Calculating improper integral $intlimits_0^1 frac{arcsin(x)}{sqrt{1-x^2}},dx$












0












$begingroup$


Calculate improper integral $displaystyle int_0^1dfrac{arcsin(x)}{sqrt{1-x^2}},dx$



We had the following equation to calculate improper integrals (2nd style):

Let f in$left(a,bright]$ unbounded, but $forall varepsilon >0$ in every subinterval $left[a+varepsilon,bright]$ is bounded, we define:



$displaystyle int_a^b f(x),dx:=lim_{varepsilonto0^+}displaystyle int_{a+varepsilon}^b f(x), dx$



However, I only came up with this solution:
begin{align}
&displaystyle int_0^1 dfrac{arcsin(x)}{sqrt{1-x^2}},dx && mid u=arcsin x to dx=sqrt{1-x^2}, duquad u=arcsin0=0 quad u=arcsin1=pi/2\
&=displaystyle int_0^{fracpi2}u,du\
&= left[frac{u^2}{2}right]^{fracpi2}_0\
&=frac{pi^2}{8}
end{align}
My question is now, which is more accurate (and is this even correct)?






integration analysis definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Its one, I'm sorry.
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:18






  • 1




    $begingroup$
    What you did at the end is fine, because the substitution is correct and thus the improper integral becomes a proper Riemann one.
    $endgroup$
    – DonAntonio
    Jan 23 at 20:25






  • 1




    $begingroup$
    I would also have done this integral this way, very well ;)
    $endgroup$
    – Marine Galantin
    Jan 23 at 20:25










  • $begingroup$
    Thanks for your feedback!
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:28










  • $begingroup$
    Technically you have to rewrite your improper integral as a limit, then make the substitution. Since everything in sight is continuous though, it works out to be the same thing in the end.
    $endgroup$
    – Cheerful Parsnip
    Jan 23 at 20:31
















0












$begingroup$


Calculate improper integral $displaystyle int_0^1dfrac{arcsin(x)}{sqrt{1-x^2}},dx$



We had the following equation to calculate improper integrals (2nd style):

Let f in$left(a,bright]$ unbounded, but $forall varepsilon >0$ in every subinterval $left[a+varepsilon,bright]$ is bounded, we define:



$displaystyle int_a^b f(x),dx:=lim_{varepsilonto0^+}displaystyle int_{a+varepsilon}^b f(x), dx$



However, I only came up with this solution:
begin{align}
&displaystyle int_0^1 dfrac{arcsin(x)}{sqrt{1-x^2}},dx && mid u=arcsin x to dx=sqrt{1-x^2}, duquad u=arcsin0=0 quad u=arcsin1=pi/2\
&=displaystyle int_0^{fracpi2}u,du\
&= left[frac{u^2}{2}right]^{fracpi2}_0\
&=frac{pi^2}{8}
end{align}
My question is now, which is more accurate (and is this even correct)?






integration analysis definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Its one, I'm sorry.
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:18






  • 1




    $begingroup$
    What you did at the end is fine, because the substitution is correct and thus the improper integral becomes a proper Riemann one.
    $endgroup$
    – DonAntonio
    Jan 23 at 20:25






  • 1




    $begingroup$
    I would also have done this integral this way, very well ;)
    $endgroup$
    – Marine Galantin
    Jan 23 at 20:25










  • $begingroup$
    Thanks for your feedback!
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:28










  • $begingroup$
    Technically you have to rewrite your improper integral as a limit, then make the substitution. Since everything in sight is continuous though, it works out to be the same thing in the end.
    $endgroup$
    – Cheerful Parsnip
    Jan 23 at 20:31














0












0








0





$begingroup$


Calculate improper integral $displaystyle int_0^1dfrac{arcsin(x)}{sqrt{1-x^2}},dx$



We had the following equation to calculate improper integrals (2nd style):

Let f in$left(a,bright]$ unbounded, but $forall varepsilon >0$ in every subinterval $left[a+varepsilon,bright]$ is bounded, we define:



$displaystyle int_a^b f(x),dx:=lim_{varepsilonto0^+}displaystyle int_{a+varepsilon}^b f(x), dx$



However, I only came up with this solution:
begin{align}
&displaystyle int_0^1 dfrac{arcsin(x)}{sqrt{1-x^2}},dx && mid u=arcsin x to dx=sqrt{1-x^2}, duquad u=arcsin0=0 quad u=arcsin1=pi/2\
&=displaystyle int_0^{fracpi2}u,du\
&= left[frac{u^2}{2}right]^{fracpi2}_0\
&=frac{pi^2}{8}
end{align}
My question is now, which is more accurate (and is this even correct)?






integration analysis definite-integrals improper-integrals







share|cite|improve this question











$endgroup$




Calculate improper integral $displaystyle int_0^1dfrac{arcsin(x)}{sqrt{1-x^2}},dx$



We had the following equation to calculate improper integrals (2nd style):

Let f in$left(a,bright]$ unbounded, but $forall varepsilon >0$ in every subinterval $left[a+varepsilon,bright]$ is bounded, we define:



$displaystyle int_a^b f(x),dx:=lim_{varepsilonto0^+}displaystyle int_{a+varepsilon}^b f(x), dx$



However, I only came up with this solution:
begin{align}
&displaystyle int_0^1 dfrac{arcsin(x)}{sqrt{1-x^2}},dx && mid u=arcsin x to dx=sqrt{1-x^2}, duquad u=arcsin0=0 quad u=arcsin1=pi/2\
&=displaystyle int_0^{fracpi2}u,du\
&= left[frac{u^2}{2}right]^{fracpi2}_0\
&=frac{pi^2}{8}
end{align}
My question is now, which is more accurate (and is this even correct)?






integration analysis definite-integrals improper-integrals




integration analysis definite-integrals improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Feb 1 at 13:44









Martin Sleziak

44.8k10119272




44.8k10119272










asked Jan 23 at 20:14









DoesbaddelDoesbaddel

34011




34011












  • $begingroup$
    Its one, I'm sorry.
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:18






  • 1




    $begingroup$
    What you did at the end is fine, because the substitution is correct and thus the improper integral becomes a proper Riemann one.
    $endgroup$
    – DonAntonio
    Jan 23 at 20:25






  • 1




    $begingroup$
    I would also have done this integral this way, very well ;)
    $endgroup$
    – Marine Galantin
    Jan 23 at 20:25










  • $begingroup$
    Thanks for your feedback!
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:28










  • $begingroup$
    Technically you have to rewrite your improper integral as a limit, then make the substitution. Since everything in sight is continuous though, it works out to be the same thing in the end.
    $endgroup$
    – Cheerful Parsnip
    Jan 23 at 20:31


















  • $begingroup$
    Its one, I'm sorry.
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:18






  • 1




    $begingroup$
    What you did at the end is fine, because the substitution is correct and thus the improper integral becomes a proper Riemann one.
    $endgroup$
    – DonAntonio
    Jan 23 at 20:25






  • 1




    $begingroup$
    I would also have done this integral this way, very well ;)
    $endgroup$
    – Marine Galantin
    Jan 23 at 20:25










  • $begingroup$
    Thanks for your feedback!
    $endgroup$
    – Doesbaddel
    Jan 23 at 20:28










  • $begingroup$
    Technically you have to rewrite your improper integral as a limit, then make the substitution. Since everything in sight is continuous though, it works out to be the same thing in the end.
    $endgroup$
    – Cheerful Parsnip
    Jan 23 at 20:31
















$begingroup$
Its one, I'm sorry.
$endgroup$
– Doesbaddel
Jan 23 at 20:18




$begingroup$
Its one, I'm sorry.
$endgroup$
– Doesbaddel
Jan 23 at 20:18




1




1




$begingroup$
What you did at the end is fine, because the substitution is correct and thus the improper integral becomes a proper Riemann one.
$endgroup$
– DonAntonio
Jan 23 at 20:25




$begingroup$
What you did at the end is fine, because the substitution is correct and thus the improper integral becomes a proper Riemann one.
$endgroup$
– DonAntonio
Jan 23 at 20:25




1




1




$begingroup$
I would also have done this integral this way, very well ;)
$endgroup$
– Marine Galantin
Jan 23 at 20:25




$begingroup$
I would also have done this integral this way, very well ;)
$endgroup$
– Marine Galantin
Jan 23 at 20:25












$begingroup$
Thanks for your feedback!
$endgroup$
– Doesbaddel
Jan 23 at 20:28




$begingroup$
Thanks for your feedback!
$endgroup$
– Doesbaddel
Jan 23 at 20:28












$begingroup$
Technically you have to rewrite your improper integral as a limit, then make the substitution. Since everything in sight is continuous though, it works out to be the same thing in the end.
$endgroup$
– Cheerful Parsnip
Jan 23 at 20:31




$begingroup$
Technically you have to rewrite your improper integral as a limit, then make the substitution. Since everything in sight is continuous though, it works out to be the same thing in the end.
$endgroup$
– Cheerful Parsnip
Jan 23 at 20:31










3 Answers
3






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1












$begingroup$

In fact you show only one solution, because the first approach remains theoretical. The only way that I know to handle this integral is by the arc sine change of variable, which makes the "improperness" disappear.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The problem isn't at $x=0$, but instead at $x=1$.
    However, your substitution seems to have eliminated this issue, so yes, your answer is correct.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider



      $$int_0^bfrac{arcsin(x)}{sqrt{1-x^2}},dx,$$
      then take the limit as $bto 1^-.$ This will lead to the answer you found.






      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









        1












        $begingroup$

        In fact you show only one solution, because the first approach remains theoretical. The only way that I know to handle this integral is by the arc sine change of variable, which makes the "improperness" disappear.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          In fact you show only one solution, because the first approach remains theoretical. The only way that I know to handle this integral is by the arc sine change of variable, which makes the "improperness" disappear.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            In fact you show only one solution, because the first approach remains theoretical. The only way that I know to handle this integral is by the arc sine change of variable, which makes the "improperness" disappear.






            share|cite|improve this answer









            $endgroup$



            In fact you show only one solution, because the first approach remains theoretical. The only way that I know to handle this integral is by the arc sine change of variable, which makes the "improperness" disappear.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 23 at 20:25









            Yves DaoustYves Daoust

            129k676227




            129k676227























                1












                $begingroup$

                The problem isn't at $x=0$, but instead at $x=1$.
                However, your substitution seems to have eliminated this issue, so yes, your answer is correct.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  The problem isn't at $x=0$, but instead at $x=1$.
                  However, your substitution seems to have eliminated this issue, so yes, your answer is correct.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The problem isn't at $x=0$, but instead at $x=1$.
                    However, your substitution seems to have eliminated this issue, so yes, your answer is correct.






                    share|cite|improve this answer









                    $endgroup$



                    The problem isn't at $x=0$, but instead at $x=1$.
                    However, your substitution seems to have eliminated this issue, so yes, your answer is correct.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 23 at 20:28









                    AlexandrosAlexandros

                    9281412




                    9281412























                        1












                        $begingroup$

                        As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider



                        $$int_0^bfrac{arcsin(x)}{sqrt{1-x^2}},dx,$$
                        then take the limit as $bto 1^-.$ This will lead to the answer you found.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider



                          $$int_0^bfrac{arcsin(x)}{sqrt{1-x^2}},dx,$$
                          then take the limit as $bto 1^-.$ This will lead to the answer you found.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider



                            $$int_0^bfrac{arcsin(x)}{sqrt{1-x^2}},dx,$$
                            then take the limit as $bto 1^-.$ This will lead to the answer you found.






                            share|cite|improve this answer









                            $endgroup$



                            As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider



                            $$int_0^bfrac{arcsin(x)}{sqrt{1-x^2}},dx,$$
                            then take the limit as $bto 1^-.$ This will lead to the answer you found.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 23 at 20:34









                            zhw.zhw.

                            73.8k43175




                            73.8k43175






























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