If ∫₀^(100π) (sin ²x)/e^(( x/π - [ x/π ] ))dx = (α π ³)/(1 + 4π ²),α ∈ R where [x] is the greatest integer less than or equal to x, then the value of α is :

Solution

Maths sir · 4 steps

The integrand repeats every π, so do one period and multiply by 100.

1. Spot the period. x/π - [x/π] is the fractional part {x/π}, which has period π, and sin² x also has period π. The interval [0, 100π] holds 100 periods, so the integral is 100∫₀^π e^(-x/π)sin² x dx, because on [0, π) the fractional part is just x/π.
2. Split the square. Write sin² x = (1 - cos 2x)/2. The first piece gives ∫₀^π e^(-x/π) dx = π(1 - e⁻¹).

   ∫₀^π e^(-x/π)cos 2x dx = [e^(-x/π)(-1/π cos 2x + 2 sin 2x)/(1/π² + 4)]₀^π = π(1 - e⁻¹)/(1 + 4π²)

3. Combine. One period gives 1/2π(1 - e⁻¹)(1 - 1/(1 + 4π²)) = 2π³(1 - e⁻¹)/(1 + 4π²). Multiplying by 100 gives 200(1 - e⁻¹)π³/(1 + 4π²).
4. So. Matching with απ³/(1 + 4π²) gives α = 200(1 - e⁻¹), which is option (a).

The idea to keep. When you see x - [x] inside an integral, look for the period first. Then ∫₀^(nT) f = n∫₀ᵀ f, and on one period the fractional part is just a plain linear function.

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