how do i solve integral [θ=0 to θ=2pi] { dθ/(a + Sin^2 θ)^2 }?

Question

Update:

how do i solve integral [θ=0 to θ=pi/2 ] { dθ/(a Sin^2 θ)^2 }

i made mistake! theta is form 0 to pi/2

Update 2:

i wanna solve it by using

z=ae^(iθ)

dz=aie^(iθ) dθ

Update 3:

and sinθ=(z (1/z))/zi

Answer 1

Assuming that a > 0 (for convergence):

∫(θ = 0 to π/2) dθ/(a + sin²θ)²

= ∫(θ = 0 to π/2) dθ/(a + (1/2)(1 - cos(2θ))², via half angle identity

= ∫(θ = 0 to π/2) 4 dθ/(2a + 1 - cos(2θ))²

= ∫(x = 0 to π) 2 dx/(2a + 1 - cos x)², letting x = 2θ.

Now, use the Weierstrauss substitution (see link below):

http://en.wikipedia.org/wiki/Tangent_half-angle_su...

So, ∫(x = 0 to π) 2 dx/(2a + 1 - cos x)²

= ∫(t = 0 to ∞) 2 (2 dt/(1 + t²)) / [2a + 1 - (1 - t²)/(1 + t²)]², letting t = tan(x/2)

= ∫(t = 0 to ∞) 4(1 + t²) dt / [(2a + 1)(1 + t²) - (1 - t²)]²

= ∫(t = 0 to ∞) 4(1 + t²) dt / [2(a + (a + 1)t²)]²

= ∫(t = 0 to ∞) (1 + t²) dt / [a + (a + 1)t²]²

= ∫(t = 0 to ∞) (1 + t²) dt / [(a+1) (a/(a+1) + t²)]²

Using a little more algebra:

(1/(a+1)²) ∫(t = 0 to ∞) (1 + t²) dt / (a/(a+1) + t²)²

= (1/(a+1)²) ∫(t = 0 to ∞) (1/(a+1) + (a/(a+1) + t²)) dt / (a/(a+1) + t²)²

= (1/(a+1)²) ∫(t = 0 to ∞) [(1/(a+1))/(a/(a+1) + t²)² + 1/(a/(a+1) + t²)] dt

= (1/(a+1)³) ∫(t = 0 to ∞) [1/(a/(a+1) + t²)² + (a+1)/(a/(a+1) + t²)] dt

Now, let t = √(a/(a+1)) tan φ:

(1/(a+1)³) ∫(φ = 0 to π/2) [1/((a/(a+1)) sec²φ)² + (a+1)/((a/(a+1)) sec²φ)] √(a/(a+1)) sec²φ dφ

= (1/(a+1)³) √(a/(a+1)) ∫(φ = 0 to π/2) [cos²φ /((a/(a+1)))² + (a+1)/((a/(a+1))] dφ

= (1/(a+1)³) √(a/(a+1)) ∫(φ = 0 to π/2) ((a+1)²/a²) (cos²φ + a) dφ

= (a(a+1))^(-3/2) ∫(φ = 0 to π/2) (cos²φ + a) dφ

Apply the half angle identity one more time:

(a(a+1))^(-3/2) * ∫(φ = 0 to π/2) [(1/2) (1 + cos(2φ)) + a] dφ

= (a(a+1))^(-3/2) * [(1/2) (φ + sin(2φ)/2) + aφ] {for φ = 0 to π/2}

= (a(a+1))^(-3/2) * (1/2 + a) (π/2)

= π(2a+1) / [4(a(a+1))^(3/2)].

(Double checked on Wolfram Alpha in the cases of a = 1, 2.)

I hope this helps!