antiderivative of a function.. problem

[Slimsta]

1. The problem statement, all variables and given/known data
$g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt$
whats g'(x)...


2. Relevant equations



3. The attempt at a solution
how to find the antiderivative of sqrt(1-t^2)?

[rock.freak667]

Try t=sin(u) or t=cos(u)

[Mark44]

I have edited your formula slightly so that it now shows g(x), not just (x). You had extra $ characters that shouldn't have been there.1. The problem statement, all variables and given/known data
g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt
whats g'(x)...


2. Relevant equations



3. The attempt at a solution
how to find the antiderivative of sqrt(1-t^2)?
Use the Fundamental Theorem of Calculus to find g'(x). You will also need the chain rule since your integral isn't strictly a function of just x, but is a function of sin(x).

[tiny-tim]

Hi Slimsta! :smile:
how to find the antiderivative of sqrt(1-t^2)?
Use the Fundamental Theorem of Calculus to find g'(x) …

Just to add to what Mark44 :smile: says …

the beauty of using the Fundamental Theorem of Calculus is that you don't need to know the antiderivative. :wink:

[HallsofIvy]

1. The problem statement, all variables and given/known data
g(x)=\int _{2 }^{\sin x}\sqrt{1- t^2}dt
whats g'(x)...


2. Relevant equations



3. The attempt at a solution
how to find the antiderivative of sqrt(1-t^2)?

Try t=sin(u) or t=cos(u)
The problem does not ask you to find the anti-derivative nor is it necessary.
Letting y= sin(x), this is
[tex]g(y)=\int_2^y \sqrt{1- t^2} dt[/itex]
You can find dg/dy directly from the "Fundamental Theorem of Calculus" and then use the chain rule to find dg/dx.

[Slimsta]

The problem does not ask you to find the anti-derivative nor is it necessary.
Letting y= sin(x), this is
[tex]g(y)=\int_2^y \sqrt{1- t^2} dt[/itex]
You can find dg/dy directly from the "Fundamental Theorem of Calculus" and then use the chain rule to find dg/dx.

okay, that make sense but what if i have a function like this:
$\int _{3\pi /4}^{\pi }(3 \sec ^2x -\frac{6 }{\pi })dx$

this is confusing me :S

[Bohrok]

That's a definite integral, so that gives you the signed area between the limits of integration, or just a number. What's the derivative of a constant?

A variable (other than x since x would be the dummy variable) in either of the limits of integration would make it a function.

[Slimsta]

hh i already figured that out.. i just took out the 3 and then it becomes tanx - 2/pi x and then its easy..