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

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

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

$\ds \int \frac {\d x} {p + q \sin a x} = -\frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\sqrt {\dfrac {p - q} {p + q} } \map \tan {\dfrac \pi 4 - \dfrac {p x} 2} } + C$

Proof

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Sources

• 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (next): Back endpapers: A Brief Table of Integrals: $70$.