Funktion $$$\sin{\left(x \right)} \cos{\left(2 \right)} \cos{\left(x \right)}$$$ integraali

Määritä $$$\int \sin{\left(x \right)} \cos{\left(2 \right)} \cos{\left(x \right)}\, dx$$$.

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\cos{\left(2 \right)}$$$ ja $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$$$:

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

Olkoon $$$u=\sin{\left(x \right)}$$$.

Tällöin $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\cos{\left(x \right)} dx = du$$$.

Integraali muuttuu muotoon

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

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

$$\cos{\left(2 \right)} {\color{red}{\int{u d u}}}=\cos{\left(2 \right)} {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=\cos{\left(2 \right)} {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

Muista, että $$$u=\sin{\left(x \right)}$$$:

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

Näin ollen,

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

Lisää integrointivakio:

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

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