Integralen av $$$\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}}$$$

Bestäm $$$\int \frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx$$$.

Multiplicera täljaren och nämnaren med $$$\cos^{2}{\left(x \right)}$$$ och omvandla $$$\frac{\sin^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$$$ till $$$\tan^{2}{\left(x \right)}$$$:

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

Omvandla $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ till $$$\sec^{2}{\left(x \right)}$$$:

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

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

Då $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\sec^{2}{\left(x \right)} dx = du$$$.

Integralen blir

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=2$$$:

$${\color{red}{\int{u^{2} d u}}}={\color{red}{\frac{u^{1 + 2}}{1 + 2}}}={\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Kom ihåg att $$$u=\tan{\left(x \right)}$$$:

$$\frac{{\color{red}{u}}^{3}}{3} = \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3}$$

Alltså,

$$\int{\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3}$$

Lägg till integrationskonstanten:

$$\int{\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3}+C$$

$$$\int \frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx = \frac{\tan^{3}{\left(x \right)}}{3} + C$$$A