Primitive of x by Cosine of x/Proof 2

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Theorem

$\ds \int x \cos x \rd x = \cos x + x \sin x + C$


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:

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


and let:

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


Then:

\(\ds \int x \cos x \rd x\) \(=\) \(\ds x \sin x - \int \paren {\sin x} \times 1 \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x \sin x - \int \sin x \rd x + C\) simplifying
\(\ds \) \(=\) \(\ds x \sin x - \paren {-\cos x} + C\) Primitive of $\sin x$
\(\ds \) \(=\) \(\ds \cos x + x \sin x + C\) rearranging

$\blacksquare$


Sources

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