Primitive of Reciprocal of p plus q by Sine of a x/Examples/3 + 2 sin x
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Example of Use of Primitive of $\dfrac 1 {p + q \sin a x}$
- $\ds \int \dfrac {\d x} {3 + 2 \sin x} = \frac 2 {\sqrt 5} \map \arctan {\frac {3 \tan \dfrac x 2 + 2} {\sqrt 5} } + C$
Proof
From Primitive of $\dfrac 1 {p + q \sin a x}$:
- $\ds \int \frac {\d x} {p + q \sin a x} = \begin{cases}
\ds \frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\frac {p \tan \dfrac {a x} 2 + q} {\sqrt {p^2 - q^2} } } + C & : q^2 - p^2 < 0 \\ \ds \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\frac {p \tan \dfrac {a x} 2 + q - \sqrt {p^2 - q^2} } {p \tan \dfrac {a x} 2 + q + \sqrt {p^2 - q^2} } } + C & : q^2 - p^2 > 0 \\ \end{cases}$
The result follows on setting $a = 1$, $p \gets 3$ and $q \gets 2$, as follows.
We have that $2^2 - 3^2 < 0$, and so:
\(\ds \int \dfrac {\d x} {3 + 2 \sin x}\) | \(=\) | \(\ds \frac 2 {\sqrt {3^2 - 2^2} } \map \arctan {\frac {3 \tan \dfrac x 2 + 2} {\sqrt {3^2 - 2^2} } } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt 5} \map \arctan {\frac {3 \tan \dfrac x 2 + 2} {\sqrt 5} } + C\) | simplifying |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $29$.