Primitive of x by Square of Secant of a x

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Theorem

$\displaystyle \int x \sec^2 a x \ \mathrm d x = \frac {x \tan a x} a + \frac 1 {a^2} \ln \left\vert{\cos a x}\right\vert + C$


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:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle \sec^2 a x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {\tan a x} a\) Primitive of $\sec^2 a x$


Then:

\(\displaystyle \int x \sec^2 a x \ \mathrm d x\) \(=\) \(\displaystyle \frac {x \tan a x} a - \int \frac {\tan a x} a \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \tan a x} a - \frac 1 a \left({\frac {-1} a \ln \left\vert{\cos a x}\right\vert}\right) + C\) Primitive of $\tan a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \tan a x} a + \frac 1 {a^2} \ln \left\vert{\cos a x}\right\vert + C\) simplifying

$\blacksquare$


Also see


Sources