Integraal van $$$\cot{\left(x + \frac{\pi}{4} \right)}$$$

Bepaal $$$\int \cot{\left(x + \frac{\pi}{4} \right)}\, dx$$$.

Zij $$$u=x + \frac{\pi}{4}$$$.

Dan $$$du=\left(x + \frac{\pi}{4}\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

De integraal kan worden herschreven als

$${\color{red}{\int{\cot{\left(x + \frac{\pi}{4} \right)} d x}}} = {\color{red}{\int{\cot{\left(u \right)} d u}}}$$

Herschrijf de cotangens als $$$\cot\left( u \right)=\frac{\cos\left( u \right)}{\sin\left( u \right)}$$$:

$${\color{red}{\int{\cot{\left(u \right)} d u}}} = {\color{red}{\int{\frac{\cos{\left(u \right)}}{\sin{\left(u \right)}} d u}}}$$

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

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

De integraal wordt

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

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

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

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

$$\ln{\left(\left|{{\color{red}{v}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\sin{\left(u \right)}}}}\right| \right)}$$

We herinneren eraan dat $$$u=x + \frac{\pi}{4}$$$:

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

Dus,

$$\int{\cot{\left(x + \frac{\pi}{4} \right)} d x} = \ln{\left(\left|{\sin{\left(x + \frac{\pi}{4} \right)}}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{\cot{\left(x + \frac{\pi}{4} \right)} d x} = \ln{\left(\left|{\sin{\left(x + \frac{\pi}{4} \right)}}\right| \right)}+C$$

$$$\int \cot{\left(x + \frac{\pi}{4} \right)}\, dx = \ln\left(\left|{\sin{\left(x + \frac{\pi}{4} \right)}}\right|\right) + C$$$A