$$$\sin{\left(\alpha \left(\beta + x\right) \right)}$$$ 关于$$$x$$$的积分

求$$$\int \sin{\left(\alpha \left(\beta + x\right) \right)}\, dx$$$。

设$$$u=\alpha \left(\beta + x\right)$$$。

则$$$du=\left(\alpha \left(\beta + x\right)\right)^{\prime }dx = \alpha dx$$$ (步骤见»)，并有$$$dx = \frac{du}{\alpha}$$$。

所以，

$${\color{red}{\int{\sin{\left(\alpha \left(\beta + x\right) \right)} d x}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{\alpha} d u}}}$$

对 $$$c=\frac{1}{\alpha}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

正弦函数的积分为 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

回忆一下 $$$u=\alpha \left(\beta + x\right)$$$:

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

因此，

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

加上积分常数：

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

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