Primitive of x by Sine of a x

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Theorem

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

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:

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


and let:

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


Then:

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

$\blacksquare$


Also see


Sources