Primitive of x by Square of Hyperbolic Secant of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int x \operatorname{sech}^2 a x \ \mathrm d x = \frac {x \tanh a x} a - \frac 1 {a^2} \ln \left\vert{\cosh a x}\right\vert + C$


Proof

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\displaystyle \frac {\mathrm d v}{\mathrm d x}\) \(=\) \(\displaystyle \operatorname{sech}^2 a x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {\tanh a x} a\) Primitive of $\operatorname{sech}^2 a x$


Then:

\(\displaystyle \int x \operatorname{sech}^2 a x \ \mathrm d x\) \(=\) \(\displaystyle x \left({\frac {\tanh a x} a}\right) - \int \left({\frac {\tanh a x} a}\right) \times 1 \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \tanh a x} a - \frac 1 a \int \tanh a x \ \mathrm d x + C\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \tanh a x} a - \frac 1 a \frac {\ln \left\vert{\cosh a x}\right\vert} a + C\) Primitive of $\tanh a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \tanh a x} a - \frac 1 {a^2} \ln \left\vert{\cosh a x}\right\vert + C\) simplifying

$\blacksquare$

Also see


Sources