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

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Theorem

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

where $a$ and $b$ are real numbers with $a > b > 0$.


Proof

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

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

We have:

\(\ds \frac \partial {\partial a} \int_0^{2 \pi} \frac {\d x} {a + b \sin x}\) \(=\) \(\ds \int_0^{2 \pi} \frac \partial {\partial a} \paren {\frac 1 {a + b \sin x} } \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds -\int_0^{2 \pi} \frac {\d x} {\paren {a + b \sin x}^2}\) 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:

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

$\blacksquare$


Also see


Sources

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