Integral of $$$\cot{\left(x + \frac{\pi}{4} \right)}$$$

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

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

Then $$$du=\left(x + \frac{\pi}{4}\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Thus,

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

Rewrite the cotangent as $$$\cot\left( u \right)=\frac{\cos\left( u \right)}{\sin\left( u \right)}$$$:

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

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

Then $$$dv=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$ (steps can be seen »), and we have that $$$\cos{\left(u \right)} du = dv$$$.

The integral becomes

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

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Recall that $$$v=\sin{\left(u \right)}$$$:

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

Recall that $$$u=x + \frac{\pi}{4}$$$:

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

Therefore,

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

Add the constant of integration:

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

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