Help with an integration question with the substitution given?

Question

Please help, i tried to to answer it but the working is too complicated can anyone help out there, the question says:
Using substitution 5x=4cosØ to show
∫ √ (16-25x^2) dx = 4 π + 6√ 3 / 15

Answer 1

if 5x = 4cos θ, then
16 - 25x^2 = 16 - 16cos^2 θ = 16(1-cos^2 θ) = 16sin^2 θ

so, √(16-25x^2) = 4sinθ

5dx = -4sinθ dθ
dx = -4/5 sinθ dθ

∫√(16-25x^2) dx = ∫4sinθ (-4/5 sinθ) dθ
= -16/5 ∫ sin^2θ dθ

now, sin^2θ = (1-cos 2θ)/2, so we have

-8/5 ∫ 1-cos2θ dθ
= -8/5 (θ - 1/2 sin 2θ)
= -8/5 (θ - sinθ cosθ)

now, since 5x = 4cosθ
cosθ = 5/4 x
sinθ = √(1 - 25/16 x^2) = 1/4 √(16-25x^2)

and the solution is thus

-8/5 (arccos(5/4 x) - 5/4 x * 1/4 √(16-25x^2)
-8/5 (arccos(5x/4) - 5/16 √(16-25x^2))
= 1/2 √(16-25x^2) + 8/5 cos^-1 5x/4

Presumably if you use your limits of integration, that will produce the value you desire.