Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Cosine x/Proof 2

Theorem

$\ds \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$

Proof

\(\ds \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2}\)

\(=\)

\(\ds \intlimits {\frac {b \sin x} {\paren {b^2 - a^2} \paren {a + b \cos x} } } 0 {2 \pi} - \frac a {b^2 - a^2} \int_0^{2 \pi} \frac {\d x} {a + b \cos x}\)

Primitive of Reciprocal of square of p plus q by Cosine of a x

\(\ds \)

\(=\)

\(\ds \frac {b \sin 2 \pi} {\paren {b^2 - a^2} \paren {a + b \cos 2 \pi} } - \frac {b \sin 0} {\paren {b^2 - a^2} \paren {a + b \cos 0} } + \frac a {a^2 - b^2} \int_0^{2 \pi} \frac {\d x} {a + b \cos x}\)

\(\ds \)

\(=\)

\(\ds \frac a {a^2 - b^2} \int_0^{2 \pi} \frac {\d x} {a + b \cos x}\)

Sine of Integer Multiple of Pi

\(\ds \)

\(=\)

\(\ds \frac a {a^2 - b^2} \times \frac {2 \pi} {\sqrt {a^2 - b^2} }\)

Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x

\(\ds \)

\(=\)

\(\ds \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }\)

$\blacksquare$