# Trigonometric Substitution 11x^2dx/(25-x^2)^(3/2)

1. Jun 13, 2013

### FallingMan

1. The problem statement, all variables and given/known data
Integral (11x^2)/(25-x^2)^(3/2) dx from 0 to (5*sqrt(3))/2

2. Relevant equations

sin^2(θ) = 1 - cos^2(θ)

3. The attempt at a solution

1. Factor out 11 from integral for simplicity.

11 * integral (x^2)/(25-x^2)^(3/2)

2. Re-write denominator of integral to look similar to 1-cos^2(θ)

11 * integral (x^2)/(25(1-(1/25)x^2)))^(3/2)dx

3. Equate cos^2(θ) and (1/25)x^2

cos(θ) = (1/5)x
θ = arccos(1/5*x)
x = 5cos(θ)
dx = -5sin(θ)dθ

4. Substitute cos^2(θ) = (1/25)x^2 into integral, Substitute -5sin(θ)dθ = dx

11*(-5) * integral (x^2)(sin(θ)/(25(1-cos^2(θ)))^(3/2)

5. Substitute sin^2 for (1-cos^2)

-55 * integral (x^2)(sin(θ)/(25(sin^2(θ))^(3/2)

No idea what to do from here.

I have a feeling my approach in general is totally off. Any advice would be greatly appreciated.

2. Jun 13, 2013

### Pranav-Arora

If x^2/25=cos^2(θ), what is x^2? Substitute that in your final expression.

3. Jun 13, 2013

### FallingMan

Hi Pranav-Arora. I'll try...

-55 * integral (25cos^2(θ))(sin(θ)/(25(sin^2(θ))^(3/2)

Not sure how to proceed from there. I gotta go to classes now, but I'll be back to think more about the problem shortly.