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

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.#

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#

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 #

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#