# `int_0^0.6 x^2 /sqrt(9 - 25x^2) dx` Evaluate the integral

## Expert Answers

`int_0^0.6x^2/sqrt(9-25x^2)dx`

Let's first evaluate the indefinite integral by applying the integral substitution,

Let `x=3/5sin(u)`

`=>dx=3/5cos(u)du`

Plug in the above in the integral,

`intx^2/sqrt(9-25x^2)dx=int(3/5sin(u))^2/(sqrt(9-25(3/5sin(u))^2))(3/5cos(u))du`

`=int(27sin^2(u)cos(u))/(125sqrt(9-9sin^2(u)))du`

`=int(27sin^2(u)cos(u))/(125sqrt(9(1-sin^2(u))))du`

use the identity:`1-sin^2(x)=cos^2(x)` `<br> `

`=int(27sin^2(u)cos(u))/(125*3sqrt(cos^2(u)))du`

`=int(9sin^2(u)cos(u))/(125cos(u))du`

`=9/125intsin^2(u)du`

Now use the identity:`sin^2(x)=(1-cos(2x))/2`

`=9/125int(1-cos(2u))/2du`

`=9/250int(1-cos(2u))du`

`=9/250(int1du-intcos(2u)du)`

`=9/250(u-sin(2u)/2)`

We have taken `x=3/5sin(u)`

`=>u=arcsin(5/3x)`

Substitute back u and add a...

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`int_0^0.6x^2/sqrt(9-25x^2)dx`

Let's first evaluate the indefinite integral by applying the integral substitution,

Let `x=3/5sin(u)`

`=>dx=3/5cos(u)du`

Plug in the above in the integral,

`intx^2/sqrt(9-25x^2)dx=int(3/5sin(u))^2/(sqrt(9-25(3/5sin(u))^2))(3/5cos(u))du`

`=int(27sin^2(u)cos(u))/(125sqrt(9-9sin^2(u)))du`

`=int(27sin^2(u)cos(u))/(125sqrt(9(1-sin^2(u))))du`

use the identity:`1-sin^2(x)=cos^2(x)` `<br> `

`=int(27sin^2(u)cos(u))/(125*3sqrt(cos^2(u)))du`

`=int(9sin^2(u)cos(u))/(125cos(u))du`

`=9/125intsin^2(u)du`

Now use the identity:`sin^2(x)=(1-cos(2x))/2`

`=9/125int(1-cos(2u))/2du`

`=9/250int(1-cos(2u))du`

`=9/250(int1du-intcos(2u)du)`

`=9/250(u-sin(2u)/2)`

We have taken `x=3/5sin(u)`

`=>u=arcsin(5/3x)`

Substitute back u and add a constant C to the solution,

`=9/250(arcsin(5/3x)-1/2sin(2arcsin(5/3x)))+C`

Now let's evaluate the definite integral,

`int_0^0.6x^2/sqrt(9-25x^2)dx=[9/250(arcsin(5/3x)-1/2sin(2arcsin(5/3x)))]_0^0.6`

`=[9/250(arcsin(5/3*0.6)-1/2sin(2arcsin(5/3*0.6)))]-[9/250(arcsin(5/3*0)-1/2sin(2arcsin(5/3*0)))]`

`=[9/250(arcsin(1)-1/2sin(2arcsin(1)))]-[9/250(arcsin(0)-1/2sin(2arcsin(0)))]`

`=[9/250(pi/2-1/2sin(2*pi/2))]-[0]`

`=[9/250(pi/2-1/2sin(pi))]`

`=[9/250(pi/2-1/2*0)]`

`=(9pi)/500`

`=0.05655`

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You need to solve the definite integral, such that:

`sqrt(9 - 25x^2) = sqrt (9(1 - 25x^2/9))`

`sqrt (9(1 - 25x^2/9)) = 3sqrt(1 - 25x^2/9)`

You need to use the following substitution `(5x/3) = sin u => (5/3)dx = cos u du`

`x^2 = (9/25)*sin^2 u`

`int_0^(0.6) x^2/(3sqrt(1 - 25x^2/9)) dx = int_(u_1)^(u_2) ((9/25)*sin^2 u)/(3sqrt(1 - sin^2 u)) du`

Using the trigonometric formula` 1 - sin^2 u = cos ^2 u` yields:

`int_(u_1)^(u_2) ((3/25)*sin^2 u)/(sqrt(cos ^2 u)) du`

`int_(u_1)^(u_2) ((3/25)*sin^2 u)/(cos u) du = int_(u_1)^(u_2) ((3/25))tan u*sin u du`

Replace `tan u = (2u)/(1-u^2)` and `sin u = (2u)/(1+u^2), ` where `u = (tan(a/2))  `

`int_(u_1)^(u_2) ((3/25))tan u*sin u du = int_(u_1)^(u_2) ((3/25)) *(4u^2)/(1-u^4) * (1 + u^2) du = (12)/(25)*int_(u_1)^(u_2) u^2/(1 - u^2)`

`(12)/(25)*int_(u_1)^(u_2) u^2/(1 - u^2) = - (12)/(25)*int_(u_1)^(u_2) (1 - u^2 - 1)/(1 - u^2) `

`- (12)/(25)*int_(u_1)^(u_2) (1 - u^2 - 1)/(1 - u^2) = - (12)/(25)*(u - ln|(1-u)/(1+u)|)|_(u_1)^(u_2)`

`int_0^(0.6) x^2/(3sqrt(1 - 25x^2/9)) dx = - (12)/(25)*(arcsin(5x/3) - ln|(1-arcsin(5x/3))/(1+arcsin(5x/3))|)|_(0)^(0.6)`

`int_0^(0.6) x^2/(3sqrt(1 - 25x^2/9)) dx = - (12)/(25)*(arcsin 1) =  (12)/(25)*(pi/2) = (6pi/25)`

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