# One last integration problem

1. Feb 20, 2008

### Ravenatic20

This is what I have so far:

$$\int (\frac{x^2}{\sqrt{4 + x^2}}) dx$$

$$x = 2 tan u$$
$$dx = (2 sec^2u) du$$

$$x^2 = 4 tan^2 u$$

$$\sqrt{4 + x^2} = \sqrt{4 + 4 tan^2 u}$$

$$\sqrt{4 + x^2} = \sqrt{4 (1 + tan^2 u}$$

$$\sqrt{4 + x^2} = \sqrt{4 sec^2 u}$$

$$\sqrt{4 + x^2} = 2 sec u$$

Then I have:

$$\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = \int \frac{4 tan^2 u}{2 sec u} 2 sec^2 u du$$

$$\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = \int (4 tan^2 u)(sec u) du$$

I keep getting stuck on this part, any ideas?

2. Feb 20, 2008

### rocomath

Assuming your work is correct ...

$$4\int\tan^{2}\theta\sec \theta d\theta$$

$$4\int(\sec \theta \tan \theta) \tan \theta d\theta$$

Or go straight to this ...

$$4\int(\sec^{3}\theta-\sec \theta) d\theta$$

Either way, integration by parts!

Last edited: Feb 20, 2008
3. Feb 20, 2008

### Ravenatic20

Taking that...

$$4\int(\sec u \tan u) \tan u du$$

And knowing that:

$$\int(\sec u \tan u) du = \sec u + C$$

and...

$$\int\sec u du = ln |\sec u + \tan u|)+ C$$

Can we just do it by parts that way?

So...

$$\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = 4(\sec u + ln |\sec u + \tan u|) + C$$

(crosses figures)

4. Feb 20, 2008

### rocomath

You must be drunk!

$$4\int(\sec \theta \tan \theta) \tan \theta d\theta$$

Usually, you want to reserve "u" so I'm using theta in it's place.

$$u=\tan \theta$$
$$du=\sec^{2} \theta d\theta$$

$$dV=\sec \theta \tan \theta d\theta$$
$$V=\sec \theta$$

$$4\int(\sec \theta \tan \theta) \tan \theta d\theta=4(\sec \theta \tan \theta-\int\sec^{3}\theta d\theta)$$

Do parts again, you will actually have to do it 3x, I think. Take $$\int\sec^{3}\theta d\theta$$ to the side and you will notice that you get it back. Just move it to the other side and divide by the constant, and it's solved.

Last edited: Feb 20, 2008
5. Feb 20, 2008

### Ravenatic20

No I'm not drunk, I'm just new to this material. However I think were just about done:

I understand how you did everything, so now...

$$4\int(\sec \theta \tan \theta) \tan \theta d\theta=4(\sec \theta \tan \theta-\int\sec^{3}\theta d\theta)$$
Where... $$\int\sec^{3}\theta d\theta) = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} ln|sec \theta + tan \theta| + C$$

This gives us:
$$\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = 4 (\sec x \tan x)-\frac{1}{2} \sec x \tan x + \frac{1}{2} ln|sec x + tan x| + C$$

?

6. Feb 20, 2008

### rocomath

To be honest, I did not work it out myself. Looks good though! Btw, hopefully you weren't angry with my comment. I didn't mean anything by it in terms of you just having learned this :p But um, your answer could be right, all depends if your teacher wants you to re-substitute ... this would be the most painful part, lol.

7. Feb 20, 2008

### Ravenatic20

Haha no offense taken and thanks for your help!

8. Feb 20, 2008

### HallsofIvy

Staff Emeritus
Actually, he meant it helps to be drunk when doing calculus!

9. Feb 20, 2008

### ssd

A much faster way to learn these type of integrals is to go through a standard text book step by step, and use paper and pen.

10. Feb 20, 2008

### rocomath

Really? Mr. Obvious.

11. Feb 20, 2008

### coomast

Another way of doing this integral is by using the substitution:

$$x=2\cdot sinh(t)$$

after setting:

$$x^2=x^2+4-4$$

in the numerator. You will end up with:

$$I=\frac{x}{2}\sqrt{4+x^2}-2\cdot argsinh\left( \frac{x}{2}\right)+C$$

Btw: I'm sober

12. Feb 21, 2008

### ssd

Some times people has to be more obvious than needed for some people who are reluctent to see in text books the standard result like

integral [sqrt(a^2 + x^2)] dx = (x/2) sqrt(a^2 + x^2)+ [(a^2)/2] ln|(x+sqrt(a^2+x^2)|+c
(I leave the derivation of this result for obvious reasons)

and cannot see x^2 = x^2 +a^2-a^2, as shown in the previous post by COOMAST. Then the remaining term is easily (obviously?) undone by substituting x=2tan(u).
After COOMAST's method, I would prefer this method as it will be easier to go back in terms of the original veriable. Is it obv....?

Last edited: Feb 21, 2008
13. Feb 21, 2008

### unplebeian

Don't drink and derive.