$$$- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}$$$の積分

$$$\int \left(- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{2} d x} - \int{\sin{\left(x \right)} d x} + \int{\cos{\left(x \right)} d x}\right)}}$$

$$$c=\frac{1}{2}$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$- \int{\sin{\left(x \right)} d x} + \int{\cos{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{2} d x}}} = - \int{\sin{\left(x \right)} d x} + \int{\cos{\left(x \right)} d x} + {\color{red}{\left(\frac{x}{2}\right)}}$$

正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:

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

余弦の積分は$$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$:

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

したがって、

$$\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}\right)d x} = \frac{x}{2} + \sin{\left(x \right)} + \cos{\left(x \right)}$$

簡単化せよ:

$$\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}\right)d x} = \frac{x}{2} + \sqrt{2} \sin{\left(x + \frac{\pi}{4} \right)}$$

積分定数を加える:

$$\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}\right)d x} = \frac{x}{2} + \sqrt{2} \sin{\left(x + \frac{\pi}{4} \right)}+C$$

$$$\int \left(- \sin{\left(x \right)} + \cos{\left(x \right)} + \frac{1}{2}\right)\, dx = \left(\frac{x}{2} + \sqrt{2} \sin{\left(x + \frac{\pi}{4} \right)}\right) + C$$$A