Question

How do you evaluate the integral `int x^3/sqrt(1-9x^2)`?

Answer 1

`-1/243*sqrt(1-9x^2)(2+9x^2)+C.`

Explanation:

Let, `I=intx^3/sqrt(1-9x^2)dx.`

We substitute `t^2=1-9x^2 rArr 9x^2=1-t^2.`

`:. x^2=1/9(1-t^2), and, 2xdx=1/9(-2t)dt,`

`" or, xdx=-1/9tdt.`

`:. I=int(x^2*xdx)/sqrt(1-9x^2),`

`=int1/9(1-t^2)*(-1/9)t*1/t*dt,`

`=1/81int(t^2-1)dt,`

`=1/81[t^3/3-t],`

`=1/243[t^3-3t],`

`=t/243[t^2-3],`

`=1/243*sqrt(1-9x^2)[1-9x^2-3],`

`rArr I= -1/243*sqrt(1-9x^2)(2+9x^2)+C.`

Answer 2

`int \ x^3/(sqrt(1-9x^2)) \ dx = - 1/243 (2+9x^2)sqrt(1-9x^2) + c`

Explanation:

We want to evaluate:

`I = int \ x^3/(sqrt(1-9x^2)) \ dx`

Let `u = 1-9x^2 => (du)/dx = -18x`; `x^2=1/9(1-u)`

Substituting into the integral we get:

`I = -1/18 \ int \ x^2/(sqrt(1-9x^2)) \ (-18x) \ dx`
`\ \ = -1/18 \ int \ (1/9(1-u))/(sqrt(u)) \ du`
`\ \ = -1/162 \ int \ 1/(sqrt(u)) - sqrt(u) \ du`
`\ \ = -1/162 ( u^(1/2)/(1/2) - u^(3/2)/(3/2)) + c`
`\ \ = - 1/162 ( 2sqrt(u) - 2/3usqrt(u) ) + c`
`\ \ = - 1/162 ( 2/3(3-u)sqrt(u) ) + c`
`\ \ = - 1/243 (3-u)sqrt(u) + c`

And, if we restore the substitution we get:

`I = - 1/243 (3-1+9x^2)sqrt(1-9x^2) + c`
`\ \ = - 1/243 (2+9x^2)sqrt(1-9x^2) + c`

Answer 3

`int \ x^3/(sqrt(1-9x^2)) \ dx = -1/243 (2+9x^2)sqrt(1-9x^2) + c`

Explanation:

Alternatively, using a trigonometric substitution:

We want to evaluate:

`I = int \ x^3/(sqrt(1-9x^2)) \ dx`

Let `sin theta = 3x => cos theta (d theta)/dx =3`; `\ \x=1/3sin theta`

Substituting into the integral we get:

`I = int \ (1/3 sin theta)^3/(sqrt(1-sin^2 theta)) \ (cos theta/3) \ d theta`

`\ \ = 1/81 int \ (sin^3 theta cos theta)/(sqrt(cos^2 theta)) \ d theta`

`\ \ = 1/81 int \ sin^3 theta \ d theta`

`\ \ = 1/81 (1/3cos^3 theta -cos theta ) + c`

`\ \ = -1/243 (3-cos^2 theta)costheta + c`

Using the identity `sin^2 theta + cos^2 theta -= 1 => cos^2 theta = 1-9x^2`

And, if we restore the substitution, using the above relationship, we get:

`I = -1/243 (3-1+9x^2)sqrt(1-9x^2) + c`
`\ \ = -1/243 (2+9x^2)sqrt(1-9x^2) + c`

Answer 4

`int x^3/sqrt(1-9x^2)dx =-((2+9x^2)sqrt(1-9x^2))/243+C`

Explanation:

Evaluate:

`int x^3/sqrt(1-9x^2)dx`

Substitute:

`x = 1/3 sint`

as the integrand is defined only for `x in (-1/3,1/3)` `t` varies in `(-pi/2,pi/2)`.

`int x^3/sqrt(1-9x^2)dx = int ((1/3sint)^3 d(1/3sint))/sqrt(1-9(1/3sint)^2)`

`int x^3/sqrt(1-9x^2)dx = 1/81 int (sin^3t cost)/sqrt(1-sin^2t)dt`

Now:

`sqrt(1-sin^2t) = sqrt(cos^2t) = cost`

as for `t in (-pi/2,pi/2)`, `cost` is positive.

`int x^3/sqrt(1-9x^2)dx = 1/81 int (sin^3t cost)/costdt =1/81 int sin^3tdt`

`int x^3/sqrt(1-9x^2)dx = 1/81 int (1-cos^2t)sintdt =1/81 int (cos^2t-1)d(cost)`

`int x^3/sqrt(1-9x^2)dx =(cos^3t-3cost)/243+C`

To undo the substitution:

`cost= sqrt(1-sin^2t) = sqrt(1-9x^2)`

so:

`int x^3/sqrt(1-9x^2)dx =cost(cos^2t-3)/243+C`

`int x^3/sqrt(1-9x^2)dx =cost(1-sin^2t-3)/243+C`

`int x^3/sqrt(1-9x^2)dx =-((2+9x^2)sqrt(1-9x^2))/243+C`