Evaluate integral of cos( natural log of x) with respect to x

$\int \cos (\ln (x)) d x$

Let $u = \ln (x)$ . Then $d u = \frac{1}{x} d x$ , so $x d u = d x$ . Rewrite using $u$ and $d$ $u$ .

$\int e^{u} \cos (u) d u$

Reorder $e^{u}$ and $\cos (u)$ .

$\int \cos (u) e^{u} d u$

Integrate by parts using the formula $\int w d v = w v - \int v d w$ , where $w = \cos (u)$ and $dv = e^{u}$ .

$\cos (u) e^{u} - \int e^{u} (- \sin (u)) d u$

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$\cos (u) e^{u} - - \int e^{u} (\sin (u)) d u$

Simplify the expression.

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Multiply $- 1$ by $- 1$ .

$\cos (u) e^{u} + 1 \int e^{u} (\sin (u)) d u$

Multiply $\int e^{u} (\sin (u)) d u$ by $1$ .

$\cos (u) e^{u} + \int e^{u} (\sin (u)) d u$

Reorder $e^{u}$ and $\sin (u)$ .

$\cos (u) e^{u} + \int \sin (u) e^{u} d u$

Integrate by parts using the formula $\int w d v = w v - \int v d w$ , where $w = \sin (u)$ and $dv = e^{u}$ .

$\cos (u) e^{u} + \sin (u) e^{u} - \int e^{u} \cos (u) d u$

Solving for $\int e^{u} \cos (u) d u$ , we find that $\int e^{u} \cos (u) d u$ = $\frac{\cos (u) e^{u} + \sin (u) e^{u}}{2}$ .

$\frac{\cos (u) e^{u} + \sin (u) e^{u}}{2} + C$

Replace all occurrences of $u$ with $\ln (x)$ .

$\frac{\cos (\ln (x)) e^{\ln (x)} + \sin (\ln (x)) e^{\ln (x)}}{2} + C$

Simplify the answer.

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Exponentiation and log are inverse functions.

$\frac{\cos (\ln (x)) x + \sin (\ln (x)) e^{\ln (x)}}{2} + C$

Exponentiation and log are inverse functions.

$\frac{\cos (\ln (x)) x + \sin (\ln (x)) x}{2} + C$

Evaluate integral of cos( natural log of x) with respect to x

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