Integraal van $$$\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}}$$$

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

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

Dan $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sin{\left(x \right)} dx = - du$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

We herinneren eraan dat $$$u=\cos{\left(x \right)}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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