Primitive of x by Cosine of x

Theorem

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

where $C$ is an arbitrary constant.

Proof 1

From Primitive of $x \cos a x$:

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

The result follows on setting $a = 1$.

$\blacksquare$

Proof 2

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

$\blacksquare$

Proof 3

We have:

\(\ds \map {\dfrac \d {\d x} } {\cos x + x \sin x}\)

\(=\)

\(\ds \map {\dfrac \d {\d x} } {\cos x} + \map {\dfrac \d {\d x} } {x \sin x}\)

Sum Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \map {\dfrac \d {\d x} } {\cos x} + x \map {\dfrac \d {\d x} } {\sin x} + \map {\dfrac \d {\d x} } x \sin x\)

Product Rule for Derivatives

\(\ds \)

\(=\)

\(\ds -\sin x + x \cos x + \sin x\)

Derivative of Cosine Function, Derivative of Sine Function, Derivative of Identity Function

\(\ds \)

\(=\)

\(\ds x \cos x\)

simplifying

Hence the result by definition of primitive.

$\blacksquare$

Warning

It is a mistake to use Primitive of Constant Multiple of Function here:

$\ds \int x \cos x \rd x = x \int \cos x \rd x$

and thence to deduce:

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

because $x$ is not a constant in the expression $x \cos x$.