Integral of $$$\frac{\sin{\left(\sqrt{x} \right)}}{\sqrt{x}}$$$

Find $$$\int \frac{\sin{\left(\sqrt{x} \right)}}{\sqrt{x}}\, dx$$$.

Let $$$u=\sqrt{x}$$$.

Then $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

So,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Recall that $$$u=\sqrt{x}$$$:

$$- 2 \cos{\left({\color{red}{u}} \right)} = - 2 \cos{\left({\color{red}{\sqrt{x}}} \right)}$$

Therefore,

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

Add the constant of integration:

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

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