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

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

Integroi termi kerrallaan:

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

Kerro osoittaja ja nimittäjä yhdellä sinillä ja ilmaise kaikki muu kosinin avulla, käyttäen kaavaa $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ parametrilla $$$\alpha=x$$$:

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

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

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

Näin ollen,

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

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

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

Koska osoittajan aste ei ole pienempi kuin nimittäjän aste, suorita polynomien jakokulma (vaiheet voidaan nähdä »):

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

Integroi termi kerrallaan:

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

Sovella vakiosääntöä $$$\int c\, du = c u$$$ käyttäen $$$c=1$$$:

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

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

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

Integroi termi kerrallaan:

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

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

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

Olkoon $$$v=u + 1$$$.

Tällöin $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$du = dv$$$.

Siis,

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

Funktion $$$\frac{1}{v}$$$ integraali on $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- u + \int{\sin{\left(x \right)} d x} - \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = - u + \int{\sin{\left(x \right)} d x} - \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Muista, että $$$v=u + 1$$$:

$$- u + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - \int{\frac{1}{2 \left(u - 1\right)} d u} = - u + \frac{\ln{\left(\left|{{\color{red}{\left(u + 1\right)}}}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - \int{\frac{1}{2 \left(u - 1\right)} d u}$$

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

$$- u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{1}{2 \left(u - 1\right)} d u}}} = - u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{u - 1} d u}}{2}\right)}}$$

Olkoon $$$v=u - 1$$$.

Tällöin $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$du = dv$$$.

Integraali muuttuu muotoon

$$- u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\frac{1}{u - 1} d u}}}}{2} = - u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2}$$

Funktion $$$\frac{1}{v}$$$ integraali on $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = - u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Muista, että $$$v=u - 1$$$:

$$- u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} = - u + \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x}$$

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

$$- \frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - {\color{red}{u}} = - \frac{\ln{\left(\left|{-1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} + \int{\sin{\left(x \right)} d x} - {\color{red}{\cos{\left(x \right)}}}$$

Sinifunktion integraali on $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$- \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \cos{\left(x \right)} + {\color{red}{\int{\sin{\left(x \right)} d x}}} = - \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \cos{\left(x \right)} + {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

Näin ollen,

$$\int{\left(\sin{\left(x \right)} - \frac{\cos^{2}{\left(x \right)}}{\sin{\left(x \right)}}\right)d x} = - \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - 2 \cos{\left(x \right)}$$

Lisää integrointivakio:

$$\int{\left(\sin{\left(x \right)} - \frac{\cos^{2}{\left(x \right)}}{\sin{\left(x \right)}}\right)d x} = - \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - 2 \cos{\left(x \right)}+C$$

$$$\int \left(\sin{\left(x \right)} - \frac{\cos^{2}{\left(x \right)}}{\sin{\left(x \right)}}\right)\, dx = \left(- \frac{\ln\left(\left|{\cos{\left(x \right)} - 1}\right|\right)}{2} + \frac{\ln\left(\left|{\cos{\left(x \right)} + 1}\right|\right)}{2} - 2 \cos{\left(x \right)}\right) + C$$$A