Integral of $$$x e^{3} \cos{\left(x e^{3} \right)}$$$
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Find $$$\int x e^{3} \cos{\left(x e^{3} \right)}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=e^{3}$$$ and $$$f{\left(x \right)} = x \cos{\left(x e^{3} \right)}$$$:
$${\color{red}{\int{x e^{3} \cos{\left(x e^{3} \right)} d x}}} = {\color{red}{e^{3} \int{x \cos{\left(x e^{3} \right)} d x}}}$$
For the integral $$$\int{x \cos{\left(x e^{3} \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=\cos{\left(x e^{3} \right)} dx$$$.
Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\cos{\left(x e^{3} \right)} d x}=\frac{\sin{\left(x e^{3} \right)}}{e^{3}}$$$ (steps can be seen »).
The integral can be rewritten as
$$e^{3} {\color{red}{\int{x \cos{\left(x e^{3} \right)} d x}}}=e^{3} {\color{red}{\left(x \cdot \frac{\sin{\left(x e^{3} \right)}}{e^{3}}-\int{\frac{\sin{\left(x e^{3} \right)}}{e^{3}} \cdot 1 d x}\right)}}=e^{3} {\color{red}{\left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \int{\frac{\sin{\left(x e^{3} \right)}}{e^{3}} d x}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=e^{-3}$$$ and $$$f{\left(x \right)} = \sin{\left(x e^{3} \right)}$$$:
$$e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - {\color{red}{\int{\frac{\sin{\left(x e^{3} \right)}}{e^{3}} d x}}}\right) = e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - {\color{red}{\frac{\int{\sin{\left(x e^{3} \right)} d x}}{e^{3}}}}\right)$$
Let $$$u=x e^{3}$$$.
Then $$$du=\left(x e^{3}\right)^{\prime }dx = e^{3} dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{e^{3}}$$$.
The integral becomes
$$e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \frac{{\color{red}{\int{\sin{\left(x e^{3} \right)} d x}}}}{e^{3}}\right) = e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{e^{3}} d u}}}}{e^{3}}\right)$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=e^{-3}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{e^{3}} d u}}}}{e^{3}}\right) = e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \frac{{\color{red}{\frac{\int{\sin{\left(u \right)} d u}}{e^{3}}}}}{e^{3}}\right)$$
The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{e^{6}}\right) = e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{e^{6}}\right)$$
Recall that $$$u=x e^{3}$$$:
$$e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} + \frac{\cos{\left({\color{red}{u}} \right)}}{e^{6}}\right) = e^{3} \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} + \frac{\cos{\left({\color{red}{x e^{3}}} \right)}}{e^{6}}\right)$$
Therefore,
$$\int{x e^{3} \cos{\left(x e^{3} \right)} d x} = \left(\frac{x \sin{\left(x e^{3} \right)}}{e^{3}} + \frac{\cos{\left(x e^{3} \right)}}{e^{6}}\right) e^{3}$$
Simplify:
$$\int{x e^{3} \cos{\left(x e^{3} \right)} d x} = x \sin{\left(x e^{3} \right)} + \frac{\cos{\left(x e^{3} \right)}}{e^{3}}$$
Add the constant of integration:
$$\int{x e^{3} \cos{\left(x e^{3} \right)} d x} = x \sin{\left(x e^{3} \right)} + \frac{\cos{\left(x e^{3} \right)}}{e^{3}}+C$$
Answer
$$$\int x e^{3} \cos{\left(x e^{3} \right)}\, dx = \left(x \sin{\left(x e^{3} \right)} + \frac{\cos{\left(x e^{3} \right)}}{e^{3}}\right) + C$$$A