1. ## Help with Cos(x^2)dx

integral of [cos(t^2)dt] on the interval [0,3x^2]

2. Originally Posted by Enigma
integral of [cos(t^2)dt] on the interval [0,3x^2]
$\displaystyle \frac{d}{dx}\int_a^{f(x)}g(t)dt=g(f(x))\cdot{f'(x) }$

So

$\displaystyle \frac{d}{dx}\int_0^{3x^2}\cos\left(t^2\right)=\cos \left(\left(3x^2\right)^2\right)\cdot\left(3x^2\ri ght)'=6x\cos\left(9x^4\right)$

3. Originally Posted by Mathstud28
$\displaystyle \frac{d}{dx}\int_a^{f(x)}g(t)dt=g(f(x))\cdot{f'(x) }$

So

$\displaystyle \frac{d}{dx}\int_0^{3x^2}\cos\left(t^2\right)=\cos \left(\left(3x^2\right)^2\right)\cdot\left(3x^2\ri ght)'=6x\cos\left(9x^4\right)$
How does that help the OP, you have replaced one tricky integral with another, or are you answering what you think they meant to ask, then tell us.

CB

4. Originally Posted by Enigma
integral of [cos(t^2)dt] on the interval [0,3x^2]
Please post the entire question (which I suspect is something like find the derivative of $\displaystyle \int_0^{3x^2} \cos (t^2) \, dt$. Correct ....? In which case you're barking up the wrong tree by trying to first integrate to get an expression to differentiate. The integral has no elementary closed form solution. The correct approach is shown in post #2 but you might want clarification ....).