Primitive of Reciprocal of p by Sine of a x plus q by 1 plus Cosine of a x

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Theorem

$\ds \int \frac {\rd x} {p \sin a x + q \paren {1 + \cos a x} } = \frac 1 {a p} \ln \size {q + p \tan \frac {a x} 2} + C$


Proof

Let $z = a x$.

Then $\d x = \dfrac {\d z} a$ and so:

$(1): \quad \ds \int \frac {\rd x} {p \sin a x + q \paren {1 + \cos a x} } = \dfrac 1 a \int \frac {\rd z} {p \sin z + q \paren {1 + \cos z} }$


Then:

\(\ds u\) \(=\) \(\ds \tan \frac z 2\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {\rd z} {p \sin z + q \paren {1 + \cos z} }\) \(=\) \(\ds \int \frac {\frac {2 \rd u} {\paren {1 + u^2} } } {p \frac {2 u} {1 + u^2} + q \paren {1 + \frac {1 - u^2} {1 + u^2} } }\) Weierstrass Substitution
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {2 p u + q \paren {1 + u^2 + 1 - u^2} }\) multiplying top and bottom by $1 + u^2$
\(\ds \) \(=\) \(\ds \int \frac {\d u} {p u + q}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 p \ln \size {p u + q} + C\) Primitive of $\dfrac 1 {a x + b}$
\(\ds \) \(=\) \(\ds \frac 1 p \ln \size {q + p \tan \frac z 2} + C\) substituting for $u$
\(\ds \) \(=\) \(\ds \frac 1 {a p} \ln \size {q + p \tan \frac {a x} 2} + C\) from $(1)$

$\blacksquare$


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