Use `x = 4/5 sin(theta)` so

`sqrt(16-25x^2) = sqrt(16-25(16/25sin(theta))) = 4 cos(theta)`

And our integral becomes

`int x^2/sqrt(16-25x^2) dx = int (16/25sin^2(theta))/(4cos(theta)) 4/5 cos(theta) d theta`

Simplifying we get

`= 16/125 int sin^2(theta) d theta = 8/125(theta - sin(theta)cos(theta)) + C`

Since `sin(theta) = 5/4x` and `cos(theta) = sqrt(1-25/16x^2)` we get

`int x^2/(sqrt(16-25x^2)) dx = 8/125(arcsin((5x)/4) - 5/4x(sqrt(1-25/16x^2))) + C`

So our integral becomes

`int^0.8_0 x^2/(sqrt(16-25x^2)) dx = [8/125(arcsin((5x)/4) - 5/4x(sqrt(1-25/16x^2))]^0.8_0`

Since 0.8 is 4/5 we can evaluate to get

`8/125(arcsin(1)-sqrt(1-1))-8/125(arcsin(0)+5/4(0)(1-25/16(0)))`

`=8/125arcsin(1) = 8/125pi/2=(4pi)/125`

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