Primitive of Reciprocal of p plus q by Cosine of a x/p^2 less than q^2/Also presented as

Primitive of $\frac 1 {p + q \cos a x}$ for $p^2 < q^2$: Also presented as

The result for $p^2 < q^2$ is also seen presented in the form:

$\ds \int \frac {\d x} {p + q \cos a x} = \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\dfrac {q + p \cos a x + \sqrt {q^2 - p^2} \sin a x} {p + q \cos a x} } + C$

Proof

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Sources

• 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Back endpapers: A Brief Table of Integrals: $75$.