Primitive of Reciprocal of p squared plus square of q by Cosine of a x

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Theorem

$\ds \int \frac {\d x} {p^2 + q^2 \cos^2 a x} = \frac 1 {a p \sqrt{p^2 + q^2} } \arctan \frac {p \tan a x} {\sqrt {p^2 + q^2} } + C$


where $C$ is an arbitrary constant.


Proof

\(\ds \int \frac {\d x} {p^2 + q^2 \cos^2 a x}\) \(=\) \(\ds \int \frac {\csc^2 a x \rd x} {p^2 \csc^2 a x + q^2 \cot^2 a x}\) multiplying the numerator and the denominator by $\csc^2 a x$
\(\ds \) \(=\) \(\ds \int \frac {\csc^2 a x \rd x} {p^2 + \paren {p^2 + q^2} \cot^2 a x}\) Difference of Squares of Cosecant and Cotangent
\(\ds \) \(=\) \(\ds \int \frac {-\paren {\cot a x}' \rd x} {a p^2 + a \paren {p^2 + q^2} \cot^2 a x}\) Derivative of Cotangent Function
\(\ds \) \(=\) \(\ds \int \frac {-\d t} {a p^2 + a \paren {p^2 + q^2} t^2}\) putting $t = \cot a x$
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 + q^2} } \int \frac {\d t} {\paren {\tfrac p {\sqrt {p^2 + q^2} } }^2 + t^2}\)
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 + q^2} } \frac {\sqrt {p^2 + q^2} } p \, \map \arctan {\frac {\sqrt {p^2 + q^2} } p t } + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C\)
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} } p \cot a x} + C\) substituting $t = \cot a x$
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} } {p \tan a x} } + C\)
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 + q^2} } \map \arccot {\frac {p \tan a x} {\sqrt {p^2 + q^2} } } + C\) Arctangent of Reciprocal equals Arccotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {p^2 + q^2} } \paren {\map \arctan {\frac {p \tan a x} {\sqrt {p^2 + q^2} } } - \frac \pi 2} + C\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {p \tan a x} {\sqrt {p^2 + q^2} } } + C\) absorbing into the arbitrary constant

$\blacksquare$


Also see


Sources