# evaluate the integral. (x^2)/(sqrt(16-25x^2)) dx (upper limit 0.8, lower 0) please explain as you solve, thanks so much!make sure you don't forget to include all the information when you start to...

evaluate the integral. (x^2)/(sqrt(16-25x^2)) dx (upper limit 0.8, lower 0)

please explain as you solve, thanks so much!

make sure you don't forget to include all the information when you start to solve the problem, the numerator!

*print*Print*list*Cite

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.` **

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`