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

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

Herschrijf de integraand:

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

Vermenigvuldig de teller en de noemer met één cosinus en druk de rest uit in termen van de sinus, met behulp van de formule $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ met $$$\alpha=x$$$:

$${\color{red}{\int{\frac{\sin^{3}{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sin^{3}{\left(x \right)} \cos{\left(x \right)}}{1 - \sin^{2}{\left(x \right)}} d x}}}$$

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

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

Dus,

$${\color{red}{\int{\frac{\sin^{3}{\left(x \right)} \cos{\left(x \right)}}{1 - \sin^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{u^{3}}{1 - u^{2}} d u}}}$$

Aangezien de graad van de teller niet kleiner is dan die van de noemer, voer een staartdeling van polynomen uit (stappen zijn te zien »):

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

Integreer termgewijs:

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

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

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

Zij $$$v=1 - u^{2}$$$.

Dan $$$dv=\left(1 - u^{2}\right)^{\prime }du = - 2 u du$$$ (de stappen zijn te zien »), en dan geldt dat $$$u du = - \frac{dv}{2}$$$.

Dus,

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

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

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

De integraal van $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

We herinneren eraan dat $$$v=1 - u^{2}$$$:

$$- \frac{u^{2}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = - \frac{u^{2}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(1 - u^{2}\right)}}}\right| \right)}}{2}$$

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

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

Dus,

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

Vereenvoudig:

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

Voeg de constante van integratie toe (en verwijder de constante uit de uitdrukking):

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

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