What is # int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx #?

Question

Charles Anctil · Accepted Answer

# int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx = 11/405 sqrt(3) #

We seek: # I = int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx # We can perform a substitution of the form: # u = 4-9x^2 iff 9x^2=4-u#, and #(du)/dx = -18x # This substitution will require an associated change of limits from #x# to #u#, and we have: When #x= { (1/3), (2/3) :} => u= { (3), (0) :}# And we manipulate the integral as follows: # I = int_(1/3)^(2/3) \ (9x^2)/9(-18x)/(-18)sqrt(4-9x^2) \ dx # # \ \ = -1/162 \ int_(1/3)^(2/3) \ (9x^2)(-18x)sqrt(4-9x^2) \ dx # # \ \ = -1/162 \ int_(3)^(0) \ (4-u)sqrt(u) \ du # # \ \ = 1/162 \ int_(0)^(3) \ (4-u)u^(1/2) \ du # # \ \ = 1/162 \ int_(0)^(3) \ 4u^(1/2) - u^(3/2) \ du # # \ \ = 1/162 \ [ (4u^(3/2))/(3/2) - (u^(5/2))/(5/2) ]_(0)^(3) # # \ \ = 2/162 [ (4u^(3/2))/(3) - (u^(5/2))/(5) ]_(0)^(3) # # \ \ = 1/81 (4/3 (3)^(3/2) - 1/5 (3)^(5/2) ) # # \ \ = 1/81 (4/3 (3sqrt(3)) - 1/5 (9sqrt(3)) ) # # \ \ = 1/81 (12/3-9/5) sqrt(3) # # \ \ = 1/81 (11/5) sqrt(3) # # \ \ = 11/405 sqrt(3) # So the last answer is correct.

William Abdalla · Answer

undefined To solve the integral ∫(1/3)^(2/3) x^3√(4-9x^2) dx, we can use trigonometric substitution. Let's let x = (2/3)sin(θ). Then dx = (2/3)cos(θ)dθ.

Now, substitute x and dx into the integral:
∫(1/3)^(2/3) x^3√(4-9x^2) dx = ∫sin^3(θ)√(4-9(2/3)^2sin^2(θ))(2/3)cos(θ)dθ.

This simplifies to:
(2/3)∫sin^3(θ)√(4-4sin^2(θ))cos(θ)dθ.

Using the identity sin^2(θ) + cos^2(θ) = 1, we have sin^2(θ) = 1 - cos^2(θ). Substitute this into the integral:
(2/3)∫sin^3(θ)√(4-4(1-cos^2(θ)))cos(θ)dθ = (2/3)∫sin^3(θ)√(4cos^2(θ))cos(θ)dθ.

Now, simplify inside the square root:
(2/3)∫sin^3(θ)√(4cos^2(θ))cos(θ)dθ = (2/3)∫sin^3(θ)(2cos(θ))dθ = (4/3)∫sin^3(θ)cos(θ)dθ.

Now, we can use the trigonometric identity ∫sin^m(θ)cos^n(θ)dθ = -(cos^(n-1)(θ)sin^(m+1)(θ))/(m+1) + (n-1)/(m+1)∫sin^m(θ)cos^(n-2)(θ)dθ when m is odd.

Applying this identity with m = 3 and n = 1:
(4/3)∫sin^3(θ)cos(θ)dθ = (4/3) * [-(cos^0(θ)sin^4(θ))/4 + 3/4∫sin^3(θ)cos(θ)dθ].

Now, we have a simpler integral: ∫sin^3(θ)cos(θ)dθ. Using the same trigonometric identity, we can solve this integral. After solving for θ, we'll need to go back to x using the substitution x = (2/3)sin(θ) to obtain the final answer in terms of x.