If ∫π/6^(π/3) √(1-sin 2 x) d x=α+β √2+γ √3, where α, β and γ are rational numbers, then 3 α+4 β-γ is equal to _________.

Solution

Maths sir · 4 steps

Turn the root into |sin x - cos x|, then split the integral where the sign flips, at π/4.

1. Remove the root. Write 1 = sin² x + cos² x and sin 2x = 2 sin x cos x. Then 1 - sin 2x = (sin x - cos x)², so √(1-sin 2x) = |sin x - cos x|, never just sin x - cos x.
2. Split at pi/4. On [π/6, π/4] we have cos x ≥ sin x. On [π/4, π/3] we have sin x ≥ cos x.

   I = ∫_(π/6)^(π/4) (cos x - sin x) dx + ∫_(π/4)^(π/3) (sin x - cos x) dx

3. Evaluate both parts. The first part is [sin x + cos x]_(π/6)^(π/4) = √2 - 1/2 - √3/2. The second is [-cos x - sin x]_(π/4)^(π/3) = √2 - 1/2 - √3/2. Adding gives I = -1 + 2√2 - √3.
4. So. α = -1, β = 2, γ = -1, and 3α + 4β - γ = -3 + 8 + 1 = 6.

The common slip: 0. Writing √(1-sin 2x) as cos x - sin x over the whole interval, with no modulus, makes the integral come out as 0, so α = β = γ = 0 and the answer becomes 0.

The idea to keep. Whenever √(something)² appears, write it as a modulus. Then find where that something changes sign inside the limits and split the integral there.

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