Integral de $$$\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$
Calculadora relacionada: Calculadora de Integrais Definidas e Impróprias
Sua entrada
Encontre $$$\int \frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx$$$.
Solução
Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\sin{\left(1 \right)}$$$ e $$$f{\left(x \right)} = \frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$:
$${\color{red}{\int{\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}} = {\color{red}{\sin{\left(1 \right)} \int{\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}}$$
Reescreva em termos da cotangente:
$$\sin{\left(1 \right)} {\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}} = \sin{\left(1 \right)} {\color{red}{\int{\cot^{2}{\left(x \right)} d x}}}$$
Seja $$$u=\cot{\left(x \right)}$$$.
Então $$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (veja os passos »), e obtemos $$$\csc^{2}{\left(x \right)} dx = - du$$$.
Assim,
$$\sin{\left(1 \right)} {\color{red}{\int{\cot^{2}{\left(x \right)} d x}}} = \sin{\left(1 \right)} {\color{red}{\int{\left(- \frac{u^{2}}{u^{2} + 1}\right)d u}}}$$
Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=-1$$$ e $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$:
$$\sin{\left(1 \right)} {\color{red}{\int{\left(- \frac{u^{2}}{u^{2} + 1}\right)d u}}} = \sin{\left(1 \right)} {\color{red}{\left(- \int{\frac{u^{2}}{u^{2} + 1} d u}\right)}}$$
Reescreva e separe a fração:
$$- \sin{\left(1 \right)} {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}} = - \sin{\left(1 \right)} {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}$$
Integre termo a termo:
$$- \sin{\left(1 \right)} {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}} = - \sin{\left(1 \right)} {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$
Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:
$$- \sin{\left(1 \right)} \left(- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{1 d u}}}\right) = - \sin{\left(1 \right)} \left(- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{u}}\right)$$
A integral de $$$\frac{1}{u^{2} + 1}$$$ é $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:
$$- \sin{\left(1 \right)} \left(u - {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}\right) = - \sin{\left(1 \right)} \left(u - {\color{red}{\operatorname{atan}{\left(u \right)}}}\right)$$
Recorde que $$$u=\cot{\left(x \right)}$$$:
$$- \sin{\left(1 \right)} \left(- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}}\right) = - \sin{\left(1 \right)} \left(- \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} + {\color{red}{\cot{\left(x \right)}}}\right)$$
Portanto,
$$\int{\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \left(\cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)}$$
Simplifique:
$$\int{\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = \left(- \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)}$$
Adicione a constante de integração:
$$\int{\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = \left(- \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)}+C$$
Resposta
$$$\int \frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx = \left(- \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)} + C$$$A