Question about trig substitution intervals

[vande060]

Homework Statement

my professor tell me that when looking at the case ∫ √ (a^2 - x^2) , the trig substitution of course is asinϑ where -pi/2 ≤ ϑ ≤ pi/2. What I don't understand is why my professor tells me that when this term, √ (a^2 - x^2), is in the denomenator of the integrand that we must say -pi/2 < ϑ < pi/2 to avoid dividing by zero, but I don't see how you cna be dividing by zero in this case

here is an example we can look at

∫ x^2/√(9-25x^2)

Homework Equations

The Attempt at a Solution

∫ x^2/√(9-25x^2)

x = sinϑ
dx = cosϑ
√(9-25x^2) = 3cosϑ

∫ (9/25sin^2ϑ cosϑ dϑ) / 3cosϑ

I don't see how having -pi/2 ≤ ϑ ≤ pi/2 is any different from having -pi/2 < ϑ < pi/2 in this case, or any case for that matter. since you always have the constant in front of sine in the term (a^2 - x^2), the worst that can happen in the interval -pi/2 ≤ ϑ ≤ pi/2 is ( a^2 - 0) in the denominator.

 

[vela]

x = sinϑ
dx = cosϑ
√(9-25x^2) = 3cosϑ

This isn't correct.

∫ (9/25sin^2ϑ cosϑ dϑ) / 3cosϑ

I don't see how having -pi/2 ≤ ϑ ≤ pi/2 is any different from having -pi/2 < ϑ < pi/2 in this case, or any case for that matter. since you always have the constant in front of sine in the term (a^2 - x^2), the worst that can happen in the interval -pi/2 ≤ ϑ ≤ pi/2 is ( a^2 - 0) in the denominator.

If you let x=±π/2, you have sin x=±1 and a²-a²sin² x = 0.

 

[vande060]

This isn't correct.

If you let x=±π/2, you have sin x=±1 and a²-a²sin² x = 0.

Okay I understand now thank you