Primitive of x by Cosine of a x

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Theorem

$\displaystyle \int x \cos a x \rd x = \frac {\cos a x} {a^2} + \frac {x \sin a x} a + C$

where $C$ is an arbitrary constant.


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \cos a x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {\sin a x} a\) Primitive of $\cos a x$


Then:

\(\displaystyle \int x \cos \left({a x}\right) \rd x\) \(=\) \(\displaystyle x \left({\frac {\sin a x} a}\right) - \int \left({\frac {\sin a x} a}\right) \times 1 \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \sin a x} a - \frac 1 a \int \sin a x \rd x + C\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \sin a x} a - \frac 1 a \left({\frac {-\cos a x} a}\right) + C\) Primitive of $\sin a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\cos a x} {a^2} + \frac {x \sin a x} a + C\) simplification

$\blacksquare$


Also see


Sources