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

How to find integral of

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

I wrote $\sin(3x)=\sin(8x-5x)$ but it generated $\frac{\sin(8x) \cos(5x)}{\sin(5x)}$.

How should I proceed?

• Just wondering, is that an exercise from a textbook? – imranfat Feb 22 '16 at 16:32
• One approach is to convert this to an integral of a rational function using the substitution $z=e^{ix}$. Then, $$\int \frac{\sin(3x)}{\sin(5x)}\,dx=-i\int \frac{z(z^6-1)}{z^{10}-1}\,dz=-\frac i2 \left(\int \frac{z}{z^4+z^3+z^2+z+1}\,dz+\int \frac{z}{z^4-z^3+z^2-z+1}\,dz\right)$$It looks ugly, but perhaps tractable. – Mark Viola Feb 22 '16 at 16:47
• @Dr.MV Any method to solve it in terms of elementary functions? – Ananya Feb 22 '16 at 16:49
• Yes. It does have a solution in terms of elementary functions. – Mark Viola Feb 22 '16 at 16:50
• Even though it didn't work out, your idea of writing $8 = 5-3$ was a good one. Did you solve similar problems or did you just get the idea suddenly ? – Saikat Feb 22 '16 at 17:08

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

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

Now, observe that

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

and now perhaps integrating by parts will help.

• After IBP, one is left with the task of evaluating $\int \cos(2x) \log(\sin(5x))\,dx$. – Mark Viola Feb 22 '16 at 16:41
• @Dr.MV True, thank you. It doesn't make things easier (I thought incorrectly that there'd be $\;\cos 5x\;$ inside that last integral). I'm erasing this, and I think this integral is increidibly difficult, although perhaps I missed some trick. – DonAntonio Feb 22 '16 at 17:02

Using Chebyshev polynomials (that can be used also for sine when you have odd multipliers for the angle) we have: $$\sin3x = 3\sin x-4\sin^3x \\ \sin5x = 16\sin^5x-20\sin^3x+5\sin x$$ So: $$\frac{\sin3x}{\sin5x}=\frac{3-4\sin^2x}{16\sin^4x-20\sin^2x+5}=\frac{-\left (4\sin^2x-\frac{5}{2} \right )+\frac{1}{2}}{\left (4\sin^2x-\frac{5}{2} \right )^2-\frac{5}{4}}$$ 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: $$I(a^2, b^2)=\int {\frac{1}{a^2cos^2x-b^2}}dx=\int {\frac{1+\tan^2x}{a^2-b^2-b^2\tan^2x}}dx$$ using $u=\tan x$, $du=(1+\tan^2x)dx$: $$I(a^2, b^2)=\int {\frac{1}{a^2-b^2-b^2u^2}}du=\frac{1}{b^2}\int {\frac{1}{\frac{a^2-b^2}{b^2}-u^2}}du=\frac{1}{b^2}\frac{b}{\sqrt{a^2-b^2}}\tanh^{-1}\frac{bu}{\sqrt{a^2-b^2}}=\frac{1}{b\sqrt{a^2-b^2}}\tanh^{-1}\frac{b\tan x}{\sqrt{a^2-b^2}}$$ Carefully doing the substitutions I found: $$\int {\frac{\sin3x}{\sin5x}}dx= I\left (4, \frac{3+\sqrt{5}}{2}\right ) + I\left (4, \frac{3-\sqrt{5}}{2}\right )= \frac{\sqrt{5}+1}{4\sqrt{5}}\frac{1}{\sqrt{3+\sqrt{5}}\sqrt{5-\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3+\sqrt{5}}}{\sqrt{5-\sqrt{5}}}\tan x+\frac{\sqrt{5}-1}{4\sqrt{5}}\frac{1}{\sqrt{3-\sqrt{5}}\sqrt{5+\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3-\sqrt{5}}}{\sqrt{5+\sqrt{5}}}\tan x + C$$

Express

\begin{align} \sin(3x)& =2\sin x \left(\cos(2x)- \cos\frac{2\pi}{3}\right) \\ \sin(5x)& =4\sin x \left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right) \end{align}

Then, decompose the integrand as follows \begin{align}\frac{\sin (3x)}{\sin(5x)} =&\frac{\cos(2x)- \cos\frac{2\pi}{3}} {2\left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)}\\ =&\frac{\frac1{\sqrt5}\cos\frac{2\pi}{5} }{\cos(2x)- \cos\frac{4\pi}{5} } -\frac{ \frac1{\sqrt5} \cos\frac{4\pi}{5}}{\cos(2x)- \cos\frac{2\pi}{5} } \\ =&\frac{\frac1{2\sqrt5}\cos\frac{2\pi}{5} }{\sin(\frac{2\pi}5+x)\sin(\frac{2\pi}5-x) }-\frac{\frac1{2\sqrt5} \cos\frac{4\pi}{5} }{\sin(\frac{\pi}5+x)\sin(\frac{\pi}5-x) }\\ =&\frac15\sin\frac\pi5\left(\cot(\frac{2\pi}5+x)+\cot(\frac{2\pi}5-x) \right)\\ &\hspace{1cm}+\frac15\sin\frac{2\pi}5\left(\cot(\frac{\pi}5+x)+\cot(\frac{\pi}5-x) \right) \end{align} Apply $$(\ln \sin t)’= \cot t$$ to integrate $$\int \frac{\sin (3x)}{\sin(5x)}dx= \frac15\sin\frac{\pi}5\>\ln \frac{\sin(\frac{2\pi}5+x) }{\sin(\frac{2\pi}5-x) } +\frac15\sin\frac{2\pi}5\>\ln \frac{\sin(\frac{\pi}5+x) }{\sin(\frac\pi5-x) }+C$$