** 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!

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- Mar 1st 2009, 07:52 PMKroogerHow To integrate this?

** 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, 07:59 PMsleepiiee
I don't know if it works out or not but try x = 3tany

- Mar 1st 2009, 08:00 PMKrooger
Yea i have tried that over and over but Im getting no where :(

- Mar 1st 2009, 08:13 PMsleepiiee
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, 09:27 PMKrooger
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:

At this point its clearly wrong what did I do? - Mar 1st 2009, 10:05 PMsbcd90
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, 10:08 PMKrooger
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 1st 2009, 11:33 PMredsoxfan325

Integration by parts:

Let and

Thus, and

Your integral is now:

Use and .

Now you have:

Let

Integration by parts again:

Let and .

Thus and and the integral becomes: .

This becomes: .

Thus and .

Using a table of integrals, we find that .

Thus .

Hence, .

Plugging back into the equation a few lines up, we have .

Remember though, that since , .

Thus, becomes .

Using the triangle, if the tangent of an angle is , the secant of that angle is . Plugging everything makes that nasty equation become: .

Simplifying that yields: , 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, 08:48 PMredsoxfan325
I just realized I made a mistake at the end. When I said , I was wrong. It's actually , making the final answer:

Sorry for the mistake. - Mar 2nd 2009, 08:57 PMKrooger
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