Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem : 

Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$

My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$,
$$
\begin{align}
\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx &= \frac{105}{19}\int^{\pi/2}_0 \frac{\sin (4\pi -8x)}{\cos x}\ \text dx\\
&= \frac{105}{19}\int^{\pi/2}_0 -\frac{\sin 8x}{\cos x}\ \text dx
\end{align}
$$
But it seems it won't work please help thanks

Using formulas $\sin2a = 2\sin a\cos a,\quad \cos a \cos b = \frac12(\cos(a+b)-\cos(a-b))$, have:
$$\dfrac{\sin 8x}{\sin x} = 8\cos 4x\cos 2x\cos x = 4\cos4x(\cos3x+\cos x) = 2(\cos7x+\cos 5x+\cos 3x +\cos x),$$so
\begin{align}
\dfrac {105}{19}\int_0^\limits\dfrac\pi2\dfrac{\sin 8x}{\sin x} &= \dfrac {210}{19}\int_0^\limits\dfrac\pi2(\cos7x+\cos 5x+\cos 3x +\cos x)\,dx =\\
&= \dfrac {210}{19}\left(\frac17\sin7x+\frac15\sin 5x+\frac13\sin 3x +\sin x\right)\biggr|_0^\frac\pi2 =
\\&= \frac{210}{19}\left(-\frac17+\frac15-\frac13+1\right) = \frac{210}{19}\cdot\frac{76}{105} = 8.
\end{align}
