Primitive of x by Square of Hyperbolic Cosecant of a x

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Theorem

$\displaystyle \int x \csch^2 a x \rd x = \frac {-x \coth a x} a + \frac 1 {a^2} \ln \size {\sinh a x} + C$


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:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \csch^2 a x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle -\frac {\coth a x} a\) Primitive of $\csch^2 a x$


Then:

\(\displaystyle \int x \csch^2 a x \rd x\) \(=\) \(\displaystyle x \paren {-\frac {\coth a x} a} - \int \paren {-\frac {\coth a x} a} \times 1 \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-x \coth a x} a + \frac 1 a \int \coth a x \rd x + C\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle \frac {-x \coth a x} a + \frac 1 a \frac {\ln \size {\sinh a x} } a + C\) Primitive of $\coth a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-x \coth a x} a + \frac 1 {a^2} \ln \size {\sinh a x} + C\) simplifying

$\blacksquare$

Also see


Sources