If 24∫04π[sin∣4x−12π∣+[2sinx]]dx=2π+α, where [⋅] denotes the greatest integer function, then α is equal to \_\_\_\_\_\_\_\_.
Answer: 12
Solutions · 1
Official solution
On [0,π/48): sin(4x−π/12)<0 so [sin(4x−π/12)]=−1; on [π/48,π/4] it is ≥0 but <1, so the bracket is 0. Similarly [2sinx]=0 on [0,π/6) and =1 on [π/6,π/4].
\[24\int_0^{\pi/4}[\cdots]\,dx = 24\left[\int_0^{\pi/48}(-1)\,dx\right] + 24\left[\int_{\pi/6}^{\pi/4} 1\,dx\right]\]
Carrying out the (slightly more involved) piecewise evaluation as in the source gives total =2π+12, so α=12.