Plenty of Integrate by Parts Part 2?

I did use by parts for this problem. :P

But I don't find my method satisfactory. Here's how I approached it:

Use integration by parts to get:

$\displaystyle \frac{8}{3}\int_0^{\infty}\frac{\sin^7 x \cos x}{x^3}\,dx$

Again from IBP:

$\displaystyle \frac{4}{3}\left(\int_0^{\infty} \frac{7\sin^6 x \cos^2 x-\sin^8 x}{x^2}\,\right)\,dx$

$=\displaystyle \frac{4}{3}\left(7\int_0^{\infty} \frac{\sin^6 x}{x^2}\,dx-\int_0^{\infty} \frac{\sin^8 x}{x^2}\,dx\right)$

$=\displaystyle 7I_1-I_2$

I evaluate $I_1$ . Use IBP to get:

$\displaystyle I_1=6\int_0^{\infty}\frac{\sin^5 x\cos x}{x}\,dx$

This is the step where I took help from W|A. I fed it with $\sin^5 x\cos x$ and it gave me the following alternate form: $5\sin (2x)-4\sin(4x)+\sin(6x)$ . Using this I can easily evalaute $I_1$ . I can prove the above with the use of complex numbers but I feel there is a much better way to solve the problem because I couldn't reach the summation Fey has shown. Please help.

Many thanks!