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An antiderivative of $\sin x$ is $-\cos x$. The fundamental theorem of calculus gives

The product rule and chain rule give $y^{\prime }=\sin (x^2)+2x^2\cos (x^2)$. At zero the slope is zero and the curve passes through $(0,1)$. Consequently the tangent line is

For independent fair dice, the probability of avoiding one on each die is $5/6$. ThereforeThe complement of an event gives

For this separable differential equation, away from $x=1$, separation gives $\ln |y+1|=\ln |x-1|+C$, or $y+1=K(x-1)$ on a connected interval. The initial condition gives $2=-K$, henceThe maximal interval containing $x=0$ on which this solves the printed differential equation is $\boxed{-\infty <x<1}$. Although the linear formula extends through one, the right-hand side there is $0/0$ and the original equation is undefined. The equilibrium solution $y=-1$, lost if one divides by $y+1$, does not satisfy the initial condition.

Put $h=x-\pi /3$. The values $f(\pi /3)=1/2$, $f^{\prime }(\pi /3)=-\sqrt{3}/2$, and $f^{\prime \prime }(\pi /3)=-1/2$ give the Taylor expansionThe factorial in the quadratic term accounts for its coefficient $-1/4$.

Multiply numerator and denominator by the complex conjugate $5+4i$:The real part and imaginary part are therefore

The characteristic polynomial isIts roots give the two eigenvaluesTheir sum is zero, agreeing with the trace of the matrix.

The first partial derivatives factor asBoth vanish at the origin, verifying that it is a stationary point. Since $2x^2+1>0$, every stationary point has $y=0$; the first equation then gives $x=0,\pm 1$. Thus one requested additional point is $\boxed{(1,0)}$; the other is $\boxed{(-1,0)}$.

The integrand is an even function, since both $x$ and $\sin x$ are odd functions. Using integration by parts,The endpoint cosine terms vanish, leaving the sine contribution.