Primitive of x cubed by Sine of a x

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Theorem

$\ds \int x^3 \sin a x \rd x = \paren {\frac {3 x^2} {a^2} - \frac 6 {a^4} } \sin a x + \paren {\frac {6 x} {a^3} - \frac {x^3} a} \cos a x + C$

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:

\(\ds u\) \(=\) \(\ds x^3\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 3 x^2\) Derivative of Power


and let:

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


Then:

\(\ds \int x^3 \sin a x \rd x\) \(=\) \(\ds x^3 \paren {-\frac {\cos a x} a} - \int 3 x^2 \paren {-\frac {\cos a x} a} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds -\frac {x^3} a \cos a x + \frac 3 a \int x^2 \cos a x \rd x + C\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds -\frac {x^3} a \cos a x + \frac 3 a \paren {\frac {2 x} {a^2} \cos a x + \paren {\frac {x^2} a - \frac 2 {a^3} } \sin a x} + C\) Primitive of $x^2 \cos a x$
\(\ds \) \(=\) \(\ds \paren {\frac {3 x^2} {a^2} - \frac 6 {a^4} } \sin a x + \paren {\frac {6 x} {a^3} - \frac {x^3} a} \cos a x + C\) simplifying

$\blacksquare$


Also see


Sources

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