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

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Using the result from the previous page we get

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

Now if

$J = \frac{i}{2}{\sin ^{ - 1}}x$

then

$\sin \frac{{2J}}{i} = x$

Now since

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

we get

$x = \sin \frac{{2J}}{i} = \frac{{{e^{ - 2J}} - {e^{2J}}}}{{2i}}$

$2ix = {e^{ - 2J}} - {e^{2J}} = \frac{1}{w} - w,\quad w = {e^{2J}}$

and thus

$2ixw = 1 - {w^2} \Rightarrow {w^2} + 2ixw - 1 = 0$

This gives us

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

$\begin{gathered} J = \frac{1}{2}\ln \left( {x \pm \sqrt {{x^2} - 1} } \right) \hfill \\ \quad = \pm \frac{1}{2}\ln \left( {x + \sqrt {{x^2} - 1} } \right) \hfill \\ \end{gathered}$

Then we have

$\begin{gathered} K = \frac{i}{2}x\sqrt {1 - {x^2}} \hfill \\ \quad = \frac{1}{2}\sqrt { - 1} \sqrt {1 - {x^2}} \hfill \\ \quad = \frac{1}{2}\sqrt {{x^2} - 1} \hfill \\ \end{gathered}$

Together this gives us

$\begin{gathered} \int {\sqrt {{x^2} - 1} \,dx = } \hfill \\ \quad = \frac{i}{2}{\sin ^{ - 1}}x + \frac{i}{2}x\sqrt {1 - {x^2}} + C \hfill \\ \quad = J + K + C = K + J + C \hfill \\ \quad = \frac{1}{2}x\sqrt {{x^2} - 1} \pm \frac{1}{2}\ln \left( {x + \sqrt {{x^2} - 1} } \right) + C \hfill \\ \end{gathered}$

So what sign to use? If we look at the original function between 1 and 2 we have an increasing function starting at 0 and ending at

$\sqrt {{2^2} - 1} = \sqrt 3$

So the integral from 1 to 2 should be positive and less than the above. But our integral is

$\begin{gathered} \int_1^2 {\sqrt {{x^2} - 1} } = \hfill \\ \quad = \frac{1}{2}2\sqrt 3 \pm \frac{1}{2}\ln \left( {2 + \sqrt 3 } \right) \hfill \\ \quad = \sqrt 3 \pm \frac{1}{2}\ln \left( {2 + \sqrt 3 } \right) \hfill \\ \end{gathered}$

We must thus have

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

A graph of this (but whit the slight change of adding an absolute function to include the negative x-values) is seen below.

Generalizing a bit we get

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

Here the first leading a comes from the 1/1/a we get as we integrate over a linear inner function, and we then need to divide by the x-coefficient (the gradient of the liner term x/a). Then it is “just” a bunch of algebra (it took me three trials to get all the a´s right).

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Previous page : The integral of sqrt(1-x^2) dx
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Last modified: May 18, 2019 @ 13:08