Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x/Proof 1
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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
Let $u = p^2 + q^2$ and $v = q^2 - p^2$.
Then:
\(\text {(1)}: \quad\) | \(\ds u + v\) | \(=\) | \(\ds 2 q^2\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds u - v\) | \(=\) | \(\ds 2 p ^2\) |
Also:
\(\ds u^2 - v^2\) | \(=\) | \(\ds \paren {u + v} \paren {u - v}\) | ||||||||||||
\(\ds u^2 - v^2\) | \(=\) | \(\ds \paren {2 q^2} \paren {2 p^2}\) | from $\paren 1$ and $\paren 2$ | |||||||||||
\(\text {(3)}: \quad\) | \(\ds u^2 - v^2\) | \(=\) | \(\ds 4 p^2 q^2\) |
Therefore:
\(\ds \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x}\) | \(=\) | \(\ds \int \frac {\d x} {p^2 \paren {\frac {1 - \cos 2 a x} 2} + q^2 \cos^2 a x}\) | Square of Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d x} {p^2 \paren {\frac {1 - \cos 2 a x} 2} + q^2 \paren {\frac {1 + \cos 2 a x} 2} }\) | Square of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {2 \d x} {p^2 - p^2 \cos 2 a x + q^2 + q^2 \cos 2 a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {2 \d x} {p^2 + q^2 + \paren {q^2 - p^2} \cos 2 a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren { \frac 2 {\paren {2 a} \sqrt{u^2 - v^2} } } \map \arctan {\sqrt{ \frac {u - v} {u + v} } \tan \frac {\paren {2 a} x} 2} + C\) | Primitive of $\dfrac 1 {p + q \cos a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {\frac 1 {a \sqrt{4 p^2 q^2} } } \map \arctan {\sqrt{ \frac {2 p^2} {2 q^2} } \tan a x} + C\) | from $\paren 3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac 2 {2 a p q} } \map \arctan {\sqrt{ \frac {p^2} {q^2} } \tan a x} + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac 1 {a p q} } \map \arctan {\frac p q \tan a x} + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C\) |
$\blacksquare$