Primitive of Cosine of a x over x squared

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Theorem

$\ds \int \frac {\cos a x \rd x} {x^2} = \frac {-\cos a x} x - a \int \frac {\sin a x \rd x} x$


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:

\(\ds u\) \(=\) \(\ds \cos a x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds -a \sin a x\) Derivative of $\cos a x$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {x^2}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds - \frac 1 x\) Primitive of Power


Then:

\(\ds \int \frac {\cos a x \rd x} {x^2}\) \(=\) \(\ds \int \cos a x \frac 1 {x^2} \rd x\)
\(\ds \) \(=\) \(\ds \cos a x \paren {-\frac 1 x} - \int \paren {-\frac 1 x} \paren {-a \sin a x} \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {-\cos a x} x - a \int \frac {\sin a x \rd x} x\) simplifying

$\blacksquare$


Also see


Sources

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