Primitive of Inverse Hyperbolic Secant of x over a

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Theorem

$\displaystyle \int \sech^{-1} \frac x a \rd x = \begin{cases} x \sech^{-1} \dfrac x a + a \arcsin \dfrac x a + C & : \sech^{-1} \dfrac x a > 0 \\ x \sech^{-1} \dfrac x a - a \arcsin \dfrac x a + C & : \sech^{-1} \dfrac x a < 0 \end{cases}$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle \sech^{-1} \frac x a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle \frac {-a} {x \sqrt {a^2 - x^2} }\) Derivative of $\sech^{-1} \dfrac x a$


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle x\) Primitive of Constant


Then:

\(\displaystyle \int \sech^{-1} \frac x a \rd x\) \(=\) \(\displaystyle x \sech^{-1} \frac x a - \int x \paren {\frac {-a} {x \sqrt {a^2 - x^2} } } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle x \sech^{-1} \frac x a + a \int \frac {\d x} {\sqrt {a^2 - x^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle x \sech^{-1} \frac x a + \arcsin \frac x a + C\) Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$



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