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
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): integration by parts
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): integration by parts