25-9x2=52-32x2=32((5/3)2-x2).
Let x=(5/3)sinθ, then dx=(5/3)cosθdθ, and (5/3)2-x2=(5/3)2-(5/3)2sin2θ=(5/3)2(1-sin2θ)=((5/3)cosθ)2 (trig identity).
So √(25-9x2)dx=[√(32((5/3)cosθ)2)][(5/3)cosθdθ]=
5cosθ.(5/3)cosθdθ=(25/3)cos2θdθ.
cos(2θ)=2cos2θ-1, so cos2θ=½(1+cos(2θ)) (trig identity).
(25/3)cos2θdθ=(25/6)(1+cos(2θ))dθ. Therefore, (25/6)∫(1+cos(2θ))dθ=(25/6)(θ+sin(2θ)/2).
Now to convert and apply the limits.
When x=-5/3, -5/3=(5/3)sinθ, sinθ=-1, θ=3π/2 (lower limit); x=5/3, sinθ=1, θ=π/2 (upper limit). However, 3π/2>π/2, so the limits need to be reversed.
Plug these revised limits into the integrated expression:
(25/6)[(3π/2+sin(3π)/2)-(π/2+sin(π)/2)]=(25/6)[π]=25π/6.