$$$y \sin{\left(y^{2} \right)}$$$ 的积分
您的输入
求$$$\int y \sin{\left(y^{2} \right)}\, dy$$$。
解答
设$$$u=y^{2}$$$。
则$$$du=\left(y^{2}\right)^{\prime }dy = 2 y dy$$$ (步骤见»),并有$$$y dy = \frac{du}{2}$$$。
积分变为
$${\color{red}{\int{y \sin{\left(y^{2} \right)} d y}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$
正弦函数的积分为 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$
回忆一下 $$$u=y^{2}$$$:
$$- \frac{\cos{\left({\color{red}{u}} \right)}}{2} = - \frac{\cos{\left({\color{red}{y^{2}}} \right)}}{2}$$
因此,
$$\int{y \sin{\left(y^{2} \right)} d y} = - \frac{\cos{\left(y^{2} \right)}}{2}$$
加上积分常数:
$$\int{y \sin{\left(y^{2} \right)} d y} = - \frac{\cos{\left(y^{2} \right)}}{2}+C$$
答案
$$$\int y \sin{\left(y^{2} \right)}\, dy = - \frac{\cos{\left(y^{2} \right)}}{2} + C$$$A