Primitive of x by Square of Secant of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

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


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:

\(\ds u\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 1\) Derivative of Identity Function


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:

\(\ds \int x \sec^2 a x \rd x\) \(=\) \(\ds \frac {x \tan a x} a - \int \frac {\tan a x} a \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x \tan a x} a - \frac 1 a \paren {\frac {-1} a \ln \size {\cos a x} } + C\) Primitive of $\tan a x$
\(\ds \) \(=\) \(\ds \frac {x \tan a x} a + \frac 1 {a^2} \ln \size {\cos a x} + C\) simplifying

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_x_by_Square_of_Secant_of_a_x&oldid=503629"