$$$\frac{i n t^{2} u}{2} + \cos{\left(x \right)}$$$ 對 $$$x$$$ 的積分
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求$$$\int \left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{\frac{i n t^{2} u}{2} d x} + \int{\cos{\left(x \right)} d x}\right)}}$$
配合 $$$c=\frac{i n t^{2} u}{2}$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$\int{\cos{\left(x \right)} d x} + {\color{red}{\int{\frac{i n t^{2} u}{2} d x}}} = \int{\cos{\left(x \right)} d x} + {\color{red}{\left(\frac{i n t^{2} u x}{2}\right)}}$$
餘弦函數的積分為 $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$：
$$\frac{i n t^{2} u x}{2} + {\color{red}{\int{\cos{\left(x \right)} d x}}} = \frac{i n t^{2} u x}{2} + {\color{red}{\sin{\left(x \right)}}}$$
因此，
$$\int{\left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)d x} = \frac{i n t^{2} u x}{2} + \sin{\left(x \right)}$$
加上積分常數：
$$\int{\left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)d x} = \frac{i n t^{2} u x}{2} + \sin{\left(x \right)}+C$$
答案
$$$\int \left(\frac{i n t^{2} u}{2} + \cos{\left(x \right)}\right)\, dx = \left(\frac{i n t^{2} u x}{2} + \sin{\left(x \right)}\right) + C$$$A