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On the disc $z=0$, the upward normal selects only the third component, which equals $e^0/(1+0)=1$. Hence the surface integral is the area of the unit disk:

The two orthogonal directions $(1,1{)}^T$ and $(1,-1{)}^T$ are preserved by this symmetric matrix. Direct multiplication, using the hyperbolic function identities $\cosh u\pm \sinh u=e^{\pm u}$, gives the eigenvalue–eigenvector pairsEach eigenvector can be multiplied by any nonzero scalar. For $u\ne 0$ these give the two one-dimensional eigenspaces. When $u=0$, the matrix is the identity: the only eigenvalue is $1$, with multiplicity two, and every nonzero vector is an eigenvector.

The tangent half-angle substitution gives $\sin \theta =2t/(1+t^2)$ and $\cos \theta =(1-t^2)/(1+t^2)$. Dividing and then taking the reciprocal square of the cosine givesThese formulas apply where $t$ is defined and $t\ne \pm 1$; at $t=\pm 1$ the original tangent and secant function are undefined because $\cos \theta =0$.

Use integration by substitution with $u=x^4$, so $du=4x^{3}dx$. ThenDifferentiating the answer by the chain rule gives $x^3{e}^{-x^4}$, confirming the antiderivative. The real constant of integration $C$ is arbitrary.

Let $p(x)=x^3-5x-1$. Its values at useful points areSince a polynomial is a continuous function, the intermediate value theorem gives one polynomial root in each of the disjoint intervals $(-3,-2)$, $(-1,0)$ and $(2,3)$. A nonzero cubic polynomial has at most three distinct polynomial roots. Hence it hasThese three distinct polynomial roots already exhaust its degree of a polynomial, so each has multiplicity of a root one.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

For $f(x)=x{e}^{-2x^2}$, the product rule and chain rule giveThe positive exponential factor makes $f$ increasing for $0<x<1/2$ and decreasing for $1/2<x<1$. Its only interior stationary point is therefore a maximum at $x=1/2$. Checking it together with both endpoints,ConsequentlyChecking the endpoints is necessary for extrema on a closed interval.

Both sums are geometric series. For first term $a$ and common ratio $r$ with $|r|<1$, the sum is $a/(1-r)$. Taking the initial term to be $2$ givesandBoth have absolute convergence because the series of absolute values has ratio $1/2$.

Using the Pythagorean trigonometric identity and the double-angle formula,The required value $1/2$ therefore occurs when ${\sin }^2(2\theta )=1$. Within $0\le \theta \le \pi$, the only possibilities are $2\theta =\pi /2$ and $3\pi /2$. HenceEquivalently equality requires ${\sin }^2\theta ={\cos }^2\theta =1/2$.