Primitive of x by Inverse Hyperbolic Secant of x over a

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Theorem

$\ds \int x \arsech \frac x a \rd x = \dfrac {x^2} 2 \arsech \dfrac x a - \dfrac {a \sqrt {a^2 - x^2} } 2 + C$

where $\arsech$ denotes the real area hyperbolic secant.


Corollary

$\ds \int x \paren {-\sech^{-1} \frac x a} \rd x = -\dfrac {x^2} 2 \paren {-\sech^{-1} \frac x a} + \dfrac {a \sqrt {a^2 - x^2} } 2 + C$

where $-\sech^{-1}$ denotes the negative branch of the real inverse hyperbolic secant multifunction.


Proof

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {x^2} 2\) Primitive of Power


Then:

\(\ds \int x \arsech \frac x a \rd x\) \(=\) \(\ds \frac {x^2} 2 \arsech \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {a^2 - x^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arsech \frac x a + \frac a 2 \int \frac {x \rd x} {\sqrt {a^2 - x^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arsech \frac x a + \frac a 2 \paren {-\sqrt {a^2 - x^2} } + C\) Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \arsech \frac x a - \frac {a \sqrt {a^2 - x^2} } 2 + C\) simplifying

$\blacksquare$


Also see


Sources

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