Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

Factor out the dominant cubic term and use a geometric series in $1/(3x^2)$:Thus the requested leading term and absolute error order areThe relative error of retaining only the leading term is $O(x^{-2})$; it differs from the absolute error order $O(x^{-5})$.

For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.

The expression assumes $Y$ is invertible. By the cyclic property of the trace,For a real or complex antisymmetric matrix, $X^T=-X$ forces every diagonal entry to satisfy $x_{ii}=-x_{ii}$, hence $x_{ii}=0$. ThereforeThis also demonstrates invariance of the matrix trace under matrix similarity. Antisymmetry of $Z$ itself is not needed.

Let $H(u)={\sin }^{2}u+\ln u$. The chain rule gives $H^{\prime }(u)=\sin 2u+1/u$, so the two given summands combine into its logarithmic derivative, $f(u)=H^{\prime }(u)/H(u)$. For $u>1$, $H(u)>0$, so the real logarithm is defined throughout the integration path. Since $H(\pi )=\ln \pi$,This satisfies $F(\pi )=0$, and differentiation recovers both terms of the integrand.

The fundamental theorem of calculus, using continuity of $f$ near $x$, givesThe lower endpoint $a$ is constant, so it contributes no additional term.