here it is: http://img266.imageshack.us/img266/435/untitledvm3.jpg

any help please.

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- Feb 18th 2007, 06:25 PMrcmangoantiderivative.
here it is: http://img266.imageshack.us/img266/435/untitledvm3.jpg

any help please. - Feb 18th 2007, 06:41 PMAfterShock
- Feb 18th 2007, 07:06 PMJameson
I think I can read it correctly.

In latex (when it works again) $\displaystyle \int_{0}{1} \sqrt{9+4t^2}dx$.

In easier to read type: int sqrt(9+4t^2)dt, evaluated from 0 to 1.

Before I do this, have you studied trig substitution, because if you haven't this might be slightly difficult to get at first? - Feb 18th 2007, 07:31 PMrcmango
i know some trig identities if you mean that is trig substitution?

- Feb 19th 2007, 06:59 AMJameson
Alright. Well draw a right triangle with the horizontal leg being 3 and the vertical leg being 2t. So it follows that the hypotenuse is sqrt(9+4^2).

Now let's make some substitutions. Call the angle between the horizontal leg and the hypotenuse theta.

tan(theta) = (2t)/3, and taking the derivative...

sec^2(theta)d(theta) = (2/3)dt.

So t = [3tan(theta)]/2 and dt = [3sec^2(theta)]/2

Now plug this back into the original integral.

int sqrt(9+4t^2)dt = int sqrt[(9/4)tan^2(theta)+9] *(3/2)sec^2(theta)d(theta).

Is this making sense?