Can anyone explain antiderivatives of square roots to me?

[ManBearPig]

I've got no idea what im doing lol

 

[Leros]

sqrt(x) = x^(1/2)

The rest should work itself out with the standard methods.

 

[mrkun]

Originally posted by: Leros
sqrt(x) = x^(1/2)

The rest should work itself out with the standard methods.

What he said.

The little shortcut is that you just add the numerator and denominator for integrals and subtract them for derivatives.

e.g. int[x^(1/2)]dx = (2/3)x^(3/2)

d/dx[x^(5/2)] = (5/2)x^(3/2)


Edit: Here's an online calculus calculator.

 

[DrPizza]

I'm assuming you're referring to the problems I label "tricky trig".

If you have something like integral(1/sqrt(9-x^2))dx, you're kinda screwed. u-substitution won't work. integration by parts (u dv) doesn't work either.

However, look at the values that x can take on... x has to be between -3 and 3.

Now, for the tricky trig part:
You have your pythagorean identities:
sin^2 + cos^2 =1
tan^2 + 1 = sec^2

watch what happens to the square root if I make this tricky trig substitution: let x = 3sinu
Shouldn't be a problem, because the value of sin of an angle vary between -1 and 1. 3 times these values makes it between -3 and 3. Thus, we're not "stepping out of bounds, running up the sideline, and stepping back in bounds." Such a weird substitution ought to be allowed. Seems weird, but watch...

x^2 becomes 9sin^2(u)
and 9 - 9sin^2(u) can be factored into
9(1-sin^2(u)) and, heyyy, 1-sin^2(u) is the same as cos^2(u) --- it's a pythagorean trig identity.

So, suddenly our ugly square root has transformed into
sqrt(9 cos^2(u) ) that's the square root of a perfect square.
= 3cos(u)

Now, when you're making a substitution like this, you'll have to remember to substitute for dx as well.
Since x = 3sinu
dx = 3cosu du

And, if there are other x's in the integral, substitute for them as well.
--------------------
if you run into (a+x^2) inside a square root, take advantage of another identity:
1+tan^2 = sec^2
So, for x you'll substitute sqrt(a) tan u
then, when substituting for x^2, you'll have (a + a tan^2(u) )
factor out the a, substitute for 1+tan^2, and you're all set to proceed.
(also substitute for dx and any other x's.)
----
There's a 3rd case: x^2 - a
you'll take advantage of sec^2 - 1 = tan^2
in a similar fashion.

Note, you don't necessarily have to have square roots present to take advantage of these substitutions. i.e. if you let x = sqrt(a) * tan u, the value of the right side can be anything (if the domain of x also allows it to take on any values and isn't restricted by a square root.)


Helpful?

 

[DrPizza]

note: it's nearly December. I wouldn't think any schools are just now getting to things like the integral of the square root of x. Those integrals are among the very first integrals that are taught, either toward the end of Calc I or the beginning of calc II

 

[mrkun]

Originally posted by: DrPizza
note: it's nearly December. I wouldn't think any schools are just now getting to things like the integral of the square root of x. Those integrals are among the very first integrals that are taught, either toward the end of Calc I or the beginning of calc II

I think when I had Calc I we started integration with about three weeks left in the semester.

Edit: It could also be for AP Calc, right?