Primitive of Reciprocal of p plus q by Cotangent of a x

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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

$\blacksquare$


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Sources