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.
$\displaystyle \int_{-1}^x \sin(t^2) \cdot 2t \, dt$
substitution ... let $\displaystyle u = t^2$
$\displaystyle du = 2t \, dt$
substitute and reset the limits of integration ...
$\displaystyle \int_1^{x^2} \sin(u) \, du = \left[-\cos(u)\right]_{1}^{x^2}$
$\displaystyle \cos(1) - \cos(x^2)$
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. $\displaystyle \int \sin x dx = \textcolor{red}{-}\cos x +C$