Evalúa: $\int \frac {\sin (x)}{\sin (x-\alpha) } dx$

Question

Evalúa: $$\int \frac {\sin (x)}{\sin (x-\alpha) } dx$$

Mi intento: $$=\int \dfrac {\sin (x)}{\sin (x-\alpha) }dx$$ $$=-\int \dfrac {-\sin (x)}{\sin (x).\cos (\alpha) - \cos (x).\sin (\alpha )} dx$$ Déjalo, $\cos (x)=t$ $$-\sin (x).dx=dt$$ Ahora, $$=-\int \dfrac {dt}{\sqrt {1-t^2}.\cos (\alpha)-t.\sin (\alpha)}$$

Answer 1

Razonamiento aproximado: $$\begin{align*} \int\dfrac{\sin x}{\sin(x-\alpha)}dx&=\int\dfrac{\sin(u+\alpha)}{\sin u}du\\ &=\int\dfrac{\sin u\cos\alpha+\cos u\sin\alpha}{\sin u}du\\ &=\int\cos\alpha du+\sin\alpha\int\cot udu\\ &=\cdots, \end{align*}$$ donde $u=x-\alpha$ .

Answer 2

pista

Sustituir $z=x-\alpha$ para obtener $\sin(\alpha +z)$ en el numerador

Entonces $$I=\int \frac {\sin(z+\alpha)} {\sin(z)}dz$$ $$I=\int \frac {\sin(z)\cos(\alpha)+\sin(\alpha)\cos(z)} {\sin(z)}dz$$ $$I=\sin(\alpha)\int \cot(z)dz+z\cos(\alpha)+K$$ $$\boxed{I=\sin(\alpha)\ln|\sin(x-\alpha)|+x\cos(\alpha)+K}$$

Answer 3

$\int{\frac{\sin(x)}{\sin(x-a)}dx}=\int{\frac{\sin(u+a)}{\sin(u)}du}$ donde $u=x-a$ . A partir de aquí puede utilizar la misma identidad de pecado de su intento de obtener $$\int{\frac{\sin(u+a)}{\sin(u)}du}=\int{\frac{\sin(u)\cos(a)+\sin(a)\cos(u)}{\sin(u)}du}=\int{\cos(a)+\sin(a)\cot(u)du}=u\cos(a)+\sin(a)\ln(\sin(u))+C=x\cos(a)+\sin(a)\ln(\sin(x-a))+C$$ donde C es una constante.