Absolute value in indefinite integral












4















I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$




I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$



I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?



Thanks in advance.






indefinite-integrals absolute-value







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  • 1




    Cause u is between $-pi/2, pi/2$
    – dmtri
    Dec 21 at 8:58






  • 2




    Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
    – Claude Leibovici
    Dec 21 at 9:22
















4















I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$




I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$



I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?



Thanks in advance.






indefinite-integrals absolute-value







share|cite|improve this question




















  • 1




    Cause u is between $-pi/2, pi/2$
    – dmtri
    Dec 21 at 8:58






  • 2




    Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
    – Claude Leibovici
    Dec 21 at 9:22














4












4








4








I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$




I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$



I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?



Thanks in advance.






indefinite-integrals absolute-value







share|cite|improve this question
















I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$




I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$



I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?



Thanks in advance.






indefinite-integrals absolute-value




indefinite-integrals absolute-value






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share|cite|improve this question













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edited Dec 21 at 17:40









trancelocation

9,0901521




9,0901521










asked Dec 21 at 8:53









user121

273




273








  • 1




    Cause u is between $-pi/2, pi/2$
    – dmtri
    Dec 21 at 8:58






  • 2




    Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
    – Claude Leibovici
    Dec 21 at 9:22














  • 1




    Cause u is between $-pi/2, pi/2$
    – dmtri
    Dec 21 at 8:58






  • 2




    Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
    – Claude Leibovici
    Dec 21 at 9:22








1




1




Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58




Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58




2




2




Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22




Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22










2 Answers
2






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3














When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$






share|cite|improve this answer





























    2














    Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.



    Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      3














      When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$






      share|cite|improve this answer


























        3














        When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$






        share|cite|improve this answer
























          3












          3








          3






          When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$






          share|cite|improve this answer












          When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 at 11:18









          Mostafa Ayaz

          13.7k3836




          13.7k3836























              2














              Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.



              Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?






              share|cite|improve this answer


























                2














                Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.



                Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?






                share|cite|improve this answer
























                  2












                  2








                  2






                  Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.



                  Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?






                  share|cite|improve this answer












                  Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.



                  Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 at 9:00









                  KM101

                  4,173417




                  4,173417






























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