroup$
They used the identity $sin(2u)=2sin(u)cos(u)$
$endgroup$
– coreyman317
Jan 21 at 9:13






$begingroup$
They used the identity $sin(2u)=2sin(u)cos(u)$
$endgroup$
– coreyman317
Jan 21 at 9:13






4




4




$begingroup$
The answers differ by a constant (the "C" that you omitted).
$endgroup$
– David Mitra
Jan 21 at 9:13




$begingroup$
The answers differ by a constant (the "C" that you omitted).
$endgroup$
– David Mitra
Jan 21 at 9:13




2




2




$begingroup$
$sin^2(ax+b)=dfrac{1-cos(2(ax+b))}2$; the $1/2$ gets absorbed in the arbitrary constant
$endgroup$
– Shubham Johri
Jan 21 at 9:16






$begingroup$
$sin^2(ax+b)=dfrac{1-cos(2(ax+b))}2$; the $1/2$ gets absorbed in the arbitrary constant
$endgroup$
– Shubham Johri
Jan 21 at 9:16






3




3




$begingroup$
Possible duplicate of Integral of $sin x cos x$ using two methods differs by a constant?
$endgroup$
– Hans Lundmark
Jan 21 at 9:18




$begingroup$
Possible duplicate of Integral of $sin x cos x$ using two methods differs by a constant?
$endgroup$
– Hans Lundmark
Jan 21 at 9:18










3 Answers
3






active

oldest

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5












$begingroup$

Note that $$1-cos2(ax+b)=2sin^2(ax+b)$$



So the textbook gives almost the same as your outcome and the difference is constant.



That is allowed because to be found is an antiderivative (not an integral).






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The other method probably is that $int sin (ax+b) cos (ax+b) dx=frac{1}{2}int sin (2ax+2b) dx= frac{1}{2} cdot frac{1}{2a}(-cos(2ax+2b))+c=frac{1}{4a}(2 sin^2(ax+b)-1)+c)= frac{1}{2a} sin^2(ax+b)+c$



    using $cos(2x)=1-2sin^2(x)$ and putting $-frac{1}{4a}$ inside the constant






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      You can solve with the substitution
      u=sin(ax+b) then,
      du=acos(ax+b)dx so,
      ∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
      which is same with your answer-1/4a






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        Note that $$1-cos2(ax+b)=2sin^2(ax+b)$$



        So the textbook gives almost the same as your outcome and the difference is constant.



        That is allowed because to be found is an antiderivative (not an integral).






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Note that $$1-cos2(ax+b)=2sin^2(ax+b)$$



          So the textbook gives almost the same as your outcome and the difference is constant.



          That is allowed because to be found is an antiderivative (not an integral).






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Note that $$1-cos2(ax+b)=2sin^2(ax+b)$$



            So the textbook gives almost the same as your outcome and the difference is constant.



            That is allowed because to be found is an antiderivative (not an integral).






            share|cite|improve this answer









            $endgroup$



            Note that $$1-cos2(ax+b)=2sin^2(ax+b)$$



            So the textbook gives almost the same as your outcome and the difference is constant.



            That is allowed because to be found is an antiderivative (not an integral).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 21 at 9:15









            drhabdrhab

            102k545136




            102k545136























                2












                $begingroup$

                The other method probably is that $int sin (ax+b) cos (ax+b) dx=frac{1}{2}int sin (2ax+2b) dx= frac{1}{2} cdot frac{1}{2a}(-cos(2ax+2b))+c=frac{1}{4a}(2 sin^2(ax+b)-1)+c)= frac{1}{2a} sin^2(ax+b)+c$



                using $cos(2x)=1-2sin^2(x)$ and putting $-frac{1}{4a}$ inside the constant






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  The other method probably is that $int sin (ax+b) cos (ax+b) dx=frac{1}{2}int sin (2ax+2b) dx= frac{1}{2} cdot frac{1}{2a}(-cos(2ax+2b))+c=frac{1}{4a}(2 sin^2(ax+b)-1)+c)= frac{1}{2a} sin^2(ax+b)+c$



                  using $cos(2x)=1-2sin^2(x)$ and putting $-frac{1}{4a}$ inside the constant






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The other method probably is that $int sin (ax+b) cos (ax+b) dx=frac{1}{2}int sin (2ax+2b) dx= frac{1}{2} cdot frac{1}{2a}(-cos(2ax+2b))+c=frac{1}{4a}(2 sin^2(ax+b)-1)+c)= frac{1}{2a} sin^2(ax+b)+c$



                    using $cos(2x)=1-2sin^2(x)$ and putting $-frac{1}{4a}$ inside the constant






                    share|cite|improve this answer









                    $endgroup$



                    The other method probably is that $int sin (ax+b) cos (ax+b) dx=frac{1}{2}int sin (2ax+2b) dx= frac{1}{2} cdot frac{1}{2a}(-cos(2ax+2b))+c=frac{1}{4a}(2 sin^2(ax+b)-1)+c)= frac{1}{2a} sin^2(ax+b)+c$



                    using $cos(2x)=1-2sin^2(x)$ and putting $-frac{1}{4a}$ inside the constant







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 21 at 9:16









                    user289143user289143

                    1,002313




                    1,002313























                        -1












                        $begingroup$

                        You can solve with the substitution
                        u=sin(ax+b) then,
                        du=acos(ax+b)dx so,
                        ∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
                        which is same with your answer-1/4a






                        share|cite|improve this answer









                        $endgroup$


















                          -1












                          $begingroup$

                          You can solve with the substitution
                          u=sin(ax+b) then,
                          du=acos(ax+b)dx so,
                          ∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
                          which is same with your answer-1/4a






                          share|cite|improve this answer









                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            You can solve with the substitution
                            u=sin(ax+b) then,
                            du=acos(ax+b)dx so,
                            ∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
                            which is same with your answer-1/4a






                            share|cite|improve this answer









                            $endgroup$



                            You can solve with the substitution
                            u=sin(ax+b) then,
                            du=acos(ax+b)dx so,
                            ∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
                            which is same with your answer-1/4a







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 21 at 9:24









                            kongmyeongkongmyeong

                            111




                            111






























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