Compute the Integral?

**masterofcheese** · June 6th 2009, 12:23 PM

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.

**Spec** · June 6th 2009, 12:31 PM

$\int_{-1}^x 2t\sin t^2 dt$

Note that $\frac{d}{dt}\left(-\cos t^2 \right)=2t\sin t^2$

**skeeter** · June 6th 2009, 12:35 PM

Originally Posted by masterofcheese

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.

$\int_{-1}^x \sin(t^2) \cdot 2t \, dt$

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

$du = 2t \, dt$

substitute and reset the limits of integration ...

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

$\cos(1) - \cos(x^2)$

**calc101** · June 6th 2009, 01:22 PM

Skeeter, I understand your solution except for the new limits: 1 and $x^2$ .

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

Thanks

**Spec** · June 6th 2009, 01:33 PM

He made the substitution $u=t^2$ , and since $-1 \leq t \leq x$ we get $1 \leq u \leq x^2$

His final answer is correct, but he made a typo in the process by the way. $\int \sin x dx = \textcolor{red}{-}\cos x +C$

**Plato** · June 6th 2009, 01:35 PM

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

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