# Homework Help: Trigonometric substitution dx / (x^2(9-4x^2)^(1/2))

1. Mar 19, 2012

### ex81

The problem is to solve the integral of :

dx / (x^2(9-4x^2)^(1/2)) using trig substituition, which I really don't understand.

Formula's that are useful:

The integral table
Integral 1/(u^2(a^2-u^2)^(1/2))=

-((a^2-u^2)^(1/2)) / (ua^2)

Work so far:

Figuratively banging my head against this problem for 4 days...
And so the answer I have is -((9-4x^2)^(1/2))/18x. Where-as my professor's answer is -((9-4x^2)^(1/2))/9x

So I am not even sure how to solve this correctly, nor am I sure of my answer...

2. Mar 19, 2012

### scurty

Your professor has the correct answer! Let me try to guess where you got messed up.

So you have the solution to the general integral $\displaystyle\int \frac{dx}{x^2 \sqrt{a^2-x^2}} = -\frac{\sqrt{a^2-x^2}}{ua^2}$. The problem is, the integral $\displaystyle\int \frac{dx}{x^2 \sqrt{9-4x^2}}$ is not in that form. Note that the coefficient of both $x^2$ terms must be 1. Your first step should be to rewrite $\displaystyle\int \frac{dx}{x^2 \sqrt{9-4x^2}}$ as $\displaystyle\int \frac{c}{x^2 \sqrt{d^2-x^2}} dx = c \cdot \int \frac{dx}{x^2 \sqrt{d^2-x^2}}$ where $c$ and $d^2$ are constants. Then use your formula and simplify. You will get your professor's answer then!

Also, if you can, try to not use the trig tables and compute the integral yourself. You will see it come out as your professor's answer as well.

3. Mar 21, 2012

### ex81

I am not sure what you are referring to by trig table, by chance are you referring to the integral table?

As to solving it
a=3
u= x^2
du = 2
u^2 = 4x^2
(u^2)/4 =x^2

even pulling the constants generated by the du, and the (u^2)/4 doesn't make it work.

I mostly was trying to use the integral table to check what I was doing. I still have to figure out trig substitution.

4. Mar 22, 2012

### scurty

I meant the integral table, sorry.

Okay, $\frac{d}{dx} (x^2) = 2x$, not 2. so you can't solve this by u-substitution, at least not easily. Did you get to trigonometric substitution yet?

Forget about what I said for you to try to solve it. Have you tried rewriting $\displaystyle\int \frac{dx}{x^2 \sqrt{9-4x^2}}$ as $\displaystyle\int \frac{c}{x^2 \sqrt{d^2-x^2}} dx = c \cdot \int \frac{dx}{x^2 \sqrt{d^2-x^2}}$? When you do this you can use your formula you posted in the first post and everything will work out when it is simplified. The problem, in your integral, is that the coefficient of both $x^2$ terms is not 1! You need to fix that by factoring out of the square root. Does that make sense?

5. Mar 26, 2012

### ex81

no, figured out trig substitution yet....

the only way I know to factor the sqrt out of the bottom is to put it on top.
∫ √(9-4x^2)/(9x^2-4x^4). Dx

that looks equally ugly,

6. Mar 26, 2012

### scurty

I meant factor out of a term from the square root, not get rid of the square root!

Try factoring a 4 out from the square root. :)

7. Mar 26, 2012

### ex81

haha, well that makes this fun. :D

∫ 1/(x2√(4(9/4-x2)) dx

∫ 1/(2x2√(9/4-x2) dx

d^2 = 9/4

and that 1/4 =c so it pulls out

8. Mar 26, 2012

### scurty

Yep! Do you see how to finish off the problem now?

9. Apr 7, 2012

### ex81

I saw how to solve it the way I was not supposed to solve it.

Still trying to figure out trig substitution...

10. Apr 7, 2012

### SammyS

Staff Emeritus
Sketch a right triangle with hypothenuse of length 3, (b/c 32 = 9) and one leg of length 2x, (b/c (2x)2 = 4x2). The other leg will have length of √(9-4x2).

Let θ be one of the acute angles of the triangle.

If θ is the angle opposite the leg of length 2x, then sin(θ) = 2x/3, so x = (3/2)sin(θ).

What are dx, and √(9-4x2) ?

11. Apr 8, 2012

### HallsofIvy

Presumably you know that $sin^2(\theta)+ cos^2(\theta)= 1$ so that $1- sin^2(\theta)= cos^2(\theta)$. If you have $a- bx^2$ in an integrand you can use the substitution $x= \sqrt{a/b}sin(\theta)$ so that $a- bx^2= a- b(a/b)sin^2(\theta)= acos^2(\theta)$.

Of course, $dx= \sqrt{a/b}cos(\theta)d\theta$ and $x^2= (a/b)sin^2(/theta)$.