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
 A: 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 $n=4$.
A: 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}
A: HINT:
If $I_n=\int_0^{\pi/2}\dfrac{\sin nx}{\sin x}dx$
Using Prosthaphaeresis Formulas,
$$I_{m+2}-I_m=2\int_0^{\pi/2}\cos(m+1)x\ dx$$
A: Using approaches from here it seems to be that 
$$\int \frac{\sin(8x)}{\sin{x}}\mathrm{d}x=2\sin(x)+\frac{2}{3}\sin(3x)+\frac{2}{5}\sin(5x)+\frac{2}{7}\sin(7x)+C$$
Then doing the calculus for your integral we get $152/105$ so the answer is $8$.
A: Method $2$
Here are hints to another approach.
$1.$ You can observe from the De Moivre's formula that
$$\sin (8x) = 128\sin \left( x \right)\cos^7 {\left( x \right)} - 192\sin \left( x \right)\cos^5 {\left( x \right)} + 80\sin \left( x \right)\cos^3 {\left( x \right)} - 8\sin \left( x \right)\cos \left( x \right)$$
and hence
$$\frac{\sin(8x)}{\sin(x)}=128 \cos^7 {\left( x \right)} - 192 \cos^5 {\left( x \right)} + 80\cos^3 {\left( x \right)} - 8\cos \left( x \right)$$
$2.$ Next use this trick
$$\begin{align}
\cos^7 x= \cos^6 x \cos x = (1-\sin^2 x)^3 \cos x \\
\cos^5 x= \cos^4 x \cos x = (1-\sin^2 x)^2 \cos x \\
\cos^3 x= \cos^2 x \cos x = (1-\sin^2 x)^1 \cos x
\end{align}$$
$3.$ Finally, use the substitution
$$\sin x = u$$
