The integral of 1/sqrt(x^2-1) dx

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Now to

\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  

This can be done using various substitutions. We will look at a hyperbolic substitution in the end of this page. It can also be done using the substitution

w = x + \sqrt {{x^2} - 1}  

But that is begging the question, because we basically have to know the answer to come up with that idea. We could also do the substitution

x = \sec \theta

But that will require us to know, or figure out, that

\sqrt {{{\sec }^2}\theta  - 1}  = {\tan } \theta  

which is quite easy to find. You can get it from

\sin^2 \theta + \cos^2 \theta =1

in a couple of steps. Worse is that after the substitution we end up with the necessity to  find the integral of sec θ, and that is not all too strait forward.

The above can be found in various sources, so we will not go through tat here,  but instead we will  find the answer to our integral using (simple) complex analysis.

Using complex analysis

We will use complex analysis, or rather just complex numbers and a bit of calculus in a rather straight forward way. Let us first pull out a factor of -1 from the square root to get

\begin{gathered}   I = \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  = \int {\frac{{dx}}{{\sqrt {( - 1)(1 - {x^2})} }}}  =  \hfill \\   \quad  = \int {\frac{{dx}}{{i\sqrt {(1 - {x^2})} }}}  = \frac{1}{i}{\sin ^{ - 1}}x + C \hfill \\ \end{gathered}  

Technically we are done, but it is in a not all too useful form.  To fix this we need to look at the arcsine part. We want it to be able to “absorb” the i.

To start with, if

I = \frac{1}{i}{\sin ^{ - 1}}x

then

\sin (iI) = x

(We put the constant of integration back later on.) From a previous section we have that

\sin \theta  = \frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}

So

x = \sin iI = \frac{{{e^{ - I}} - {e^I}}}{{2i}}

or

2ix = {e^{ - I}} - {e^I} = \frac{1}{w} - w

where

w = {e^I}

this gives us

2ixw = 1 - {w^2}

or

{w^2} + 2ixw - 1 = 0

This we can solve using the quadratic formula to get

\begin{gathered}   w = \frac{{ - 2ix \pm \sqrt { - 4{x^2} - 4 \cdot 1 \cdot ( - 1)} }}{{2 \cdot 1}} \hfill \\   \quad  = \frac{{ - 2ix \pm \sqrt { - 4{x^2} + 4} }}{2} \hfill \\   \quad  = \frac{{ - 2ix \pm \sqrt { - 4} \sqrt {{x^2} - 1} }}{2} \hfill \\   \quad  = \frac{{ - 2ix \pm 2i\sqrt {{x^2} - 1} }}{2} \hfill \\   \quad  = i\left( {x \pm \sqrt {{x^2} - 1} } \right) \hfill \\  \end{gathered}  

Substituting back eI and adding back the constant of integration this will give us

\begin{gathered}   I = \ln \left( {i\left( {x \pm \sqrt {{x^2} - 1} } \right)} \right) + {C_1} \hfill \\   \quad  = \ln (i) + \ln \left( {x \pm \sqrt {{x^2} - 1} } \right) + {C_1} \hfill \\   \quad  = \ln \left( {x \pm \sqrt {{x^2} - 1} } \right) + C \hfill \\  \end{gathered}  

Here we let the constant absorb the ln(i). Next comes the question of which sign to use. We may to start with note that

\left( {x + \sqrt {{x^2} - 1} } \right) = {\left( {x - \sqrt {{x^2} - 1} } \right)^{ - 1}}

and thus that

\ln \left( {x \pm \sqrt {{x^2} - 1} } \right) =  \pm \ln \left( {x + \sqrt {{x^2} - 1} } \right)

If z is larger than one the logarithm is not negative, since

x \geqslant 1 \Rightarrow x + \sqrt {{x^2} - 1}  \geqslant 1

and we can also see that this function is increasing, but less and less fast. The function

\frac{1}{{\sqrt {{x^2} - 1} }}

should be the derivative of the above logarithmic expression. We can see that is s positive, and decreasing as x>0, and the only option is thus that the integral is the positive variant of the logarithm. So we get

\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  = \ln \left( {x + \sqrt {{x^2} - 1} } \right) + C,\quad x \geqslant 1

Of symmetry reasons we also have

\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  = \ln \left| {x + \sqrt {{x^2} - 1} } \right| + C,\quad \left| x \right| \geqslant 1

as shown in the graph below.

The inverse of the answer

Let us look at the inverse function

y = \ln \left( {x + \sqrt {{x^2} - 1} } \right)

We get

x = \ln \left( {y + \sqrt {{y^2} - 1} } \right)

and thus

\begin{gathered}   {e^x} = y + \sqrt {{y^2} - 1} \quad \quad \quad \quad // - y \hfill \\   {e^x} - y = \sqrt {{y^2} - 1} \quad \quad \quad \quad /{/^2} \hfill \\   {e^{2x}} - 2y{e^x} + {y^2} = {y^2} - 1\quad // - {y^2} \hfill \\   {e^{2x}} - 2y{e^x} =  - 1\quad \quad \quad \quad \;// \cdot {e^{ - x}} \hfill \\   {e^x} - 2y =  - {e^{ - x}}\quad \quad \quad \quad \;\;// + 2y + {e^{ - x}} \hfill \\   {e^x} + {e^{ - x}} = 2y\quad \quad \quad \quad \;\;\;\;// \div 2 \hfill \\   y = \frac{{{e^x} + {e^{ - x}}}}{2} = \cosh x \hfill \\  \end{gathered}  

This is one of the hyperbolic functions. You can find more on those here. So this means that

{\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)

so

\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  = {\cosh ^{ - 1}}x + C,\quad x \geqslant 1

Using hyperbolic functions directly

If we happen to know our hyperbolic functions we can do the substitution

\left| \begin{gathered}   x = \cosh y \hfill \\   dx = \sinh y\;dy \hfill \\  \end{gathered}  \right.

We also have that

{\cosh ^2}\theta  - {\sinh ^2}\theta  = 1 \Rightarrow {\cosh ^2}\theta  - 1 = {\sinh ^2}\theta  

This gives us

\begin{gathered}   \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}  = \int {\frac{{\sinh y\;dy}}{{\sqrt {{{\cosh }^2}y - 1} }}}  = \int {\frac{{\sinh y\;dy}}{{\sinh y}}}  \hfill \\   \quad  = \int {dy}  = y + C = {\cosh ^{ - 1}}x + C \hfill \\  \end{gathered}  

The above generalized a bit

Now, we get

\begin{gathered} \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}}  \hfill \\ \quad  = \int {\frac{{dx}}{{a\sqrt {{{\left( {\frac{x}{a}} \right)}^2} - 1} }}}  \hfill \\ \quad  = \frac{{\;\frac{1}{a}\;}}{{\;\frac{1}{a}\;}}{\cosh ^{ - 1}}\frac{x}{a} + C = {\cosh ^{ - 1}}\frac{x}{a} + C \hfill \\ \end{gathered}  

or

\begin{gathered} \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}}  = {\cosh ^{ - 1}}\frac{x}{a} + {C_1} \hfill \\ \quad  = \ln \left( {\frac{x}{a} + \sqrt {\frac{{{x^2}}}{{{a^2}}} - 1} } \right) + {C_1} \hfill \\ \quad  = \ln \left( {\frac{x}{a} + \frac{1}{a}\sqrt {{x^2} - {a^2}} } \right) + {C_1} \hfill \\ \quad  = \ln \left( {\frac{x}{a} + \frac{1}{a}\sqrt {{x^2} - {a^2}} } \right) + {C_1} \hfill \\ \quad  = \ln \left( {x + \sqrt {{x^2} - {a^2}} } \right) - \ln a + {C_1} \hfill \\ \quad  = \ln \left( {x + \sqrt {{x^2} - {a^2}} } \right) + C \hfill \\ \end{gathered}  

Up a level : Integrals
Previous page : The integral of 1/sqrt(1-x^2) dx
Next page : The integral of one over sqrt(x^2+1) dxLast modified: Sep 26, 2020 @ 14:18