Funktion $$$\frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}$$$ integraali

Määritä $$$\int \frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx$$$.

Kerro osoittaja ja nimittäjä luvulla $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ ja muunna $$$\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$ muotoon $$$\frac{1}{\tan^{2}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cos^{4}{\left(x \right)} \tan^{2}{\left(x \right)}} d x}}}$$

Poista kaksi kosinia ja ilmaise ne sekantin termeinä kaavan $$$\frac{1}{\cos^{2}{\left(x \right)}}=\sec^{2}{\left(x \right)}$$$ avulla.:

$${\color{red}{\int{\frac{1}{\cos^{4}{\left(x \right)} \tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} d x}}}$$

Esitä kosini tangentin funktiona käyttäen kaavaa $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$:

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

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

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

Siis,

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

Expand the expression:

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

Integroi termi kerrallaan:

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

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

$$\int{\frac{1}{u^{2}} d u} + {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2}} d u} + {\color{red}{u}}$$

Sovella potenssisääntöä $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=-2$$$:

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

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

$$- {\color{red}{u}}^{-1} + {\color{red}{u}} = - {\color{red}{\tan{\left(x \right)}}}^{-1} + {\color{red}{\tan{\left(x \right)}}}$$

Näin ollen,

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

Lisää integrointivakio:

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

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