Integral of $$$- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}$$$ with respect to $$$x$$$
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Find $$$\int \left(- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{x}{2} d x} - \int{\sin{\left(e x \right)} d x} + \int{\cos{\left(x_{1} \right)} d x}\right)}}$$
Let $$$u=e x$$$.
Then $$$du=\left(e x\right)^{\prime }dx = e dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{e}$$$.
The integral becomes
$$- \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} - {\color{red}{\int{\sin{\left(e x \right)} d x}}} = - \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} - {\color{red}{\int{\frac{\sin{\left(u \right)}}{e} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=e^{-1}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$- \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} - {\color{red}{\int{\frac{\sin{\left(u \right)}}{e} d u}}} = - \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} - {\color{red}{\frac{\int{\sin{\left(u \right)} d u}}{e}}}$$
The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$- \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{e} = - \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{e}$$
Recall that $$$u=e x$$$:
$$- \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} + \frac{\cos{\left({\color{red}{u}} \right)}}{e} = - \int{\frac{x}{2} d x} + \int{\cos{\left(x_{1} \right)} d x} + \frac{\cos{\left({\color{red}{e x}} \right)}}{e}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = x$$$:
$$\frac{\cos{\left(e x \right)}}{e} + \int{\cos{\left(x_{1} \right)} d x} - {\color{red}{\int{\frac{x}{2} d x}}} = \frac{\cos{\left(e x \right)}}{e} + \int{\cos{\left(x_{1} \right)} d x} - {\color{red}{\left(\frac{\int{x d x}}{2}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$\frac{\cos{\left(e x \right)}}{e} + \int{\cos{\left(x_{1} \right)} d x} - \frac{{\color{red}{\int{x d x}}}}{2}=\frac{\cos{\left(e x \right)}}{e} + \int{\cos{\left(x_{1} \right)} d x} - \frac{{\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{2}=\frac{\cos{\left(e x \right)}}{e} + \int{\cos{\left(x_{1} \right)} d x} - \frac{{\color{red}{\left(\frac{x^{2}}{2}\right)}}}{2}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=\cos{\left(x_{1} \right)}$$$:
$$- \frac{x^{2}}{4} + \frac{\cos{\left(e x \right)}}{e} + {\color{red}{\int{\cos{\left(x_{1} \right)} d x}}} = - \frac{x^{2}}{4} + \frac{\cos{\left(e x \right)}}{e} + {\color{red}{x \cos{\left(x_{1} \right)}}}$$
Therefore,
$$\int{\left(- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}\right)d x} = - \frac{x^{2}}{4} + x \cos{\left(x_{1} \right)} + \frac{\cos{\left(e x \right)}}{e}$$
Add the constant of integration:
$$\int{\left(- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}\right)d x} = - \frac{x^{2}}{4} + x \cos{\left(x_{1} \right)} + \frac{\cos{\left(e x \right)}}{e}+C$$
Answer
$$$\int \left(- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}\right)\, dx = \left(- \frac{x^{2}}{4} + x \cos{\left(x_{1} \right)} + \frac{\cos{\left(e x \right)}}{e}\right) + C$$$A