Integral of $$$\left(10 - 10 x\right) \sin{\left(10 x \right)}$$$

Find $$$\int \left(10 - 10 x\right) \sin{\left(10 x \right)}\, dx$$$.

Simplify the integrand:

$${\color{red}{\int{\left(10 - 10 x\right) \sin{\left(10 x \right)} d x}}} = {\color{red}{\int{10 \left(1 - x\right) \sin{\left(10 x \right)} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=10$$$ and $$$f{\left(x \right)} = \left(1 - x\right) \sin{\left(10 x \right)}$$$:

$${\color{red}{\int{10 \left(1 - x\right) \sin{\left(10 x \right)} d x}}} = {\color{red}{\left(10 \int{\left(1 - x\right) \sin{\left(10 x \right)} d x}\right)}}$$

For the integral $$$\int{\left(1 - x\right) \sin{\left(10 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=1 - x$$$ and $$$\operatorname{dv}=\sin{\left(10 x \right)} dx$$$.

Then $$$\operatorname{du}=\left(1 - x\right)^{\prime }dx=- dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\sin{\left(10 x \right)} d x}=- \frac{\cos{\left(10 x \right)}}{10}$$$ (steps can be seen »).

Therefore,

$$10 {\color{red}{\int{\left(1 - x\right) \sin{\left(10 x \right)} d x}}}=10 {\color{red}{\left(\left(1 - x\right) \cdot \left(- \frac{\cos{\left(10 x \right)}}{10}\right)-\int{\left(- \frac{\cos{\left(10 x \right)}}{10}\right) \cdot \left(-1\right) d x}\right)}}=10 {\color{red}{\left(- \frac{\left(1 - x\right) \cos{\left(10 x \right)}}{10} - \int{\frac{\cos{\left(10 x \right)}}{10} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{10}$$$ and $$$f{\left(x \right)} = \cos{\left(10 x \right)}$$$:

$$- \left(1 - x\right) \cos{\left(10 x \right)} - 10 {\color{red}{\int{\frac{\cos{\left(10 x \right)}}{10} d x}}} = - \left(1 - x\right) \cos{\left(10 x \right)} - 10 {\color{red}{\left(\frac{\int{\cos{\left(10 x \right)} d x}}{10}\right)}}$$

Let $$$u=10 x$$$.

Then $$$du=\left(10 x\right)^{\prime }dx = 10 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{10}$$$.

The integral becomes

$$- \left(1 - x\right) \cos{\left(10 x \right)} - {\color{red}{\int{\cos{\left(10 x \right)} d x}}} = - \left(1 - x\right) \cos{\left(10 x \right)} - {\color{red}{\int{\frac{\cos{\left(u \right)}}{10} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{10}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- \left(1 - x\right) \cos{\left(10 x \right)} - {\color{red}{\int{\frac{\cos{\left(u \right)}}{10} d u}}} = - \left(1 - x\right) \cos{\left(10 x \right)} - {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{10}\right)}}$$

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

$$- \left(1 - x\right) \cos{\left(10 x \right)} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{10} = - \left(1 - x\right) \cos{\left(10 x \right)} - \frac{{\color{red}{\sin{\left(u \right)}}}}{10}$$

Recall that $$$u=10 x$$$:

$$- \left(1 - x\right) \cos{\left(10 x \right)} - \frac{\sin{\left({\color{red}{u}} \right)}}{10} = - \left(1 - x\right) \cos{\left(10 x \right)} - \frac{\sin{\left({\color{red}{\left(10 x\right)}} \right)}}{10}$$

Therefore,

$$\int{\left(10 - 10 x\right) \sin{\left(10 x \right)} d x} = - \left(1 - x\right) \cos{\left(10 x \right)} - \frac{\sin{\left(10 x \right)}}{10}$$

Simplify:

$$\int{\left(10 - 10 x\right) \sin{\left(10 x \right)} d x} = \left(x - 1\right) \cos{\left(10 x \right)} - \frac{\sin{\left(10 x \right)}}{10}$$

Add the constant of integration:

$$\int{\left(10 - 10 x\right) \sin{\left(10 x \right)} d x} = \left(x - 1\right) \cos{\left(10 x \right)} - \frac{\sin{\left(10 x \right)}}{10}+C$$

$$$\int \left(10 - 10 x\right) \sin{\left(10 x \right)}\, dx = \left(\left(x - 1\right) \cos{\left(10 x \right)} - \frac{\sin{\left(10 x \right)}}{10}\right) + C$$$A