$intfrac{sin(x)}{sqrt{1-cos^2(x)}}$ Where did I mess up the domain?












1












$begingroup$


So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$



I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?






indefinite-integrals trigonometric-integrals







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  • $begingroup$
    What makes you think that it is wrong ? And why do you incriminate the domain ?
    $endgroup$
    – Yves Daoust
    Jan 29 at 12:56


















1












$begingroup$


So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$



I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?






indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    What makes you think that it is wrong ? And why do you incriminate the domain ?
    $endgroup$
    – Yves Daoust
    Jan 29 at 12:56
















1












1








1


0



$begingroup$


So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$



I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?






indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$




So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$



I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?






indefinite-integrals trigonometric-integrals




indefinite-integrals trigonometric-integrals






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edited Jan 29 at 13:00









Thomas Shelby

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4,5862727










asked Jan 29 at 12:49









Ibrahim ElsadekIbrahim Elsadek

112




112












  • $begingroup$
    What makes you think that it is wrong ? And why do you incriminate the domain ?
    $endgroup$
    – Yves Daoust
    Jan 29 at 12:56




















  • $begingroup$
    What makes you think that it is wrong ? And why do you incriminate the domain ?
    $endgroup$
    – Yves Daoust
    Jan 29 at 12:56


















$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56






$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56












3 Answers
3






active

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2












$begingroup$

You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
    $endgroup$
    – Ibrahim Elsadek
    Jan 29 at 13:02





















2












$begingroup$

We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$



and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.



Can you take it from here ?






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The integrand is a square wave. Its antiderivative is a triangle wave.



    enter image description here






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
        $endgroup$
        – Ibrahim Elsadek
        Jan 29 at 13:02


















      2












      $begingroup$

      You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
        $endgroup$
        – Ibrahim Elsadek
        Jan 29 at 13:02
















      2












      2








      2





      $begingroup$

      You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.






      share|cite|improve this answer









      $endgroup$



      You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 29 at 12:53









      5xum5xum

      91.9k394161




      91.9k394161












      • $begingroup$
        OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
        $endgroup$
        – Ibrahim Elsadek
        Jan 29 at 13:02




















      • $begingroup$
        OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
        $endgroup$
        – Ibrahim Elsadek
        Jan 29 at 13:02


















      $begingroup$
      OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
      $endgroup$
      – Ibrahim Elsadek
      Jan 29 at 13:02






      $begingroup$
      OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
      $endgroup$
      – Ibrahim Elsadek
      Jan 29 at 13:02













      2












      $begingroup$

      We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$



      and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.



      Can you take it from here ?






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$



        and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.



        Can you take it from here ?






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$



          and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.



          Can you take it from here ?






          share|cite|improve this answer









          $endgroup$



          We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$



          and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.



          Can you take it from here ?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 12:54









          FredFred

          48.9k11849




          48.9k11849























              1












              $begingroup$

              The integrand is a square wave. Its antiderivative is a triangle wave.



              enter image description here






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The integrand is a square wave. Its antiderivative is a triangle wave.



                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The integrand is a square wave. Its antiderivative is a triangle wave.



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  The integrand is a square wave. Its antiderivative is a triangle wave.



                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 29 at 13:03









                  Yves DaoustYves Daoust

                  131k676229




                  131k676229






























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