Integral de $$$\cot{\left(x + \frac{\pi}{4} \right)}$$$
Calculadora relacionada: Calculadora de Integrais Definidas e Impróprias
Sua entrada
Encontre $$$\int \cot{\left(x + \frac{\pi}{4} \right)}\, dx$$$.
Solução
Seja $$$u=x + \frac{\pi}{4}$$$.
Então $$$du=\left(x + \frac{\pi}{4}\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.
A integral torna-se
$${\color{red}{\int{\cot{\left(x + \frac{\pi}{4} \right)} d x}}} = {\color{red}{\int{\cot{\left(u \right)} d u}}}$$
Reescreva a cotangente como $$$\cot\left( u \right)=\frac{\cos\left( u \right)}{\sin\left( u \right)}$$$:
$${\color{red}{\int{\cot{\left(u \right)} d u}}} = {\color{red}{\int{\frac{\cos{\left(u \right)}}{\sin{\left(u \right)}} d u}}}$$
Seja $$$v=\sin{\left(u \right)}$$$.
Então $$$dv=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$ (veja os passos »), e obtemos $$$\cos{\left(u \right)} du = dv$$$.
Portanto,
$${\color{red}{\int{\frac{\cos{\left(u \right)}}{\sin{\left(u \right)}} d u}}} = {\color{red}{\int{\frac{1}{v} d v}}}$$
A integral de $$$\frac{1}{v}$$$ é $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$${\color{red}{\int{\frac{1}{v} d v}}} = {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$
Recorde que $$$v=\sin{\left(u \right)}$$$:
$$\ln{\left(\left|{{\color{red}{v}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\sin{\left(u \right)}}}}\right| \right)}$$
Recorde que $$$u=x + \frac{\pi}{4}$$$:
$$\ln{\left(\left|{\sin{\left({\color{red}{u}} \right)}}\right| \right)} = \ln{\left(\left|{\sin{\left({\color{red}{\left(x + \frac{\pi}{4}\right)}} \right)}}\right| \right)}$$
Portanto,
$$\int{\cot{\left(x + \frac{\pi}{4} \right)} d x} = \ln{\left(\left|{\sin{\left(x + \frac{\pi}{4} \right)}}\right| \right)}$$
Adicione a constante de integração:
$$\int{\cot{\left(x + \frac{\pi}{4} \right)} d x} = \ln{\left(\left|{\sin{\left(x + \frac{\pi}{4} \right)}}\right| \right)}+C$$
Resposta
$$$\int \cot{\left(x + \frac{\pi}{4} \right)}\, dx = \ln\left(\left|{\sin{\left(x + \frac{\pi}{4} \right)}}\right|\right) + C$$$A