1. ## trig substitution problem

can someone solve

∫ square root of (9-4x^2dx)

2. Originally Posted by cummings15
can someone solve

∫ square root of (9-4x^2dx)
$\int\sqrt{9-4x^2}\, dx=3\int \sqrt{1-\left(\tfrac{2}{3}x\right)^2}\, dx=\ldots$

Substituting $z := \tfrac{2}{3}x$, thus $dz = \tfrac{2}{3}dx$ and $dx = \tfrac{3}{2}dz$. This gives you

$=3\int \sqrt{1-z^2}\cdot \tfrac{3}{2}\,dz=\tfrac{9}{2}\int \sqrt{1-z^2}\, dz$

and, maybe, you kow how to solve this last integral...

3. i think the final answer is

(2/3)(9-4x^2)(3/2) + C

4. Originally Posted by cummings15
i think the final answer is

(2/3)(9-4x^2)(3/2) + C
No, I don't think so, because you have

$\int \sqrt{1-z^2}\, dz=\frac{1}{2}\left(\arcsin(z)+z\sqrt{1-z^2}\right)+C$

which is perhaps a little more complicated than you thought.

5. ok so i tried solving that and i got

(9/4)arcsin(2x/3)+2x(square root of 9-4x^2)+c

6. Originally Posted by cummings15
ok so i tried solving that and i got

(9/4)arcsin(2x/3)+2x(square root of 9-4x^2)+c
Why don't you just use Wolfram Mathematica Online Integrator to check your result? - If it is not quite correct, then at least it is close...

7. here are my choices

2/3(9-4x^2)^3/2 + C

1/3tan^-1(2x/3)+c

square root of (9-4x^2)^3/2 divided by 8x

9/4arcsin(2x/3) + 2x square root of (9-4x^2) + C

i think it is either the first or last choice

it didn't show up in wolfram

8. Originally Posted by cummings15
here are my choices

2/3(9-4x^2)^3/2 + C

1/3tan^-1(2x/3)+c

square root of (9-4x^2)^3/2 divided by 8x

9/4arcsin(2x/3) + 2x square root of (9-4x^2) + C

i think it is either the first or last choice
I get (as does Wolfram's Integrator): $\int \sqrt{9-4x^2}\, dx=\tfrac{9}{4}\arcsin\left(\tfrac{2x}{3}\right)+{ \color{red}\tfrac{x}{2}}\sqrt{9-4x^2}+C$

it didn't show up in wolfram
Well, you have to enter the integrand first: I could not get a link to the result because that particular link got mangled when I tried to insert it into my post.

9. Originally Posted by cummings15
can someone solve

∫ square root of (9-4x^2dx)
hi

or make a substitution of x=3/2 sin theta