$$$\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}$$$의 적분

$$$\int \left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}\, dx$$$을(를) 구하시오.

피적분함수를 단순화하세요.:

$${\color{red}{\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2} d x}}} = {\color{red}{\int{\frac{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}}{4} d x}}}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{1}{4}$$$와 $$$f{\left(x \right)} = \left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}$$$에 적용하세요:

$${\color{red}{\int{\frac{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}}{4} d x}}} = {\color{red}{\left(\frac{\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2} d x}}{4}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{\left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2} d x}}}}{4} = \frac{{\color{red}{\int{\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} \cos{\left(x \right)} + \cos^{2}{\left(x \right)}\right)d x}}}}{4}$$

각 항별로 적분하십시오:

$$\frac{{\color{red}{\int{\left(\sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} \cos{\left(x \right)} + \cos^{2}{\left(x \right)}\right)d x}}}}{4} = \frac{{\color{red}{\left(- \int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x} + \int{\sin^{2}{\left(x \right)} d x} + \int{\cos^{2}{\left(x \right)} d x}\right)}}}{4}$$

멱 감소 공식 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$를 $$$\alpha=x$$$에 적용하세요:

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

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$에 적용하세요:

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

각 항별로 적분하십시오:

$$- \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{8} = - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$

상수 법칙 $$$\int c\, dx = c x$$$을 $$$c=1$$$에 적용하십시오:

$$- \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{\int{1 d x}}}}{8} = - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{x}}}{8}$$

$$$u=2 x$$$라 하자.

그러면 $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dx = \frac{du}{2}$$$임을 얻습니다.

따라서,

$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8}$$

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$에 적용하세요:

$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{8} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{8}$$

코사인의 적분은 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{16} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{16}$$

다음 $$$u=2 x$$$을 기억하라:

$$\frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{16} = \frac{x}{8} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{\int{\sin^{2}{\left(x \right)} d x}}{4} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{16}$$

멱 감소 공식 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$를 $$$\alpha=x$$$에 적용하세요:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin^{2}{\left(x \right)} d x}}}}{4} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}}{4}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(x \right)} = 1 - \cos{\left(2 x \right)}$$$에 적용하세요:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}}{4} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}{2}\right)}}}{4}$$

각 항별로 적분하십시오:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\int{1 d x} - \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$

상수 법칙 $$$\int c\, dx = c x$$$을 $$$c=1$$$에 적용하십시오:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{\int{1 d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{x}}}{8}$$

이미 계산된 적분 $$$\int{\cos{\left(2 x \right)} d x}$$$:

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

따라서,

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{8}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=2$$$와 $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$$$에 적용하세요:

$$\frac{x}{4} - \frac{{\color{red}{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\left(2 \int{\sin{\left(x \right)} \cos{\left(x \right)} d x}\right)}}}{4}$$

$$$v=\sin{\left(x \right)}$$$라 하자.

그러면 $$$dv=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$\cos{\left(x \right)} dx = dv$$$임을 얻습니다.

따라서,

$$\frac{x}{4} - \frac{{\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}}{2} = \frac{x}{4} - \frac{{\color{red}{\int{v d v}}}}{2}$$

멱법칙($$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=1$$$에 적용합니다:

$$\frac{x}{4} - \frac{{\color{red}{\int{v d v}}}}{2}=\frac{x}{4} - \frac{{\color{red}{\frac{v^{1 + 1}}{1 + 1}}}}{2}=\frac{x}{4} - \frac{{\color{red}{\left(\frac{v^{2}}{2}\right)}}}{2}$$

다음 $$$v=\sin{\left(x \right)}$$$을 기억하라:

$$\frac{x}{4} - \frac{{\color{red}{v}}^{2}}{4} = \frac{x}{4} - \frac{{\color{red}{\sin{\left(x \right)}}}^{2}}{4}$$

따라서,

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

간단히 하시오:

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

적분 상수를 추가하세요:

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

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