# antiderivative.

• Feb 18th 2007, 06:25 PM
rcmango
antiderivative.
• Feb 18th 2007, 06:41 PM
AfterShock
Quote:

Originally Posted by rcmango

I can't read that; int(sqrt(7 + ____), t = 0...1)dt

Type what the integral is here please.
• Feb 18th 2007, 07:06 PM
Jameson
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 PM
rcmango
i know some trig identities if you mean that is trig substitution?
• Feb 19th 2007, 06:59 AM
Jameson
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?