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

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

Voor de integraal $$$\int{x \sec^{2}{\left(x \right)} d x}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Zij $$$\operatorname{u}=x$$$ en $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Dan $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (de stappen zijn te zien »).

De integraal wordt

$${\color{red}{\int{x \sec^{2}{\left(x \right)} d x}}}={\color{red}{\left(x \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 1 d x}\right)}}={\color{red}{\left(x \tan{\left(x \right)} - \int{\tan{\left(x \right)} d x}\right)}}$$

Herschrijf de raaklijn als $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

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

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 $$$\sin{\left(x \right)} dx = - du$$$.

De integraal wordt

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

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

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

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

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

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

$$x \tan{\left(x \right)} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x \tan{\left(x \right)} + \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

Dus,

$$\int{x \sec^{2}{\left(x \right)} d x} = x \tan{\left(x \right)} + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{x \sec^{2}{\left(x \right)} d x} = x \tan{\left(x \right)} + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$$

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