# Integrals of sqrt (simple, but I am getting frustrated!)

1. Mar 20, 2005

### Meghan

I have been studying all day, and this little bugger keeps bothering me.
How do I start this integral.
int{sqrt(1-4x^2)}
And what method do I use to solve it. Substitution or by Parts?

Last edited: Mar 20, 2005
2. Mar 20, 2005

### Data

can you think of some function $$f(u)$$ such that making the substitution $$x = f(u)$$ will make $$1-4x^2$$ into a perfect square?

Think trig.

3. Mar 20, 2005

### whozum

Try using trig substitutions, if youve learned that already. Substitute x = (sinu)/4, and the integral will become int{sqrt(1-sinu^2)}dx,

From there translate 1-sinu^2 to cosu^2 and go ahead and simplify sqrt(cos^2u).

Dont forget since you made a change of variable you must change dx to du, do so by deriving your new x equation (x = sinu/4) and putting it into your integral. dx = (cosu/4)*du, du = 4dx/cosu

So your integral will become int{cosu*4/cosu*du}. Your cosins will cancel and your final integral should become int{4*du}

Last edited: Mar 20, 2005
4. Mar 20, 2005

### Data

pft, sure, don't give her the chance to think about my question on her own!

5. Mar 20, 2005

### whozum

Sorry you weren't there when I posted my reply, you beat me to it :)

6. Mar 20, 2005

### Data

And since you've posted that reply already, I should point out that you actually want to sub $$x = \frac{\sin{u}}{2}$$, not $$x = \frac{\sin{u}}{4}$$!

7. Mar 20, 2005

### Meghan

But how does that help me, because I still don't understand what process to use or where to start with the sqrt?

8. Mar 20, 2005

### whozum

By making the trig substitution, you will eliminate the square root because you will be square rooting cos^2(x). The square root of cos^2(x) is cos(x). After you are done with that part, you will have to convert the integral from dx to du by the steps above. After solving the integral you will have an expression in terms of u which you can solve for x by the relationship x = sinu/2

9. Mar 20, 2005

### Data

It's a method called "trigonometric substitution."

Let's see what happens when we sub in $x = \frac{\sin{u}}{2} \Longrightarrow dx = \frac{\cos{u}}{2}du$:

we get

$$\sqrt{1 - 4x^2} = \sqrt{1 - 4\frac{\sin^2{u}}{4}} = \sqrt{1 - \sin^2{u}}$$

Recalling that $\cos^2{u} = 1 - \sin^2{u}$, this just turns into

$$\sqrt{\cos^2{u}} = \cos{u}$$

$$\int \sqrt{1 - 4x^2} dx = \int \cos{u} dx$$

and substituting in for $dx$, this is just

$$\int \frac{\cos^2{u}}{2}du$$

and I'm sure you can work it out from there :)

As I mentioned, this is an example of a method called trigonometric substitution. Some other examples are integrands involving $\sqrt{a^2 + x^2}$ in which we substitute $x = a\tan{u}$ to give $\sqrt{a^2 + a^2\tan^2{u}} = \sqrt{a^2\sec^2{u}} = a\sec{u}$, or involving $\sqrt{x^2 - a^2}$ where we sub $x = a\sec{u}$ to make a similar simplification.

10. Mar 20, 2005

### Data

ahh, you beat me this time!

11. Mar 21, 2005

### dextercioby

Instead of using sec & tan,u can very nicely use $\sinh,\cosh$...

Hyperbolic trigonometry is useful & elegant.