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

Method $1$
$1.$ Use the identity
$$2\sin x \sum_{k=1}^{n}\cos(2k-1)x = \sin2nx$$
which can easily be verified by using
$$2\sin x \cos(2k-1)x = \sin 2k x - \sin 2(k-1) x$$
and the telescoping property of the sums.
$2.$ Use the above formula to get
$$\frac{\sin 2nx}{\sin x} = 2 \sum_{k=1}^{n}\cos(2k-1)x$$
$3.$ Integrate to obtain
$$\begin{align}
F(x) &=\int \frac{\sin 2nx}{\sin x} \, dx \\
&= 2 \sum_{k=1}^{n} \int \cos(2k-1)x \, dx \\
&= 2 \sum_{k=1}^{n} \frac{1}{2k-1} \sin(2k-1)x + C
\end{align}$$
$4.$ Evaluating the definite integral results in harmonic partial sums
$$\begin{align}
I &= \int_{0}^{\frac{\pi}{2}} \frac{\sin 2nx}{\sin x} \\
&= F(\frac{\pi}{2})-F(0)\\
&= 2 \sum_{k=1}^{n} \frac{\sin (2k-1) \frac{\pi}{2}}{2k-1} - 0\\
&= \boxed{2 \sum_{k=1}^{n} \frac{(-1)^{k+1}}{2k-1}}
\end{align}$$
In your case, you may choose $n=4$.
