Primitive of Cube of Secant of a x/Proof 2
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Theorem
- $\ds \int \sec^3 a x \rd x = \frac 1 {2 a} \paren {\sec a x \tan a x + \ln \size {\sec a x + \tan a x} } + C$
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:
| \(\ds u\) | \(=\) | \(\ds \sec a x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds a \sec a x \tan a x\) | Derivative of Function of Constant Multiple, Derivative of $\sec$ |
and let:
| \(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \sec^2 a x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {\tan a x} a\) | Primitive of $\sec^2 a x$ |
Then:
| \(\ds \int \sec^3 a x \rd x\) | \(=\) | \(\ds \int \sec a x \sec^2 a x \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sec a x \frac {\tan a x} a - \int \paren {\frac {\tan a x} a} \paren {a \sec a x \tan a x} \rd x + C\) | Integration by Parts | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \sec a x \tan a x - \int \tan^2 a x \sec a x \rd x + C\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \sec a x \tan a x - \int \paren {\sec^2 a x - 1} \sec a x \rd x + C\) | Difference of $\sec^2$ and $\tan^2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \sec a x \tan a x - \int \sec^3 a x \rd x + \int \sec a x \rd x + C\) | Linear Combination of Primitives | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \int \sec^3 a x \rd x\) | \(=\) | \(\ds \frac 1 a \sec a x \tan a x + \int \sec a x \rd x + C\) | simplifying | ||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \sec a x \tan a x + \frac 1 a \ln \size {\sec a x + \tan a x} + C\) | Primitive of $\sec a x$ |
The result follows.
$\blacksquare$