Primitive of x over 1 plus Cosine of a x

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Theorem

$\ds \int \frac {x \rd x} {1 + \cos a x} = \frac x a \tan \frac {a x} 2 + \frac 2 {a^2} \ln \size {\cos \frac {a x} 2} + C$


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:

\(\ds u\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 1\) Primitive of Power


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {1 + \cos a x}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac 1 a \tan \frac {a x} 2\) Primitive of $\dfrac 1 {1 + \cos a x}$


Then:

\(\ds \int \frac {x \rd x} {1 + \cos a x}\) \(=\) \(\ds \int u \rd v\)
\(\ds \) \(=\) \(\ds x \paren {\frac 1 a \tan \frac {a x} 2} - \int \paren {\frac 1 a \tan \frac {a x} 2} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac x a \tan \frac {a x} 2 - \frac 1 a \paren {\frac {-2} a \ln \size {\cos a x} } + C\) Primitive of $\tan a x$
\(\ds \) \(=\) \(\ds \frac x a \tan \frac {a x} 2 + \frac 2 {a^2} \ln \size {\cos \frac {a x} 2} + C\) simplifying

$\blacksquare$


Also see


Sources

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