Integral de$\int \frac{\sin(3x)}{\sin(5x)} \, dx$

Question

Cómo encontrar integral de

$$\int \frac{\sin(3x)}{\sin(5x)}dx$$

Escribí $\sin(3x)=\sin(8x-5x)$ pero generó $\frac{\sin(8x) \cos(5x)}{\sin(5x)}$.

¿Cómo debo proceder?

Answer 1

El uso de $\;\sin3x=\sin(5x-2x)=\sin5x\cos2x-\sin2x\cos5x$ :

$$\frac{\sin3x}{\sin5x}=\cos 2x-\sin2x\frac{\cos5x}{\sin5x}$$

Ahora, observa que

$$\int\frac{\cos kx}{\sin kx}dx=\frac1k\,\log|\sin kx|+C$$

y ahora tal vez la integración por partes será de ayuda.

Answer 2

El uso de polinomios de Chebyshev (que puede ser utilizado también para el seno cuando usted tiene impar multiplicadores para el ángulo) tenemos: $$ \sin3x = 3\sin x-4\sin^3x \\ \sin5x = 16\sin^5x-20\sin^3x+5\sin x $$ Así que: $$ \frac{\sin3x}{\sin5x}=\frac{3-4\sen^2x}{16\sin^4x-20\sen^2x+5}=\frac{-\left (4\sen^2x-\frac{5}{2} \right )+\frac{1}{2}}{\left (4\sen^2x-\frac{5}{2} \right )^2-\frac{5}{4}} $$ Let's convert sines to cosines (they look better when you integrate): $$ \frac{\sin3x}{\sin5x}=\frac{-\left (4(1-\cos^2x)-\frac{5}{2} \right )+\frac{1}{2}}{\left (4(1-\cos^2x)-\frac{5}{2} \right )^2-\frac{5}{4}}=\frac{\left (4\cos^2x-\frac{3}{2} \right )+\frac{1}{2}}{\left (4\cos^2x-\frac{3}{2} \right )^2-\frac{5}{4}} $$ Now we can decompose the denominator: $$ \frac{u+\frac{1}{2}}{u^2-\frac{5}{4}}=\frac{a}{u-\sqrt{\frac{5}{4}}}+\frac{b}{u+\sqrt{\frac{5}{4}}} \\ a=\frac{\sqrt{5}+1}{2\sqrt{5}}, b=\frac{\sqrt{5}-1}{2\sqrt{5}} $$ Then: $$ \frac{\sin3x}{\sin5x}=\frac{\frac{\sqrt{5}+1}{2\sqrt{5}}}{4\cos^2x-\frac{3}{2}-\sqrt{\frac{5}{4}}}+\frac{\frac{\sqrt{5}-1}{2\sqrt{5}}}{4\cos^2x-\frac{3}{2}+\sqrt{\frac{5}{4}}} $$ For both the denominators we have a negative constant term, so let's try to find a solution for: $$ I(a^2 b^2)=\int {\frac{1}{a^2cos^2x-b^2}}dx=\int {\frac{1+\bronceado^2x}{a^2-b^2-b^2\bronceado^2x}}dx$$ using $u=\tan x$, $du=(1+\bronceado^2x)dx$: $$ I(a^2 b^2)=\int {\frac{1}{a^2-b^2-b^2u^2}}du=\frac{1}{b^2}\int {\frac{1}{\frac{a^2-b^2}{b^2}-u^2}}du=\frac{1}{b^2}\frac{b}{\sqrt{a^2-b^2}}\tanh^{-1}\frac{bu}{\sqrt{a^2-b^2}}=\frac{1}{b\sqrt{a^2-b^2}}\tanh^{-1}\frac{b\tan x}{\sqrt{a^2-b^2}} $$ Carefully doing the substitutions I found: $$ \int {\frac{\sin3x}{\sin5x}}dx= I\left (4, \frac{3+\sqrt{5}}{2}\right ) + I\left (4, \frac{3-\sqrt{5}}{2}\right ) = \frac{\sqrt{5}+1}{4\sqrt{5}}\frac{1}{\sqrt{3+\sqrt{5}}\sqrt{5-\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3+\sqrt{5}}}{\sqrt{5-\sqrt{5}}}\tan x+\frac{\sqrt{5}-1}{4\sqrt{5}}\frac{1}{\sqrt{3-\sqrt{5}}\sqrt{5+\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3-\sqrt{5}}}{\sqrt{5+\sqrt{5}}}\tan x + C $$