Simple question regarding this trig-sub integral: integral of 2/(9-x^2)^3/2 dx

[Clandestiny]

Hey everyone, I just noticed on a worksheet that I got a problem wrong. My teacher has provided the step by step solution for it, but there's one step that doesn't make sense to me.

The problem:

integral of 2/(9-x^2)^3/2 dx.

Instructions indicate x = 3 sin theta. So dx would obviously equal 3 cos theta, d(theta).

So the step that confuses me... my teacher got the constant number 2 after simplifying/substituting. I continually pull out the constant 2/9. My answer looks 'almost' identical, with the exception of this.

I'm getting the 9 in the denominator by simplifying the original denominator this way:
[sqrt(9-9sin^2(theta))]^3 = [sqrt(9(1-sin^2(theta)))]^3 = [sqrt(9(cos^2(theta)))]^3 = (3cos(theta))^3 = 27cos(theta).

The numerator acquires a 6 after multiplying it with 3cos(theta), so 6/27 simplifies to 2/9 correct?
Is the issue that I don't have to factor out the 9 when using the Pythagorean identity for cos^2(theta)? So sqrt(9-9sin^2(theta)) = sqrt(cos^2(theta))?

I'm hoping that was clear, my apologies if it wasn't. And thanks in advance!

 

[Deleted member 4993]

Hey everyone, I just noticed on a worksheet that I got a problem wrong. My teacher has provided the step by step solution for it, but there's one step that doesn't make sense to me.

The problem:

integral of 2/(9-x^2)^3/2 dx.

Instructions indicate x = 3 sin theta. So dx would obviously equal 3 cos theta, d(theta).

So the step that confuses me... my teacher got the constant number 2 after simplifying/substituting. I continually pull out the constant 2/9. My answer looks 'almost' identical, with the exception of this.

I'm getting the 9 in the denominator by simplifying the original denominator this way:
[sqrt(9-9sin^2(theta))]^3 = [sqrt(9(1-sin^2(theta)))]^3 = [sqrt(9(cos^2(theta)))]^3 = (3cos(theta))^3 = 27cos(theta).

The numerator acquires a 6 after multiplying it with 3cos(theta), so 6/27 simplifies to 2/9 correct?
Is the issue that I don't have to factor out the 9 when using the Pythagorean identity for cos^2(theta)? So sqrt(9-9sin^2(theta)) = sqrt(cos^2(theta))?

I'm hoping that was clear, my apologies if it wasn't. And thanks in advance!

\(\displaystyle \displaystyle{\int \dfrac{2}{(9-x^2)^{\frac{3}{2}}} dx}\)

substitute:

x = 3*sin(Θ) → dx = 3 cos(Θ) dΘ

\(\displaystyle \displaystyle{\int \dfrac{2 * 3 *cos(Θ) }{[9-(3*sin(Θ))^2]^{\frac{3}{2}}}dΘ}\)

\(\displaystyle \displaystyle{\int \dfrac{2 * 3 *cos(Θ)}{9^{\frac{3}{2}}[1-sin^2(Θ)]^{\frac{3}{2}}} dΘ}\)

\(\displaystyle \displaystyle{\int \dfrac{2 * 3 *cos(Θ)}{3*9*cos^3(Θ)} dΘ}\)

\(\displaystyle \displaystyle{\int \frac{2}{9} *sec^2(Θ) dΘ}\)