Integral of $$$- x^{2} + 4 \cos{\left(2 x \right)}$$$

Find $$$\int \left(- x^{2} + 4 \cos{\left(2 x \right)}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(- x^{2} + 4 \cos{\left(2 x \right)}\right)d x}}} = {\color{red}{\left(- \int{x^{2} d x} + \int{4 \cos{\left(2 x \right)} d x}\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$\int{4 \cos{\left(2 x \right)} d x} - {\color{red}{\int{x^{2} d x}}}=\int{4 \cos{\left(2 x \right)} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\int{4 \cos{\left(2 x \right)} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$- \frac{x^{3}}{3} + {\color{red}{\int{4 \cos{\left(2 x \right)} d x}}} = - \frac{x^{3}}{3} + {\color{red}{\left(4 \int{\cos{\left(2 x \right)} d x}\right)}}$$

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,

$$- \frac{x^{3}}{3} + 4 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = - \frac{x^{3}}{3} + 4 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

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

$$- \frac{x^{3}}{3} + 4 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = - \frac{x^{3}}{3} + 4 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

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

$$- \frac{x^{3}}{3} + 2 {\color{red}{\int{\cos{\left(u \right)} d u}}} = - \frac{x^{3}}{3} + 2 {\color{red}{\sin{\left(u \right)}}}$$

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

$$- \frac{x^{3}}{3} + 2 \sin{\left({\color{red}{u}} \right)} = - \frac{x^{3}}{3} + 2 \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$

Therefore,

$$\int{\left(- x^{2} + 4 \cos{\left(2 x \right)}\right)d x} = - \frac{x^{3}}{3} + 2 \sin{\left(2 x \right)}$$

Add the constant of integration:

$$\int{\left(- x^{2} + 4 \cos{\left(2 x \right)}\right)d x} = - \frac{x^{3}}{3} + 2 \sin{\left(2 x \right)}+C$$

$$$\int \left(- x^{2} + 4 \cos{\left(2 x \right)}\right)\, dx = \left(- \frac{x^{3}}{3} + 2 \sin{\left(2 x \right)}\right) + C$$$A