Understanding the Integral
The integral we need to evaluate is
\[
\int_0^{2\pi} \{ \sin(\sin x) \sin(\cos x) \} \, dx
\]
where \(\{.\}\) denotes the fractional part of the expression.
Fractional Part Concept
- The fractional part of a number \(y\) is defined as \(y - \lfloor y \rfloor\), where \(\lfloor y \rfloor\) is the greatest integer less than or equal to \(y\).
- Since both \(\sin(\sin x)\) and \(\sin(\cos x)\) produce values in the range \([-1, 1]\), their product \(\sin(\sin x) \sin(\cos x)\) also lies within this range.
Range of the Function
- The product \(\sin(\sin x) \sin(\cos x)\) oscillates between \(-1\) and \(1\).
- Therefore, for the fractional part, we have:
\[
\{\sin(\sin x) \sin(\cos x)\} = \sin(\sin x) \sin(\cos x) + 1 \quad \text{(for negative values)}
\]
\[
\{\sin(\sin x) \sin(\cos x)\} = \sin(\sin x) \sin(\cos x) \quad \text{(for non-negative values)}
\]
Symmetry of the Integral
- The function \(\sin(\sin x)\) and \(\sin(\cos x)\) are periodic with a period of \(2\pi\).
- The integral over one full period (0 to \(2\pi\)) can exploit symmetry properties around \(\pi\).
Final Evaluation
To evaluate the integral, notice that over \(0\) to \(2\pi\):
- Contributions from \(\sin(\sin x)\) and \(\sin(\cos x)\) will balance out due to their periodic and oscillatory nature.
- Thus, the average value of \( \{\sin(\sin x) \sin(\cos x)\} \) across one complete cycle tends towards zero.
Conclusion
The value of the integral is
\[
\int_0^{2\pi} \{ \sin(\sin x) \sin(\cos x) \} \, dx = 0
\]