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