How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$

Question

• The below problem is taken from Joseph Edwards book Integral Calculus for beginners.

How does one show: $$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\} $$

• Splitting $\sin{(5x)}$ as $\sin{(4x+x)}$ doesn't seem to be of much help since then we have a big term in the denominator after expansion.

Answer 1

Wolfram Mathematica $9.0$ is able to evaluate this indefinite integral. Here is the output

\begin{equation} \sqrt{\frac{5+\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}-3\right)\tan x}{\sqrt{10-2\sqrt{5}}}\right)+\sqrt{\frac{5-\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}+3\right)\tan x}{\sqrt{10+2\sqrt{5}}}\right)+C \end{equation}

The complete proof can be downloaded here.

Using the facts \begin{align} \sin\left(\frac{\pi}{5}\right)&=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\\[10pt] \cos\left(\frac{\pi}{5}\right)&=\frac{\sqrt{5}+1}{4}\\[10pt] \sin\left(\frac{2\pi}{5}\right)&=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}\\[10pt] \cos\left(\frac{2\pi}{5}\right)&=\frac{\sqrt{5}-1}{4}\\[10pt] \sin\left(\alpha\pm\beta\right)&=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\[10pt] {\rm{artanh}}\,\theta&=\frac{1}{2}\ln\left(\frac{1+\theta}{1-\theta}\right)\,,\quad\mbox{for}\,|\theta|<1 \end{align}

We can derive the following answer \begin{equation} 5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \ln\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \ln\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\}+C \end{equation}

Answer 2

Not a complete answer, but I think the factoring technique might be helpful. Let's use $s$ to denote $\sin(x)$. You can derive $\sin(5x)=5s-20s^3+16s^5$. We know $$5s-20s^3+16s^5=16s(s-\sin(\pi/5))(s-\sin(2\pi)/5)(s-\sin(3\pi/5))(s-\sin(4\pi/5))=16s(s-\sin(\pi/5))^2(s-\sin(2\pi/5))^2$$ by factor theorem(notice $\sin(n\pi/5)$ are roots of the polynomial). You can probably do partial fraction that leads to the final answer now.

Answer 3

Note

$$\sin(5x)=4\sin x \left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)$$

and \begin{align} &\frac{\sin x}{\sin(5x)}=\frac1{4\left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)}\\=&\frac1{2\sqrt5}\left(\frac1{\cos(2x)- \cos\frac{2\pi}{5} } -\frac1{\cos(2x)- \cos\frac{4\pi}{5} } \right)\\ =&\frac1{4\sqrt5}\left(\frac1{\sin(\frac\pi5+x)\sin(\frac\pi5-x) }-\frac1{\sin(\frac{2\pi}5+x)\sin(\frac{2\pi}5-x) } \right)\\ =&\frac15\sin\frac\pi5\left(\cot(\frac\pi5+x)+\cot(\frac\pi5-x) \right)-\frac15\sin\frac{2\pi}5\left(\cot(\frac{2\pi}5+x)+\cot(\frac{2\pi}5-x) \right) \end{align} Then, apply $(\ln \sin t)’= \cot t$ to obtain $$\int \frac{\sin x}{\sin(5x)}= \frac15\sin\frac{2\pi}5\>\ln \frac{\sin(\frac{2\pi}5-x) }{\sin(\frac{2\pi}5+x) } - \frac15\sin\frac\pi5\>\ln \frac{\sin(\frac\pi5-x) }{\sin(\frac\pi5+x) }+C $$