Prove $int_0^inftyoperatorname{sech} x,dx=pi/2$, and deduce $int_0^1operatorname{sech}^{-1}x,dx$












0















Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.






improper-integrals trigonometric-integrals







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  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    2 days ago










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    2 days ago










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    2 days ago
















0















Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.






improper-integrals trigonometric-integrals







share|cite|improve this question
























  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    2 days ago










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    2 days ago










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    2 days ago














0












0








0








Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.






improper-integrals trigonometric-integrals







share|cite|improve this question
















Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.






improper-integrals trigonometric-integrals




improper-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked 2 days ago









Abec

145




145












  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    2 days ago










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    2 days ago










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    2 days ago


















  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    2 days ago










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    2 days ago










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    2 days ago
















Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago




Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago












For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago




For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago












I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago




I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago










2 Answers
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Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.






share|cite|improve this answer































    0














    When it comes to the second part of your question:



    begin{equation}
    I = int f^{-1}(t):dt
    end{equation}



    Let $x = f^{-1}(t)$ or $t = f(x)$, then



    begin{equation}
    I = int x cdot f'(x):dx
    end{equation}



    Integrate by parts:



    begin{align}
    v'(x) &= f'(x) & u(x) &= x \
    v(x) &= f(x) & u'(x) &= 1
    end{align}



    Thus,



    begin{align}
    I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
    &= x cdot f(x) - F(x)
    end{align}



    Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



    begin{align}
    I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
    end{align}



    So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



    begin{equation}
    F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
    end{equation}



    Hence,



    begin{equation}
    Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
    end{equation}



    And thus:



    begin{equation}
    int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
    end{equation}



    Where $C$ is the constant of integration.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      0














      Note that by setting $u = operatorname{sech}^{-1} x$ we have that
      begin{align}
      int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
      end{align}
      then by integration by parts you should be done.






      share|cite|improve this answer




























        0














        Note that by setting $u = operatorname{sech}^{-1} x$ we have that
        begin{align}
        int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
        end{align}
        then by integration by parts you should be done.






        share|cite|improve this answer


























          0












          0








          0






          Note that by setting $u = operatorname{sech}^{-1} x$ we have that
          begin{align}
          int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
          end{align}
          then by integration by parts you should be done.






          share|cite|improve this answer














          Note that by setting $u = operatorname{sech}^{-1} x$ we have that
          begin{align}
          int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
          end{align}
          then by integration by parts you should be done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Jacky Chong

          17.8k21128




          17.8k21128























              0














              When it comes to the second part of your question:



              begin{equation}
              I = int f^{-1}(t):dt
              end{equation}



              Let $x = f^{-1}(t)$ or $t = f(x)$, then



              begin{equation}
              I = int x cdot f'(x):dx
              end{equation}



              Integrate by parts:



              begin{align}
              v'(x) &= f'(x) & u(x) &= x \
              v(x) &= f(x) & u'(x) &= 1
              end{align}



              Thus,



              begin{align}
              I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
              &= x cdot f(x) - F(x)
              end{align}



              Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



              begin{align}
              I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
              end{align}



              So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



              begin{equation}
              F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
              end{equation}



              Hence,



              begin{equation}
              Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
              end{equation}



              And thus:



              begin{equation}
              int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
              end{equation}



              Where $C$ is the constant of integration.






              share|cite|improve this answer


























                0














                When it comes to the second part of your question:



                begin{equation}
                I = int f^{-1}(t):dt
                end{equation}



                Let $x = f^{-1}(t)$ or $t = f(x)$, then



                begin{equation}
                I = int x cdot f'(x):dx
                end{equation}



                Integrate by parts:



                begin{align}
                v'(x) &= f'(x) & u(x) &= x \
                v(x) &= f(x) & u'(x) &= 1
                end{align}



                Thus,



                begin{align}
                I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
                &= x cdot f(x) - F(x)
                end{align}



                Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



                begin{align}
                I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
                end{align}



                So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



                begin{equation}
                F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
                end{equation}



                Hence,



                begin{equation}
                Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
                end{equation}



                And thus:



                begin{equation}
                int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
                end{equation}



                Where $C$ is the constant of integration.






                share|cite|improve this answer
























                  0












                  0








                  0






                  When it comes to the second part of your question:



                  begin{equation}
                  I = int f^{-1}(t):dt
                  end{equation}



                  Let $x = f^{-1}(t)$ or $t = f(x)$, then



                  begin{equation}
                  I = int x cdot f'(x):dx
                  end{equation}



                  Integrate by parts:



                  begin{align}
                  v'(x) &= f'(x) & u(x) &= x \
                  v(x) &= f(x) & u'(x) &= 1
                  end{align}



                  Thus,



                  begin{align}
                  I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
                  &= x cdot f(x) - F(x)
                  end{align}



                  Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



                  begin{align}
                  I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
                  end{align}



                  So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



                  begin{equation}
                  F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
                  end{equation}



                  Hence,



                  begin{equation}
                  Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
                  end{equation}



                  And thus:



                  begin{equation}
                  int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
                  end{equation}



                  Where $C$ is the constant of integration.






                  share|cite|improve this answer












                  When it comes to the second part of your question:



                  begin{equation}
                  I = int f^{-1}(t):dt
                  end{equation}



                  Let $x = f^{-1}(t)$ or $t = f(x)$, then



                  begin{equation}
                  I = int x cdot f'(x):dx
                  end{equation}



                  Integrate by parts:



                  begin{align}
                  v'(x) &= f'(x) & u(x) &= x \
                  v(x) &= f(x) & u'(x) &= 1
                  end{align}



                  Thus,



                  begin{align}
                  I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
                  &= x cdot f(x) - F(x)
                  end{align}



                  Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



                  begin{align}
                  I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
                  end{align}



                  So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



                  begin{equation}
                  F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
                  end{equation}



                  Hence,



                  begin{equation}
                  Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
                  end{equation}



                  And thus:



                  begin{equation}
                  int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
                  end{equation}



                  Where $C$ is the constant of integration.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  DavidG

                  1,836619




                  1,836619






























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