Evaluate the following integral using the residue theorem

∫(0 to π/2) [a+sin^2(θ)]dθ = π/2sqrt(a^2+a) with a>0
Does anyone could help me with this exercise please?

First of all,
∫(θ = 0 to π/2) [a + sin^2(θ)] dθ
= ∫(u = 0 to π) [a + sin^2(u/2)] * du/2, letting θ = u/2
= (1/2) ∫(u = 0 to π) [a + sin^2(u/2)] du
= ∫(u = -π to π) [a + sin^2(u/2)] du, since the integrand is even
= ∫(u = -π to π) [a + (1/2)(1 - cos u)] du
= (1/2) ∫(u = -π to π) [(2a+1) - cos u] du.

Now, let z = e^(iu), dz = iz du to convert this to a contour integral about |z| = 1.
==> (1/2) ∫c [(2a+1) - (1/2)(z + 1/z)] * dz/(iz)
= (1/(4i)) ∫c [(4a+2) - (z + 1/z)] * dz/z
= (-1/(4i)) ∫c (z^2 - (4a+2)z + 1) dz/z^2.

This has a double pole at z = 0 with residue
lim(z→0) (d/dz) (z - 0)^2 * (z^2 - (4a+2)z + 1)/z^2
= lim(z→0) (d/dz) (z^2 - (4a+2)z + 1)
= lim(z→0) (2z - (4a+2))
= (-4a+2).

So, the integral equals (-1/(4i)) * 2πi * (-4a+2) = π(a - 1).
(Should your original integral have a radical in it, or be a fraction?)

I hope this helps!