If 24∫0π4(sin|4x–π12|+[2sinx])dx=2π+α, where [·] denotes the greatest integer function, then α is equal to _________. [2025]
(12)
Let I=24∫0π4(sin |4x–π12|+[2 sin x])dx
=24[∫0π/48[–sin (4x–π12)]dx+∫π/48π/4sin (4x–π12)dx+∫0π/6[0]dx+∫π/6π/4[1]dx]
=24[(1–cosπ12)–(–1–cosπ12)+π4–π64]
=24(12+π12)=12+2π
⇒ I=12+2π=2π+α [Given]
On comparing, we get α=12.