Primitive of x by Square of Hyperbolic Cosecant of a x

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

• Primitive of $x \sinh^2 a x$
• Primitive of $x \cosh^2 a x$
• Primitive of $x \tanh^2 a x$
• Primitive of $x \coth^2 a x$
• Primitive of $x \sech^2 a x$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\csch a x$: $14.642$