If ∫a01/(4+x²)dx = π/8, find the value of a.

Let I = ∫a01/(4+x²)dx
Put x = 2u ⇒ dx = 2du
=∫a01/(4+(2u)²)2du
=∫a01/(2u²+4)2du = 1/2∫a01/(u²+1)du
=1/2[tan⁻¹u]a0 + C
Substitute back u = x/2
=1/2[tan⁻¹(x/2)]a0 + C
1/2[tan⁻¹(a/2)] = π/8
tan⁻¹(a/2) = π/4
a/2 = tan(π/4)
⇒ a/2 = 1
Therefore, a = 2.