Primitive of Reciprocal of p squared minus square of q by Sine of a x
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Theorem
- $\ds \int \frac {\d x} {p^2 - q^2 \sin^2 a x} = \begin {cases}
\dfrac 1 {a p \sqrt {p^2 - q^2} } \arctan \dfrac {\sqrt {p^2 - q^2} \tan a x} p & : p^2 > q^2 \\ \dfrac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\dfrac {\sqrt {q^2 - p^2} \tan a x + p} {\sqrt {q^2 - p^2} \tan a x - p} } & : p^2 < q^2 \end {cases}$
Proof
\(\ds \int \frac {\d x} {p^2 - q^2 \sin^2 a x}\) | \(=\) | \(\ds \int \frac {\sec^2 a x \rd x} {p^2 \sec^2 a x - q^2 \tan^2 a x}\) | multiplying numerator and denominator by $\sec^2 a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\sec^2 a x \rd x} {p^2 \paren {1 + \tan^2 a x} - q^2 \tan^2 a x}\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\sec^2 a x \rd x} {p^2 + \paren {p^2 - q^2} \tan^2 a x}\) | simplifying |
Let:
\(\ds t\) | \(=\) | \(\ds \tan a x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \d t\) | \(=\) | \(\ds a \sec^2 a x \rd x\) | Derivative of Tangent Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {p^2 - q^2 \sin^2 a x}\) | \(=\) | \(\ds \frac 1 a \int \frac {\d t} {p^2 + \paren {p^2 - q^2} t^2}\) | Integration by Substitution: substituting $t = \tan a x$ | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 + \tfrac {p^2} {p^2 - q^2} }\) | rearranging into a standard form |
There are two cases to address.
First, suppose $p^2 > q^2$.
Then we have that $p^2 - q^2 > 0$, and so:
\(\ds \int \frac {\d x} {p^2 - q^2 \sin^2 a x}\) | \(=\) | \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 + \paren {\tfrac p {\sqrt {p^2 - q^2} } }^2}\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a \paren {p^2 - q^2} } \frac {\sqrt {p^2 - q^2} } p \map \arctan {\frac {\sqrt {p^2 - q^2} } p t} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a p \sqrt {p^2 - q^2} } \map \arctan {\frac {\sqrt {p^2 - q^2} } p t} + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a p \sqrt {p^2 - q^2} } \map \arctan {\frac {\sqrt {p^2 - q^2} \tan a x} p} + C\) | substituting $t = \tan a x$ |
Now suppose $p^2 < q^2$.
Then we have that $p^2 - q^2 < 0$, and so:
\(\ds \int \frac {\d x} {p^2 - q^2 \sin^2 a x}\) | \(=\) | \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 - \tfrac {p^2} {q^2 - p^2} }\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 - \paren {\tfrac p {\sqrt {q^2 - p^2} } }^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {a \paren {q^2 - p^2} } \frac {\sqrt {q^2 - p^2} } {2 p} \ln \size {\frac {t - \tfrac p {\sqrt {q^2 - p^2} } } {t + \tfrac p {\sqrt {q^2 - p^2} } } } + C\) | Primitive of $\dfrac 1 {x^2 - a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {\sqrt {q^2 - p^2} t - p} {\sqrt {q^2 - p^2} t + p} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {\sqrt {q^2 - p^2} t + p} {\sqrt {q^2 - p^2} t - p} } + C\) | Logarithm of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {\sqrt {q^2 - p^2} \tan a x + p} {\sqrt {q^2 - p^2} \tan a x - p} } + C\) | substituting $t = \tan a x$ |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$: $14.363$