Integral of $$$\frac{i n t^{2} u}{2} + \cos{\left(x \right)}$$$ with respect to $$$x$$$
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Find $$$\int \left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{\frac{i n t^{2} u}{2} d x} + \int{\cos{\left(x \right)} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=\frac{i n t^{2} u}{2}$$$:
$$\int{\cos{\left(x \right)} d x} + {\color{red}{\int{\frac{i n t^{2} u}{2} d x}}} = \int{\cos{\left(x \right)} d x} + {\color{red}{\left(\frac{i n t^{2} u x}{2}\right)}}$$
The integral of the cosine is $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$:
$$\frac{i n t^{2} u x}{2} + {\color{red}{\int{\cos{\left(x \right)} d x}}} = \frac{i n t^{2} u x}{2} + {\color{red}{\sin{\left(x \right)}}}$$
Therefore,
$$\int{\left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)d x} = \frac{i n t^{2} u x}{2} + \sin{\left(x \right)}$$
Add the constant of integration:
$$\int{\left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)d x} = \frac{i n t^{2} u x}{2} + \sin{\left(x \right)}+C$$
Answer
$$$\int \left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)\, dx = \left(\frac{i n t^{2} u x}{2} + \sin{\left(x \right)}\right) + C$$$A