Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Cosine x/Proof 1

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

From Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \cos x}$, we have:

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

We have:

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

\(=\)

\(\ds \int_0^{2 \pi} \frac \partial {\partial a} \paren {\frac 1 {a + b \cos x} } \rd x\)

Definite Integral of Partial Derivative

\(\ds \)

\(=\)

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

Quotient Rule for Derivatives

and:

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

\(=\)

\(\ds 2 \pi \paren {\frac {-2 a} {2 \sqrt {a^2 - b^2} } } \paren {\frac 1 {a^2 - b^2} }\)

Quotient Rule for Derivatives

\(\ds \)

\(=\)

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

giving:

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

$\blacksquare$