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

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

Misalkan $$$u=\frac{x}{2}$$$.

Kemudian $$$du=\left(\frac{x}{2}\right)^{\prime }dx = \frac{dx}{2}$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = 2 du$$$.

Oleh karena itu,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=2$$$ dan $$$f{\left(u \right)} = 1 - \sin{\left(2 u \right)}$$$:

$${\color{red}{\int{\left(2 - 2 \sin{\left(2 u \right)}\right)d u}}} = {\color{red}{\left(2 \int{\left(1 - \sin{\left(2 u \right)}\right)d u}\right)}}$$

Integralkan suku demi suku:

$$2 {\color{red}{\int{\left(1 - \sin{\left(2 u \right)}\right)d u}}} = 2 {\color{red}{\left(\int{1 d u} - \int{\sin{\left(2 u \right)} d u}\right)}}$$

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:

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

Misalkan $$$v=2 u$$$.

Kemudian $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = \frac{dv}{2}$$$.

Integral tersebut dapat ditulis ulang sebagai

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

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(v \right)} = \sin{\left(v \right)}$$$:

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

Integral dari sinus adalah $$$\int{\sin{\left(v \right)} d v} = - \cos{\left(v \right)}$$$:

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

Ingat bahwa $$$v=2 u$$$:

$$2 u + \cos{\left({\color{red}{v}} \right)} = 2 u + \cos{\left({\color{red}{\left(2 u\right)}} \right)}$$

Ingat bahwa $$$u=\frac{x}{2}$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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