integral of $int frac{du}{sqrt{(1-u^2)}}$












1












$begingroup$


I'm curious how my textbook got:



$$int frac{du}{sqrt{(1-u^2)}}$$



to



$$sin^{-1}(u)+ c$$



This considering that "$u$" could be simplified into $u = sin(x)$



Thus we get:



$$int frac{du}{sqrt{(1-sin^2(u))}}$$



Which is further simplified into:



$$int frac{du}{sqrt{(cos^2(u))}}$$



Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$



Meaning that we get:



$$int frac{du}{cos(u)}$$



Which is not $sin^{-1}(u)+ c$ when integrated.



What am I doing wrong?






calculus integration







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm curious how my textbook got:



    $$int frac{du}{sqrt{(1-u^2)}}$$



    to



    $$sin^{-1}(u)+ c$$



    This considering that "$u$" could be simplified into $u = sin(x)$



    Thus we get:



    $$int frac{du}{sqrt{(1-sin^2(u))}}$$



    Which is further simplified into:



    $$int frac{du}{sqrt{(cos^2(u))}}$$



    Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$



    Meaning that we get:



    $$int frac{du}{cos(u)}$$



    Which is not $sin^{-1}(u)+ c$ when integrated.



    What am I doing wrong?






    calculus integration







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm curious how my textbook got:



      $$int frac{du}{sqrt{(1-u^2)}}$$



      to



      $$sin^{-1}(u)+ c$$



      This considering that "$u$" could be simplified into $u = sin(x)$



      Thus we get:



      $$int frac{du}{sqrt{(1-sin^2(u))}}$$



      Which is further simplified into:



      $$int frac{du}{sqrt{(cos^2(u))}}$$



      Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$



      Meaning that we get:



      $$int frac{du}{cos(u)}$$



      Which is not $sin^{-1}(u)+ c$ when integrated.



      What am I doing wrong?






      calculus integration







      share|cite|improve this question











      $endgroup$




      I'm curious how my textbook got:



      $$int frac{du}{sqrt{(1-u^2)}}$$



      to



      $$sin^{-1}(u)+ c$$



      This considering that "$u$" could be simplified into $u = sin(x)$



      Thus we get:



      $$int frac{du}{sqrt{(1-sin^2(u))}}$$



      Which is further simplified into:



      $$int frac{du}{sqrt{(cos^2(u))}}$$



      Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$



      Meaning that we get:



      $$int frac{du}{cos(u)}$$



      Which is not $sin^{-1}(u)+ c$ when integrated.



      What am I doing wrong?






      calculus integration




      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 0:33









      Larry

      2,53531131




      2,53531131










      asked Dec 30 '18 at 0:22









      oxodooxodo

      322110




      322110






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:



          $$du=frac{du}{dx}dx=cos xdx$$



          Thus, the new integral becomes:



          $$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$



          Hopefully, you can take it from there.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
            $endgroup$
            – oxodo
            Dec 30 '18 at 0:37












          • $begingroup$
            @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 0:39






          • 1




            $begingroup$
            Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
            $endgroup$
            – oxodo
            Dec 30 '18 at 0:45








          • 1




            $begingroup$
            However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 0:58






          • 1




            $begingroup$
            @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 1:17



















          2












          $begingroup$

          $$I=intfrac{dx}{sqrt{1-x^2}}$$
          $$x=sin uRightarrow dx=cos u,du$$
          $$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
          Recalling that $$cos^2x=1-sin^2x$$
          We have
          $$I=intfrac{cos u,du}{cos u}$$
          $$I=int du$$
          $$I=u$$
          $$I=arcsin x+C$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:



            $$du=frac{du}{dx}dx=cos xdx$$



            Thus, the new integral becomes:



            $$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$



            Hopefully, you can take it from there.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:37












            • $begingroup$
              @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:39






            • 1




              $begingroup$
              Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:45








            • 1




              $begingroup$
              However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:58






            • 1




              $begingroup$
              @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 1:17
















            3












            $begingroup$

            So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:



            $$du=frac{du}{dx}dx=cos xdx$$



            Thus, the new integral becomes:



            $$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$



            Hopefully, you can take it from there.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:37












            • $begingroup$
              @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:39






            • 1




              $begingroup$
              Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:45








            • 1




              $begingroup$
              However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:58






            • 1




              $begingroup$
              @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 1:17














            3












            3








            3





            $begingroup$

            So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:



            $$du=frac{du}{dx}dx=cos xdx$$



            Thus, the new integral becomes:



            $$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$



            Hopefully, you can take it from there.






            share|cite|improve this answer









            $endgroup$



            So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:



            $$du=frac{du}{dx}dx=cos xdx$$



            Thus, the new integral becomes:



            $$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$



            Hopefully, you can take it from there.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 30 '18 at 0:26









            Noble MushtakNoble Mushtak

            15.3k1835




            15.3k1835












            • $begingroup$
              I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:37












            • $begingroup$
              @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:39






            • 1




              $begingroup$
              Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:45








            • 1




              $begingroup$
              However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:58






            • 1




              $begingroup$
              @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 1:17


















            • $begingroup$
              I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:37












            • $begingroup$
              @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:39






            • 1




              $begingroup$
              Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
              $endgroup$
              – oxodo
              Dec 30 '18 at 0:45








            • 1




              $begingroup$
              However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 0:58






            • 1




              $begingroup$
              @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
              $endgroup$
              – Noble Mushtak
              Dec 30 '18 at 1:17
















            $begingroup$
            I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
            $endgroup$
            – oxodo
            Dec 30 '18 at 0:37






            $begingroup$
            I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
            $endgroup$
            – oxodo
            Dec 30 '18 at 0:37














            $begingroup$
            @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 0:39




            $begingroup$
            @oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 0:39




            1




            1




            $begingroup$
            Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
            $endgroup$
            – oxodo
            Dec 30 '18 at 0:45






            $begingroup$
            Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
            $endgroup$
            – oxodo
            Dec 30 '18 at 0:45






            1




            1




            $begingroup$
            However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 0:58




            $begingroup$
            However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 0:58




            1




            1




            $begingroup$
            @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 1:17




            $begingroup$
            @oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
            $endgroup$
            – Noble Mushtak
            Dec 30 '18 at 1:17











            2












            $begingroup$

            $$I=intfrac{dx}{sqrt{1-x^2}}$$
            $$x=sin uRightarrow dx=cos u,du$$
            $$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
            Recalling that $$cos^2x=1-sin^2x$$
            We have
            $$I=intfrac{cos u,du}{cos u}$$
            $$I=int du$$
            $$I=u$$
            $$I=arcsin x+C$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              $$I=intfrac{dx}{sqrt{1-x^2}}$$
              $$x=sin uRightarrow dx=cos u,du$$
              $$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
              Recalling that $$cos^2x=1-sin^2x$$
              We have
              $$I=intfrac{cos u,du}{cos u}$$
              $$I=int du$$
              $$I=u$$
              $$I=arcsin x+C$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                $$I=intfrac{dx}{sqrt{1-x^2}}$$
                $$x=sin uRightarrow dx=cos u,du$$
                $$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
                Recalling that $$cos^2x=1-sin^2x$$
                We have
                $$I=intfrac{cos u,du}{cos u}$$
                $$I=int du$$
                $$I=u$$
                $$I=arcsin x+C$$






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                $endgroup$



                $$I=intfrac{dx}{sqrt{1-x^2}}$$
                $$x=sin uRightarrow dx=cos u,du$$
                $$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
                Recalling that $$cos^2x=1-sin^2x$$
                We have
                $$I=intfrac{cos u,du}{cos u}$$
                $$I=int du$$
                $$I=u$$
                $$I=arcsin x+C$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 30 '18 at 1:19









                clathratusclathratus

                5,1821338




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