Integral of $$$\cos{\left(\ln\left(2 x\right) \right)}$$$

Find $$$\int \cos{\left(\ln\left(2 x\right) \right)}\, dx$$$.

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

Thus,

$${\color{red}{\int{\cos{\left(\ln{\left(2 x \right)} \right)} d x}}} = {\color{red}{\int{\frac{\cos{\left(\ln{\left(u \right)} \right)}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos{\left(\ln{\left(u \right)} \right)}$$$:

$${\color{red}{\int{\frac{\cos{\left(\ln{\left(u \right)} \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}{2}\right)}}$$

For the integral $$$\int{\cos{\left(\ln{\left(u \right)} \right)} d u}$$$, use integration by parts $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$.

Let $$$\operatorname{\omega}=\cos{\left(\ln{\left(u \right)} \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{d\omega}=\left(\cos{\left(\ln{\left(u \right)} \right)}\right)^{\prime }du=- \frac{\sin{\left(\ln{\left(u \right)} \right)}}{u} du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

The integral can be rewritten as

$$\frac{{\color{red}{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}}}{2}=\frac{{\color{red}{\left(\cos{\left(\ln{\left(u \right)} \right)} \cdot u-\int{u \cdot \left(- \frac{\sin{\left(\ln{\left(u \right)} \right)}}{u}\right) d u}\right)}}}{2}=\frac{{\color{red}{\left(u \cos{\left(\ln{\left(u \right)} \right)} - \int{\left(- \sin{\left(\ln{\left(u \right)} \right)}\right)d u}\right)}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = \sin{\left(\ln{\left(u \right)} \right)}$$$:

$$\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} - \frac{{\color{red}{\int{\left(- \sin{\left(\ln{\left(u \right)} \right)}\right)d u}}}}{2} = \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} - \frac{{\color{red}{\left(- \int{\sin{\left(\ln{\left(u \right)} \right)} d u}\right)}}}{2}$$

For the integral $$$\int{\sin{\left(\ln{\left(u \right)} \right)} d u}$$$, use integration by parts $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$.

Let $$$\operatorname{\omega}=\sin{\left(\ln{\left(u \right)} \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{d\omega}=\left(\sin{\left(\ln{\left(u \right)} \right)}\right)^{\prime }du=\frac{\cos{\left(\ln{\left(u \right)} \right)}}{u} du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

Therefore,

$$\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} + \frac{{\color{red}{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}}}{2}=\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} + \frac{{\color{red}{\left(\sin{\left(\ln{\left(u \right)} \right)} \cdot u-\int{u \cdot \frac{\cos{\left(\ln{\left(u \right)} \right)}}{u} d u}\right)}}}{2}=\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} + \frac{{\color{red}{\left(u \sin{\left(\ln{\left(u \right)} \right)} - \int{\cos{\left(\ln{\left(u \right)} \right)} d u}\right)}}}{2}$$

We've arrived to an integral that we already saw.

Thus, we've obtained the following simple equation with respect to the integral:

$$\frac{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}{2} = \frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} + \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} - \frac{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}{2}$$

Solving it, we get that

$$\int{\cos{\left(\ln{\left(u \right)} \right)} d u} = \frac{u \left(\sin{\left(\ln{\left(u \right)} \right)} + \cos{\left(\ln{\left(u \right)} \right)}\right)}{2}$$

Thus,

$$\frac{{\color{red}{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}}}{2} = \frac{{\color{red}{\left(\frac{u \left(\sin{\left(\ln{\left(u \right)} \right)} + \cos{\left(\ln{\left(u \right)} \right)}\right)}{2}\right)}}}{2}$$

Recall that $$$u=2 x$$$:

$$\frac{{\color{red}{u}} \left(\sin{\left(\ln{\left({\color{red}{u}} \right)} \right)} + \cos{\left(\ln{\left({\color{red}{u}} \right)} \right)}\right)}{4} = \frac{{\color{red}{\left(2 x\right)}} \left(\sin{\left(\ln{\left({\color{red}{\left(2 x\right)}} \right)} \right)} + \cos{\left(\ln{\left({\color{red}{\left(2 x\right)}} \right)} \right)}\right)}{4}$$

Therefore,

$$\int{\cos{\left(\ln{\left(2 x \right)} \right)} d x} = \frac{x \left(\sin{\left(\ln{\left(2 x \right)} \right)} + \cos{\left(\ln{\left(2 x \right)} \right)}\right)}{2}$$

Simplify:

$$\int{\cos{\left(\ln{\left(2 x \right)} \right)} d x} = \frac{\sqrt{2} x \sin{\left(\ln{\left(x \right)} + \ln{\left(2 \right)} + \frac{\pi}{4} \right)}}{2}$$

Add the constant of integration:

$$\int{\cos{\left(\ln{\left(2 x \right)} \right)} d x} = \frac{\sqrt{2} x \sin{\left(\ln{\left(x \right)} + \ln{\left(2 \right)} + \frac{\pi}{4} \right)}}{2}+C$$

$$$\int \cos{\left(\ln\left(2 x\right) \right)}\, dx = \frac{\sqrt{2} x \sin{\left(\ln\left(x\right) + \ln\left(2\right) + \frac{\pi}{4} \right)}}{2} + C$$$A