Let [t] denote the greatest integer ≤ t. Then 2/π ∫π / 6^(5 π / 6)(8[cosec x]-5[cot x]) d x is equal to __________.

Solution

Maths sir · 4 steps

Here cosec x stays in [1, 2), so its floor is 1; split the cot part wherever cot x hits an integer.

1. The cosec part. On (π/6, 5π/6), sin x lies in (1/2, 1], so cosec x lies in [1, 2) and [cosec x] = 1. The interval has length 2π/3, so ∫ 8[cosec x] dx = 8 × 2π/3 = 16π/3.
2. The cot part. cot x falls from √3 ≈ 1.73 down to -√3. It crosses the integers 1, 0, -1 at x = π/4, π/2, 3π/4, so the floor changes value at each of these points.

   Interval	cot x lies in	[cot x]	Length	Contribution
   (π/6, π/4)	(1, √3)	1	π/12	π/12
   (π/4, π/2)	(0, 1)	0	π/4	0
   (π/2, 3π/4)	(-1, 0)	-1	π/4	-π/4
   (3π/4, 5π/6)	(-√3, -1)	-2	π/12	-π/6

3. Add the pieces. ∫ [cot x] dx = π/12 - π/4 - π/6 = -π/3. So ∫ -5[cot x] dx = -5 × (-π/3) = 5π/3.
4. So. The integral is 16π/3 + 5π/3 = 7π, and 2/π × 7π = 14.

The idea to keep. For an integral with a floor function, find where the inside function crosses each integer and split there. For negative values, the floor goes down: [-0.5] = -1, not 0.

Answer from the NTA final answer key. Solution by PrepWiser.