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As part of a problem, I'm stuck evaluating the following integral

$$ \int_0^\pi \sin\left(\left|\frac{x}{2}-\frac{y}{2}\right|\right) \sin (2y)~dy.$$ I seek assistance in evaluating it. Thank you very much.

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  • $\begingroup$ Is there any interval defined for x and y? $\endgroup$ – Ram Jan 23 '13 at 5:43
  • $\begingroup$ no Ram. there is no interval. $\endgroup$ – Paul Jan 23 '13 at 5:59
  • $\begingroup$ I mean is there any relation between x and y, I hope x is like $ 0 \le x \le \pi$ then you can reduce into two intervals, $ sin(x/2 - y/2)$ in $0 \le y \le x$ and $ sin(y/2 - x/2)$ in $x < y \le \pi$ $\endgroup$ – Ram Jan 23 '13 at 6:16
  • $\begingroup$ @Ram: yes $0\le x \le \pi$. Thanks. $\endgroup$ – Paul Jan 23 '13 at 6:24
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First, break it into two pieces, one from $0$ to $x$, the other from $x$ to $\pi$. This way, you can write an equivalent integral without absolute value signs.

Then can you do $\int\sin(a-by)\sin(cy)\,dy$? It might help to remember there's a trig identity for $\sin A\sin B$.

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  • $\begingroup$ Would this work: $$\int_0^x \sin(x-y)\sin(2y)-\int_x^\pi sin(x-y)\sin(2y) dy?$$ $\endgroup$ – Paul Jan 23 '13 at 5:58
  • $\begingroup$ It appears you've misplaced the $2$s in the denominators but, aside from that, it looks like what I had in mind. $\endgroup$ – Gerry Myerson Jan 23 '13 at 6:11

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