Primitive of x by Square of Secant of a x
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Contents
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 $\sec a x$: $14.458$