Primitive of Cube of Secant Function/Proof 2

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Theorem

$\ds \int \sec^3 x \rd x = \frac 1 2 \paren {\sec x \tan x + \ln \size {\sec x + \tan x} } + C$


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:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \sec^2 x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \tan x\) Primitive of $\sec^2 x$


Then:

\(\ds \int \sec^3 x \rd x\) \(=\) \(\ds \int \sec x \sec^2 x \rd x\)
\(\ds \) \(=\) \(\ds \sec x \tan x - \int \tan x \paren {\sec x \tan x} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \sec x \tan x - \int \tan^2 x \sec x \rd x + C\) simplifying
\(\ds \) \(=\) \(\ds \sec x \tan x - \int \paren {\sec^2 x - 1} \sec x \rd x + C\) Difference of $\sec^2$ and $\tan^2$
\(\ds \) \(=\) \(\ds \sec x \tan x - \int \sec^3 x \rd x + \int \sec x \rd x + C\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \sec x \tan x - \int \sec^3 x \rd x + \ln \size {\sec x + \tan x} + C\) Primitive of $\sec x$
\(\ds \leadsto \ \ \) \(\ds 2 \int \sec^3 a \rd x\) \(=\) \(\ds \sec x \tan x + \ln \size {\sec x + \tan x} + C\) simplifying
\(\ds \leadsto \ \ \) \(\ds \int \sec^3 a \rd x\) \(=\) \(\ds \dfrac 1 2 \paren {\sec x \tan x + \ln \size {\sec a x + \tan a x} } + C\)

$\blacksquare$


Sources

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