Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x/Proof 2
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Theorem
- $\ds \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x} = \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C$
Proof
| \(\ds \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x}\) | \(=\) | \(\ds \int \frac {\sec^2 a x \d x} {p^2 \tan^2 a x + q^2}\) | multiplying by $\dfrac {\sec^2 a x} {\sec^2 a x}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\d t} {p^2 t^2 + q^2}\) | substituting $t = \tan a x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {a p^2} \int \frac {\d t} {t^2 + \paren {\frac q p}^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {a p^2} \times \frac p q \map \arctan {\frac {p t} q} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C\) | substituting back for $t$ |
$\blacksquare$