Find$$\int_0^{\pi}\frac{\sin 5x}{\sin x}dx$$
I can solve it by involving polynomials in sine and cosine as shown in the links below, but it’s huge (doing double angle formulas twice; I noticed that using polynomials in cosine is better because the integral spits out sines which are 0 between the limits) so I want a faster method, if it exists. Please don’t use contour integration:)
The only thing I noticed is that the integrand is symmetric about the midpoint in the given interval, i.e.$$\frac{\sin 5x}{\sin x}= \frac{\sin 5(\pi-x)}{\sin(\pi- x)}.$$
Determine the indefinite integral $\int \frac{\sin x}{\sin 5x}dx$
Integral of $\int \frac{\sin(3x)}{\sin(5x)} \, dx$
Expressing $\frac {\sin(5x)}{\sin(x)}$ in powers of $\cos(x)$ using complex numbers
Best Answer
With $2\sin a \cos b = \sin (b+a)-\sin (b-a)$
$$2\sin x\ (\cos 2x + \cos 4x) = \sin {5x}-\sin x $$
Then
$$\begin{align} \int_0^\pi \frac{\sin{5x}}{\sin x}dx =&\int_0^\pi(1+2\cos 2x + 2\cos4x)\ dx= \int_0^\pi dx=\pi \end{align}$$