Primitive of x by Cosine of a x

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

$\blacksquare$