# How To integrate this?

• Mar 1st 2009, 08:52 PM
Krooger
How To integrate this?
$
\int (x^2)/(\sqrt(9+x^2)) dx
$

** Thats the best I can make this equation look...

Anyways I have been trying to get this thing for hours, I clearly do not know what I'm doing. Advice geatly apriciated!
• Mar 1st 2009, 08:59 PM
sleepiiee
I don't know if it works out or not but try x = 3tany
• Mar 1st 2009, 09:00 PM
Krooger
Yea i have tried that over and over but Im getting no where :(
• Mar 1st 2009, 09:13 PM
sleepiiee
after the trig sub, do by parts
u = tany
dv = secytany

then to integrate sec^3y you gotta do by parts again
good luck
• Mar 1st 2009, 10:27 PM
Krooger
Okay sooo I do all that and then I get to the part where it loops back to my original integral. But since i use integration by parts twice the two negatives make a positive and then you have somthing like:

A = original integral
A = stuff + A
so i clearly did somthing wrong...

Heres my step my step:
$
\int (x^2)/(\sqrt(9-x^2)) = 9\int (tan(x)^2sec(x) dx
$

$
9\int (tan(x)^2sec(x) dx = 9[tan(x)sec(x) - \int sec(x)^3
$

$
9[tan(x)sec(x) - \int sec(x)^3 = 9[tan(x)sec(x) - [sec(x)tan(x) - \int tan(x)^2sec(x)]]
$

$
9\int (tan(x)^2sec(x) dx = 9tan(x)sec(x) - 9sec(x)tan(x) + 9\int (tan(x)^2sec(x) dx
$

At this point its clearly wrong what did I do?
• Mar 1st 2009, 11:05 PM
sbcd90
integrate[x^2/sqrt.(9-x^2)] dx

now,x=3cosA.

integrate(9 cos^2 A/sqrt.(9-9cos^2 A))*-3 sinA da

=-integrate{9 cos^2 A * dA}

=-9/2 integrate{(1+cos 2A) dA}

the rest is simple!!!!!!!!!!
• Mar 1st 2009, 11:08 PM
Krooger
ooops the question is:

x^2/ sqrt(9+x^2)

Some reason I put a minus before, long day...

So tan sub needs to be used
• Mar 2nd 2009, 12:33 AM
redsoxfan325
$\int \frac{x^2}{\sqrt{9+x^2}}\,dx$

Integration by parts:

Let $x=u$ and $dv=\frac{x}{\sqrt{9+x^2}}$

Thus, $dx=du$ and $v=\sqrt{9+x^2}$

Your integral is now: $x\sqrt{9+x^2}-\int \sqrt{9+x^2}\,dx$

Use $x=3tan(t)$ and $dx=3sec^2t$.

Now you have: $x\sqrt{9+x^2}-\int \sqrt{9+9tan^2t}*3sec^2t\,dt = x\sqrt{9+x^2}-9\int sec^3t\,dt$

Let $S=\int sec^3t\,dt$

Integration by parts again:

Let $u=sec(t)$ and $dv=sec^2(t)$.

Thus $du=sec(t)tan(t)$ and $v=tan(t)$ and the integral becomes: $sec(t)tan(t)-\int tan^2(t)sec(t)\,dt$.

This becomes: $sec(t)tan(t)-\int (sec^2(t)-1)sec(t)\,dt = sec(t)tan(t)-\int sec^3(t) + \int sec(t)\,dt$.

Thus $S = sec(t)tan(t) - S + \int sec(t)\,dt$ and $2S = sec(t)tan(t) + \int sec(t)\,dt$.

Using a table of integrals, we find that $\int sec(t)\,dt = ln|sec(t)+tan(t)|$.

Thus $2S = sec(t)tan(t) + ln|sec(t)+tan(t)|$.

Hence, $\int sec^3t\,dt = S = \frac{1}{2}sec(t)tan(t) + \frac{1}{2}ln|sec(t)+tan(t)|$.

Plugging $\int sec^3t\,dt$ back into the equation a few lines up, we have $x\sqrt{9+x^2}-9(\frac{1}{2}sec(t)tan(t) + \frac{1}{2}ln|sec(t)+tan(t)|)$.

Remember though, that since $x = 3tan(t)$, $t = tan^{-1}\frac{x}{3}$.

Thus, $x\sqrt{9+x^2}-9(\frac{1}{2}sec(t)tan(t) + \frac{1}{2}ln|sec(t)+tan(t)|)$ becomes $x\sqrt{9+x^2}-9(\frac{1}{2}sec(tan^{-1}\frac{x}{3})tan(tan^{-1}\frac{x}{3}) + \frac{1}{2}ln|sec(tan^{-1}\frac{x}{3})+tan(tan^{-1}\frac{x}{3})|)$.

Using the triangle, if the tangent of an angle is $\frac{x}{3}$, the secant of that angle is $\frac{3}{\sqrt{9+x^2}}$. Plugging everything makes that nasty equation become: $x\sqrt{9+x^2}-9(\frac{1}{2}*\frac{3}{\sqrt{9+x^2}}*\frac{x}{3}) + \frac{1}{2}ln|\frac{3}{\sqrt{9+x^2}}+\frac{x}{3}|)$.

Simplifying that yields: $x\sqrt{9+x^2}-\frac{9}{2}*\frac{x}{\sqrt{9+x^2}} + \frac{9}{2}ln|\frac{3}{\sqrt{9+x^2}}+\frac{x}{3}|$, which is what I think the answer is.

(You think that's ugly, you should see what it looks like in LaTex.)
• Mar 2nd 2009, 09:48 PM
redsoxfan325
I just realized I made a mistake at the end. When I said $sec(arctan(\frac{x}{3}))=\frac{3}{\sqrt{9+x^2}}$, I was wrong. It's actually $\frac{\sqrt{9+x^2}}{3}$, making the final answer:

$\frac{1}{2}x\sqrt{9+x^2}-\frac{9}{2}ln(\frac{1}{3}\sqrt{9+x^2}+\frac{1}{3}x )$

Sorry for the mistake.
• Mar 2nd 2009, 09:57 PM
Krooger
Haha no worrys, as it turns out there is a formula in the back of my text book that goes directly from the integral to exactly what you came up with :P I really appriciate the help thank you