Integral of $$$\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right)$$$

Find $$$\int \sqrt{2} \left(1 - \sin{\left(2 x \right)}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\sqrt{2}$$$ and $$$f{\left(x \right)} = 1 - \sin{\left(2 x \right)}$$$:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

Let $$$u=2 x$$$.

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

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recall that $$$u=2 x$$$:

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

Therefore,

$$\int{\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right) d x} = \sqrt{2} \left(x + \frac{\cos{\left(2 x \right)}}{2}\right)$$

Add the constant of integration:

$$\int{\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right) d x} = \sqrt{2} \left(x + \frac{\cos{\left(2 x \right)}}{2}\right)+C$$

$$$\int \sqrt{2} \left(1 - \sin{\left(2 x \right)}\right)\, dx = \sqrt{2} \left(x + \frac{\cos{\left(2 x \right)}}{2}\right) + C$$$A