Area inside polar curves

Find the area of the region inside: r=5sin(theta) but outside: r=4

Points of intersection:
5 sin θ = 4
==> θ = arcsin(4/5) and pi - arcsin(4/5).

Between these values, note that 5 sin θ > 4.

So, the area equals (and using symmetry about π/2)
2 * ∫(θ = arcsin(4/5) to π/2) (1/2) [(5 sin θ)^2 - 4^2] dθ
= ∫(θ = arcsin(4/5) to π/2) [25 sin^2(θ) - 16] dθ
= ∫(θ = arcsin(4/5) to π/2) [(25/2) (1 - cos(2θ)) - 16] dθ
= ∫(θ = arcsin(4/5) to π/2) (-1/2) [7 + 25 cos(2θ)] dθ
= (-1/2) [7θ + (25/2) sin(2θ)] {for θ = arcsin(4/5) to π/2}
= (-1/2) [7θ + 25 sin θ cos θ] {for θ = arcsin(4/5) to π/2}
= (-1/2) [(7π/2 + 0) - (7 arcsin(4/5) + 25 * 4/5 * 3/5)], via 'sohcahtoa'
= (1/2) [12 - 7π/2 + 7 arcsin(4/5)].

I hope this helps!