Gamma function: ∫ sin⁶(x/2) cos⁶(x/2) dx

Let I = ∫₀^π sin⁶(x/2) cos⁶(x/2) dx. Put x/2 = t, dx = 2 dt; limits 0 to π/2.

I = 2 ∫₀^(π/2) sin⁶t cos⁶t dt

Use ∫₀^(π/2) sin^p t cos^q t dt = Γ((p+1)/2) Γ((q+1)/2) / (2Γ((p+q+2)/2)), with p = q = 6:

I = 2 · Γ(7/2) Γ(7/2) / (2 Γ(7))
  = Γ(7/2)² / Γ(7)

Γ(7/2) = (5/2)(3/2)(1/2)√π = 15√π/8, so Γ(7/2)² = 225π/64. Γ(7) = 720.

I = (225π/64)/720 = 225π/46080 = 5π/1024

Result: I = 5π/2¹⁰.

Note on the printed value: the question states 5π/2¹¹, but the correct value from the working above is 5π/2¹⁰ = 5π/1024 ≈ 0.01534 (a numerical check of the integral agrees). The printed 2¹¹ appears to be a misprint, or comes from a slightly different integrand. The method (substitute x/2 = t, then Beta-Gamma) is what the exam expects.