<s>[INST] 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
 [/INST] 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
$$\sin x = u$$
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