Primitive of x over 1 plus Sine of a x
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Theorem
- $\ds \int \frac {x \rd x} {1 + \sin a x} = -\frac x a \map \tan {\frac \pi 4 - \frac {a x} 2} + \frac 2 {a^2} \ln \size {\map \sin {\frac \pi 4 + \frac {a x} 2} } + C$
Proof
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds 1\) | Derivative of Identity Function |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \frac 1 {1 + \sin a x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds -\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2}\) | Primitive of $\dfrac 1 {1 + \sin a x}$ |
Then:
\(\ds \int \frac {x \rd x} {1 + \sin a x}\) | \(=\) | \(\ds x \paren {-\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2} } - \int \paren {-\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2} } \times 1 \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac x a \map \tan {\frac \pi 4 - \frac {a x} 2} + \frac 1 a \int \map \tan {\frac \pi 4 - \frac {a x} 2} \rd x + C\) | simplifying |
Then:
\(\ds z\) | \(=\) | \(\ds \frac \pi 4 - \frac {a x} 2\) | ||||||||||||
\(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds -\frac a 2\) | Derivative of Power | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 a \int \map \tan {\frac \pi 4 - \frac {a x} 2} \rd x\) | \(=\) | \(\ds -\frac 1 a \int \frac 2 a \tan z \rd z\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {-2} {a^2} \ln \size {\cos z} + C\) | Primitive of $\tan z$: Cosine Form | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \ln \size {\map \cos {\frac \pi 4 - \frac {a x} 2} } + C\) | substituting back for $z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \ln \size {\map \sin {\frac \pi 2 - \paren {\frac \pi 4 - \frac {a x} 2} } } + C\) | Sine of Complement equals Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \ln \size {\map \sin {\frac \pi 4 + \frac {a x} 2} } + C\) |
Putting it all together:
- $\ds \int \frac {x \rd x} {1 - \sin a x} = -\frac x a \map \tan {\frac \pi 4 - \frac {a x} 2} + \frac 2 {a^2} \ln \size {\map \sin {\frac \pi 4 + \frac {a x} 2} } + C$
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$: $14.357$