# Primitive of x by Square of Hyperbolic Secant of a x

## Theorem

$\ds \int x \sech^2 a x \rd x = \frac {x \tanh a x} a - \frac 1 {a^2} \ln \size {\cosh a x} + C$

## 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:

 $\ds u$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds 1$ Derivative of Identity Function

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds \sech^2 a x$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds \frac {\tanh a x} a$ Primitive of $\sech^2 a x$

Then:

 $\ds \int x \sech^2 a x \rd x$ $=$ $\ds x \paren {\frac {\tanh a x} a} - \int \paren {\frac {\tanh a x} a} \times 1 \rd x + C$ Integration by Parts $\ds$ $=$ $\ds \frac {x \tanh a x} a - \frac 1 a \int \tanh a x \rd x + C$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac {x \tanh a x} a - \frac 1 a \frac {\ln \size {\cosh a x} } a + C$ Primitive of $\tanh a x$ $\ds$ $=$ $\ds \frac {x \tanh a x} a - \frac 1 {a^2} \ln \size {\cosh a x} + C$ simplifying

$\blacksquare$