Primitive of Reciprocal of p squared plus square of q by Sine of a x
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Theorem
- $\ds \int \frac {\d x} {p^2 + q^2 \sin^2 a x} = \frac 1 {a p \sqrt {p^2 + q^2} } \arctan \frac {\sqrt {p^2 + q^2} \tan a x} p + C$
where $C$ is an arbitrary constant.
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 $\sec^2 a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\sec^2 a x \rd x} {p^2 + \paren {p^2 + q^2} \tan^2 a x}\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\paren {\tan a x}' \rd x} {a p^2 + a \paren {p^2 + q^2} \tan^2 a x}\) | Derivative of Tangent Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d t} {a p^2 + a \paren {p^2 + q^2} t^2}\) | substituting $t = \tan a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a \paren {p^2 + q^2} } \int \frac {\d t} {\paren {\tfrac p {\sqrt {p^2 + q^2} } }^2 + t^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a \paren {p^2 + q^2} } \frac {\sqrt {p^2 + q^2} } p \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} \tan a x} p} + C\) | 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 $\sin a x$: $14.362$