Primitive of Sine of a x over x squared
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Theorem
- $\ds \int \frac {\sin a x \rd x} {x^2} = -\frac {\sin a x} x + a \int \frac {\cos a x \rd x} x + C$
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:
\(\ds u\) | \(=\) | \(\ds \sin a x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds a \cos a x\) | Derivative of $\sin a x$ |
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:
\(\ds \int \frac {\sin a x \rd x} {x^2}\) | \(=\) | \(\ds \int \sin a x \frac 1 {x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin a x \paren {-\frac 1 x} - \int \paren {-\frac 1 x} a \cos a x \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {\sin a x} x + a \int \frac {\cos a x \rd x} x + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$: $14.344$