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 $(1)$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.421$