Primitive of Reciprocal of p plus q by Cotangent of a x
Jump to navigation
Jump to search
Theorem
- $\ds \int \frac {\d x} {p + q \cot a x} = \frac {p x} {p^2 + q^2} - \frac q {a \paren {p^2 + q^2} } \ln \size {p \sin a x + q \cos a x} + C$
Proof
We have:
- $\dfrac \d {\d x} \paren {p \sin a x + q \cos a x} = a p \cos a x - a q \sin a x$
Thus:
\(\ds \int \frac {\d x} {p + q \cot a x}\) | \(=\) | \(\ds \int \frac {\d x} {p + q \dfrac {\cos a x} {\sin a x} }\) | Cotangent is Cosine divided by Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\sin a x \rd x} {p \sin a x + q \cos a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \int \frac {\paren {p^2 + q^2} \sin a x \rd x} {p \sin a x + q \cos a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \int \frac {p^2 \sin a x + p q \cos a x - p q \cos a x + q^2 \sin a x} {p \sin a x + q \cos a x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \paren {\int \frac {p^2 \sin a x + p q \cos a x} {p \sin a x + q \cos a x} \rd x + \int \frac {-p q \cos a x + q^2 \sin a x} {p \sin a x + q \cos a x} \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \paren {\int p \rd x - \frac q a \int \frac {\map \d {p \sin a x + q \cos a x} } {p \sin a x + q \cos a x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p x} {p^2 + q^2} - \frac q {a \paren {p^2 + q^2} } \ln \size {p \sin a x + q \cos a x} + C\) | Primitive of Constant and Primitive of Reciprocal |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\cot a x$: $14.449$