Can You Solve This Integration Problem with a Little Help from Trigonometry?

[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?

 

[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!

 

[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)

 

[rocomath]

Can we just do it by parts that way?

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.

 

[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

?

 

[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.

 

[Ravenatic20]

Haha no offense taken and thanks for your help!

 

[HallsofIvy]

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

 

[ssd]

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

 

[rocomath]

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

Really? Mr. Obvious.

 

[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

 

[ssd]

Really? Mr. Obvious.

Some times people has to be more obvious than needed for some people who are reluctent to see in textbooks 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...?

 

[unplebeian]

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

Don't drink and derive.