your method is correct, but this question is tagged as trig substitution, in fact this can be doen very easily by using a substitution

`I = int_0^0.8(x^2/(sqrt(16-25x^2)))dx`

substitute `x = 4/5 sin(u)`

this gives, `dx = 4/5 cos(u)`

and the limits, when `x = 0, u = 0 and x = 0.8, u = pi/2` (you can check this)

so ,

`I = int_0^(pi/2)(16/25 sin^2(u))/sqrt(16-25*16/25*sin^2(u))*4/5 cos(u) du`

`I = 16/125 int_0^(pi/2)(sin^2(u))/cos(u) * cos(u) du`

`I = 16/125int_0^(pi/2)sin^2(u) du`

we know from trignometric identities,

`cos(2u) = 1 -2sin^2(u)` this gives you,

`sin^2(u) = (1-cos(2u))/2`

so,

`I = 16/125int_0^(pi/2)(1-cos(2u))/2 du`

`I = 8/125int_0^(pi/2)(1-cos(2u)) du`

`int(1-cos(2u)) du = u -sin(2u)/2`

therefore,

`I = 8/125(pi/2 - sin(pi)/2 - (0 - sin(0)/2))`

`I = 8/125(pi/2 -0 - 0 + 0)`

`I = (4pi)/125`

You need to multiply by -25 both numerator and denominator such that:

`(-1/25)int (-25x^2dx)/(sqrt(16-25x^2))`

You need to add and subtract 16 to numerator such that:

`(-1/25)(int (16 -25x^2dx)/(sqrt(16-25x^2)) - int (16dx)/(sqrt(16-25x^2)))`

You need to evaluate the first integral, hence you should write `16 -25x^2` as `sqrt(16-25x^2)*sqrt(16-25x^2) ` such that:

`int (sqrt(16-25x^2)*sqrt(16-25x^2)dx)/(sqrt(16-25x^2))`

`int (sqrt(16-25x^2)) dx = int (sqrt(25(16/25 - x^2))) dx `

`int (sqrt(25(16/25 - x^2))) dx = 5int (sqrt(16/25 - x^2)) dx `

`5int_0^(4/5) (sqrt(16/25 - x^2)) dx = (5/2)(x*sqrt(16/25 - x^2) + (16/25)arcsin 5x/4)|_0^(4/5)`

`5int_0^(4/5) (sqrt(16/25 - x^2)) dx = (5/2)(4/5*sqrt(16/25 - 16/25) + (16/25)arcsin 1 - 0- (16/25)*arcsin 0)`

`5int_0^(4/5) (sqrt(16/25 - x^2)) dx = (5/2)(pi/2) = 5pi/4`

You need to evaluate the second integral such that:

`int (16dx)/(sqrt(16-25x^2)) = 16int (dx)/(sqrt(16-25x^2))`

`(16/5)int_0^(4/5) (dx)/(sqrt(16/25-x^2)) = (16/5)arcsin(5x/4)|_0^(4/5)`

`(16/5)int_0^(4/5) (dx)/(sqrt(16/25-x^2)) = (16/5)arcsin 1 - (16/5)arcsin 0`

`(16/5)int_0^(4/5) (dx)/(sqrt(16/25-x^2)) = (16/5)(pi/2)`

`(16/5)int_0^(4/5) (dx)/(sqrt(16/25-x^2)) = (16pi/10) = 8pi/5`

`int_0^(4/5) (x^2dx)/(sqrt(16/25-x^2)) = 5pi/4 - 8pi/5`

`int_0^(4/5) (x^2dx)/(sqrt(16/25-x^2)) = (25pi - 32pi)/20`

`int_0^(4/5) (x^2dx)/(sqrt(16/25-x^2)) = -7pi/20`

**Hence, evaluating the given definite integral yields
`int_0^(4/5) (x^2dx)/(sqrt(16/25-x^2)) = -7pi/20.`**

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