# Thread: Antiderivative via FTC and Intergration by parts

1. ## Antiderivative via FTC and Intergration by parts

Define the function g(x) = anti derivative of t^2 * cos(t) dt
top value = 1/x^2 bottom value = 0

a) solve using FTC
b) solve using intergration by parts

2. Originally Posted by Lupus7874

Define the function g(x) = anti derivative of t^2 * cos(t) dt
top value = 1/x^2 bottom value = 0

a) solve using FTC
b) solve using intergration by parts
a) You want $\displaystyle \frac{dg}{dx} = \frac{d}{dx} \int_0^{1/x^2} t^2 \cos t \, dt$. Let $\displaystyle u = \frac{1}{x^2}$. Then from the chain rule:

$\displaystyle \frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du} \int_0^{u} t^2 \cos t \, dt \cdot \frac{du}{dx}$

and you should be able to do the necessary calculations.

b) Use integration by parts twice. It should be clear what choices to make (Hint: You want to end up integrating only a trig function).

3. I'm rather confused on the whole methodology of this problem, any chance someone could walk me through it?

I mean assuming I substitute what you said, I'm still lost on how to take the antiderivative of a product

4. Originally Posted by Lupus7874
I'm rather confused on the whole methodology of this problem, any chance someone could walk me through it?

I mean assuming I substitute what you said, I'm still lost on how to take the antiderivative of a product
I told you exactly how to get the solution: $\displaystyle \frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du} \int_0^{u} t^2 \cos t \, dt \cdot \frac{du}{dx}$.

$\displaystyle \frac{d}{du} \int_0^{u} t^2 \cos t \, dt$ is found using the Fundamental Theorem of Calculus (then substitute back $\displaystyle u = \frac{1}{x^2}$).

Getting $\displaystyle \frac{du}{dx}$ from $\displaystyle u = \frac{1}{x^2}$ should be trivial at the level you're apparently studying.

I don't see where the trouble can be in this ....