Integraal van $$$- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}$$$

Bepaal $$$\int \left(- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}\right)\, dx$$$.

Integreer termgewijs:

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

Herschrijf de cosinus in termen van de sinus met behulp van de formule $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ en herschrijf vervolgens de sinus met behulp van de dubbelhoekformule $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

Vermenigvuldig de teller en de noemer met $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

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

Zij $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Dan $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

De integraal kan worden herschreven als

$$- \int{\sin{\left(x \right)} d x} + {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = - \int{\sin{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

We herinneren eraan dat $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{\sin{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)} - \int{\sin{\left(x \right)} d x}$$

De integraal van de sinus is $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$\ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} - {\color{red}{\int{\sin{\left(x \right)} d x}}} = \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} - {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

Dus,

$$\int{\left(- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}\right)d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + \cos{\left(x \right)}$$

Voeg de integratieconstante toe:

$$\int{\left(- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}\right)d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + \cos{\left(x \right)}+C$$

$$$\int \left(- \sin{\left(x \right)} + \frac{1}{\cos{\left(x \right)}}\right)\, dx = \left(\ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right) + \cos{\left(x \right)}\right) + C$$$A