Funktion $$$\sinh^{2}{\left(x \right)}$$$ integraali

Määritä $$$\int \sinh^{2}{\left(x \right)}\, dx$$$.

Sovella potenssin alentamiskaavaa $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$ käyttäen $$$\alpha=x$$$:

$${\color{red}{\int{\sinh^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cosh{\left(2 x \right)}}{2} - \frac{1}{2}\right)d x}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(x \right)} = \cosh{\left(2 x \right)} - 1$$$:

$${\color{red}{\int{\left(\frac{\cosh{\left(2 x \right)}}{2} - \frac{1}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cosh{\left(2 x \right)} - 1\right)d x}}{2}\right)}}$$

Integroi termi kerrallaan:

$$\frac{{\color{red}{\int{\left(\cosh{\left(2 x \right)} - 1\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{1 d x} + \int{\cosh{\left(2 x \right)} d x}\right)}}}{2}$$

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

$$\frac{\int{\cosh{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{\int{1 d x}}}}{2} = \frac{\int{\cosh{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{x}}}{2}$$

Olkoon $$$u=2 x$$$.

Tällöin $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = \frac{du}{2}$$$.

Näin ollen,

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(u \right)} = \cosh{\left(u \right)}$$$:

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

Hyperbolisen kosinin integraali on $$$\int{\cosh{\left(u \right)} d u} = \sinh{\left(u \right)}$$$:

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

Muista, että $$$u=2 x$$$:

$$- \frac{x}{2} + \frac{\sinh{\left({\color{red}{u}} \right)}}{4} = - \frac{x}{2} + \frac{\sinh{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Näin ollen,

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

Lisää integrointivakio:

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

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