Primitive of Reciprocal of p plus q by Tangent of a x/Proof 2
Jump to navigation
Jump to search
Theorem
- $\ds \int \frac {\d x} {p + q \tan a x} = \frac {p x} {p^2 + q^2} + \frac q {a \paren {p^2 + q^2} } \ln \size {q \sin a x + p \cos a x} + C$
Proof
We have:
- $\dfrac \d {\d x} \paren {q \sin a x + p \cos a x} = a q \cos a x - a p \sin a x$
Thus:
\(\ds \int \frac {\d x} {p + q \tan a x}\) | \(=\) | \(\ds \int \frac {\d x} {p + q \dfrac {\sin a x} {\cos a x} }\) | Tangent is Sine divided by Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\cos a x \rd x} {p \cos a x + q \sin a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \int \frac {\paren {p^2 + q^2} \cos a x \rd x} {p \cos a x + q \sin a x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \int \frac {p^2 \cos a x + p q \sin a x - p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \paren {\int \frac {p^2 \cos a x + p q \sin a x} {p \cos a x + q \sin a x} \rd x + \int \frac {-p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2 + q^2} \paren {\int p \rd x + \frac q a \int \frac {\map \d {p \cos a x + q \sin a x} } {p \cos a x + q \sin a x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p x} {p^2 + q^2} + \frac q {a \paren {p^2 + q^2} } \ln \size {p \cos a x + q \sin a x} + C\) | Primitive of Constant and Primitive of Reciprocal |
$\blacksquare$