Funktion $$$- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}$$$ integraali

Määritä $$$\int \left(- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}\right)\, dx$$$.

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=- 10 \cot{\left(1 \right)}$$$ ja $$$f{\left(x \right)} = 1 - x^{3}$$$:

$${\color{red}{\int{\left(- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}\right)d x}}} = {\color{red}{\left(- 10 \cot{\left(1 \right)} \int{\left(1 - x^{3}\right)d x}\right)}}$$

Integroi termi kerrallaan:

$$- 10 \cot{\left(1 \right)} {\color{red}{\int{\left(1 - x^{3}\right)d x}}} = - 10 \cot{\left(1 \right)} {\color{red}{\left(\int{1 d x} - \int{x^{3} d x}\right)}}$$

Sovella vakiosääntöä $$$\int c\, dx = c x$$$ käyttäen $$$c=1$$$:

$$- 10 \cot{\left(1 \right)} \left(- \int{x^{3} d x} + {\color{red}{\int{1 d x}}}\right) = - 10 \cot{\left(1 \right)} \left(- \int{x^{3} d x} + {\color{red}{x}}\right)$$

Sovella potenssisääntöä $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=3$$$:

$$- 10 \cot{\left(1 \right)} \left(x - {\color{red}{\int{x^{3} d x}}}\right)=- 10 \cot{\left(1 \right)} \left(x - {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}\right)=- 10 \cot{\left(1 \right)} \left(x - {\color{red}{\left(\frac{x^{4}}{4}\right)}}\right)$$

Näin ollen,

$$\int{\left(- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}\right)d x} = - 10 \left(- \frac{x^{4}}{4} + x\right) \cot{\left(1 \right)}$$

Sievennä:

$$\int{\left(- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}\right)d x} = \frac{5 x \left(x^{3} - 4\right) \cot{\left(1 \right)}}{2}$$

Lisää integrointivakio:

$$\int{\left(- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}\right)d x} = \frac{5 x \left(x^{3} - 4\right) \cot{\left(1 \right)}}{2}+C$$

$$$\int \left(- 10 \left(1 - x^{3}\right) \cot{\left(1 \right)}\right)\, dx = \frac{5 x \left(x^{3} - 4\right) \cot{\left(1 \right)}}{2} + C$$$A