Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x} = \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C$


Proof 1

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 $\paren 3$
\(\ds \) \(=\) \(\ds \paren {\frac 2 {2 a p q} } \map \arctan {\sqrt{ \frac {p^2} {q^2} } \tan a x} + C\)
\(\ds \) \(=\) \(\ds \paren {\frac 1 {a p q} } \map \arctan {\frac p q \tan a x} + C\)
\(\ds \) \(=\) \(\ds \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C\)

$\blacksquare$


Proof 2

\(\ds \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x}\) \(=\) \(\ds \int \frac {\sec^2 a x \d x} {p^2 \tan^2 a x + q^2}\) 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 $t$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_p_squared_by_square_of_Sine_of_a_x_plus_q_squared_by_square_of_Cosine_of_a_x&oldid=544749"