Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x/Proof 2
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Theorem
- $\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$
Proof
| \(\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x}\) | \(=\) | \(\ds \int_0^\pi \frac {\d x} {a + b \cos x} + \int_\pi^{2 \pi} \frac {\d x} {a + b \cos x}\) | Sum of Integrals on Adjacent Intervals for Integrable Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 0 \pi + \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } \pi {2 \pi}\) | Primitive of $\dfrac 1 {p + q \cos a x}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} }\) | ||||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \pi} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} }\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} }\) | Tangent Function is Periodic on Reals | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren { \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} }\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} }\) | simplfying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{u \mathop \to \infty} \map \arctan {\sqrt {\frac {a - b} {a + b} } u} - \lim_{u \mathop \to -\infty} \map \arctan {\sqrt {\frac {a - b} {a + b} } u} }\) | letting $u = \tan \dfrac x 2$: Tangent Function $\to \pm \infty$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\frac \pi 2 - \paren {-\frac \pi 2} }\) | Limit to Positive and Negative Infinity of Arctangent Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \pi} {\sqrt {a^2 - b^2} }\) |
$\blacksquare$