Primitive of x squared by Cosine of a x

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Theorem

$\ds \int x^2 \cos a x \rd x = \frac {2 x \cos a x} {a^2} + \paren {\frac {x^2} a - \frac 2 {a^3} } \sin a x + C$

where $C$ is an arbitrary constant.


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:

\(\ds u\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 2 x\) Derivative of Power


and let:

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


Then:

\(\ds \int x^2 \map \cos {a x} \rd x\) \(=\) \(\ds x^2 \paren {\frac {\sin a x} a} - \int \paren {2 x \frac {\sin a x} a} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^2 \sin a x} a - \frac 2 a \int x \sin a x \rd x + C\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac {x^2 \sin a x} a - \frac 2 a \paren {\frac {\sin a x} {a^2} - \frac {x \cos a x} a} + C\) Primitive of $x \sin a x$
\(\ds \) \(=\) \(\ds \frac {2 x \cos a x} {a^2} + \paren {\frac {x^2} a - \frac 2 {a^3} } \sin a x + C\) simplification

$\blacksquare$


Also see


Sources

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