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Establish the integration formula for arcsec x

Establish that the following integration formula is correct:

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \frac{x}{|x|} \log \left| x + \sqrt{x^2-1} \right| + C. \]


Proof. First, we want to integrate by parts. Let

    \begin{align*}  u &= \operatorname{arcsec} x & du &= \frac{1}{|x| \sqrt{x^2-1} } \, dx \\  dv &= dx & v &= x. \end{align*}

Therefore, we have

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \int \frac{x}{|x| \sqrt{x^2-1}}. \]

Now, to evaluate this integral we break it into pieces (to get rid of the |x|). Since we have

    \[ \frac{x}{|x| \sqrt{x^2-1}} =  \begin{dcases} \frac{1}{\sqrt{x^2-1}} & \text{if } x > 1 \\  \frac{-1}{\sqrt{x^2-1}} & \text{if } x < -1. \end{dcases} \]

(Since \operatorname{arcsec} x is only defined for |x| \geq 1 we don’t need to worry about the cases -1 < x < 1.) We then evaluate the two pieces of the integral separately. If x > 1,

    \begin{align*}  x \operatorname{arcsec} x - \int \frac{x}{|x| \sqrt{x^2-1}} \, dx &= x \operatorname{arcsec} x - \int \frac{1}{\sqrt{x^2-1}} \, dx \\  &= x \operatorname{arcsec} x - \int \frac{x + \sqrt{x^2-1}}{(\sqrt{x^2-1})(x + \sqrt{x^2-1})} \, dx \\  &= x \operatorname{arcsec} x - \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} \right) \, dx \\  &= x \operatorname{arcsec} x - \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( 1 + \frac{x}{\sqrt{x^2-1}} \right) \, dx. \end{align*}

Now, we note that

    \[ f(x) = x + \sqrt{x^2-1} \quad \implies \quad f'(x) = 1 + \frac{x}{\sqrt{x^2-1}}. \]

Therefore, we have an integral of the form \int \frac{f'(x)}{f(x)} \, dx and so,

    \[  \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} \right) \, dx = \log \left| x + \sqrt{x^2-1} \right| + C. \]

So, for x > 1 we have

    \[ \int \operatorname{arcsec} x \, dx &= x \operatorname{arcsec} x - \log \left| x + \sqrt{x^2-1} \right| + C. \]

For the case x < -1 everything is identical except we have a negative sign (since \frac{x}{|x| \sqrt{x^2-1}} has a negative sign when x < -1) so for x < -1 we have

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x + \log \left| x + \sqrt{x^2-1} \right| + C. \]

Therefore (since \frac{x}{|x|} = 1 if x > 1 and \frac{x}{|x|} = -1 if x < -1) we have

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \frac{x}{|x|} \log \left| x + \sqrt{x^2-1} \right| + C. \qquad \blacksquare \]

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