Integral of $$$- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}$$$

The calculator will find the integral/antiderivative of $$$- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}$$$, with steps shown.

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Find $$$\int \left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)\, dx$$$.

Solution

Integrate term by term:

$${\color{red}{\int{\left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{\sin{\left(x \right)}}{12} d x} + \int{2 \cos{\left(2 x \right)} d x}\right)}}$$

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

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\int{2 \cos{\left(2 x \right)} d x}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\left(2 \int{\cos{\left(2 x \right)} d x}\right)}}$$

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,

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

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

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\int{\cos{\left(u \right)} d u}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\sin{\left(u \right)}}}$$

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

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + \sin{\left({\color{red}{u}} \right)} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{12}$$$ and $$$f{\left(x \right)} = \sin{\left(x \right)}$$$:

$$\sin{\left(2 x \right)} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{12} d x}}} = \sin{\left(2 x \right)} - {\color{red}{\left(\frac{\int{\sin{\left(x \right)} d x}}{12}\right)}}$$

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

$$\sin{\left(2 x \right)} - \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{12} = \sin{\left(2 x \right)} - \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{12}$$

Therefore,

$$\int{\left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)d x} = \sin{\left(2 x \right)} + \frac{\cos{\left(x \right)}}{12}$$

Add the constant of integration:

$$\int{\left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)d x} = \sin{\left(2 x \right)} + \frac{\cos{\left(x \right)}}{12}+C$$

Answer

$$$\int \left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)\, dx = \left(\sin{\left(2 x \right)} + \frac{\cos{\left(x \right)}}{12}\right) + C$$$A