Compute integral x on top, -1 on bottom, sin(t^2) times 2t dt

I have a calc final in three hours and I don't know how to solve this problem. :(

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- Jun 6th 2009, 12:23 PMmasterofcheeseCompute the Integral?
Compute integral x on top, -1 on bottom, sin(t^2) times 2t dt

I have a calc final in three hours and I don't know how to solve this problem. :( - Jun 6th 2009, 12:31 PMSpec
$\displaystyle \int_{-1}^x 2t\sin t^2 dt$

Note that $\displaystyle \frac{d}{dt}\left(-\cos t^2 \right)=2t\sin t^2$ - Jun 6th 2009, 12:35 PMskeeter
$\displaystyle \int_{-1}^x \sin(t^2) \cdot 2t \, dt$

substitution ... let $\displaystyle u = t^2$

$\displaystyle du = 2t \, dt$

substitute and reset the limits of integration ...

$\displaystyle \int_1^{x^2} \sin(u) \, du = \left[-\cos(u)\right]_{1}^{x^2}$

$\displaystyle \cos(1) - \cos(x^2)$ - Jun 6th 2009, 01:22 PMcalc101
Skeeter, I understand your solution except for the new limits: 1 and $\displaystyle x^2$.

Why did you introduce new limits to the integral and how did you get them?

Thanks - Jun 6th 2009, 01:33 PMSpec
He made the substitution $\displaystyle u=t^2$, and since $\displaystyle -1 \leq t \leq x$ we get $\displaystyle 1 \leq u \leq x^2$

His final answer is correct, but he made a typo in the process by the way. $\displaystyle \int \sin x dx = \textcolor{red}{-}\cos x +C$ - Jun 6th 2009, 01:35 PMPlato
I have one negative comment. I find it absolutely astounding that someone three hours away from a calculus final exam would not see at once that

$\displaystyle \int {2t\sin \left( {t^2 } \right)dt = - \cos \left( {t^2 } \right)} $.