Primitive of Reciprocal of p squared by square of Sine of a x minus q squared by square of Cosine of a x
Jump to navigation
Jump to search
Theorem
- $\ds \int \frac {\d x} {p^2 \sin^2 a x - q^2 \cos^2 a x} = \frac 1 {2 a p q} \ln \size {\frac {p \tan a x - q} {p \tan a x + q} } + C$
Proof
\(\ds \int \frac {\d x} {p^2 \sin^2 a x - q^2 \cos^2 a x}\) | \(=\) | \(\ds \int \frac {\sec^2 a x} {p^2 \tan^2 a x - q^2} \rd x\) | multiplying by $\dfrac {\sec^2 a x} {\sec^2 a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac 1 {p^2 t^2 - q^2} \rd t\) | substituting $t = \tan a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a p^2} \int \frac 1 {t^2 - \paren {\frac q p}^2} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a \frac {p^2} p q} \ln \size {\frac {t - \frac q p} {t + \frac q p} } + C\) | Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a p q} \ln \size {\frac {p t - q} {p t + q} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a p q} \ln \size {\frac {p \tan a x - q} {p \tan a x + q} } + C\) | substituting back for $t$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.424$