Primitive of x by Square of Secant of a x

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

• Primitive of $x \sin^2 a x$
• Primitive of $x \cos^2 a x$
• Primitive of $x \tan^2 a x$
• Primitive of $x \cot^2 a x$
• Primitive of $x \csc^2 a x$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sec a x$: $14.458$