I=∫−π/2π/23+[sinx]+[cosx]12(3+[x])dx
I=∫−π/2−1212(1)dx+∫−10212(2)dx+∫01312(3)dx+∫1π/2312(4)dx
I=6(2π−1)+12(0+1)+12(1−0)+16(2π−1)
I=3π−6+12+12+8π−16
I=11π+2
JEE Main 2026 — Mathematics Calculus
Let [⋅] denote the greatest integer function. Then ∫−2π2π(3+[sinx]+[cosx]12(3+[x]))dx is equal to :
Held on 28 Jan 2026 · Verified 6 Jul 2026.
13π+1
12π+5
11π+2
15π+4
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