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The numerator is the derivative of the denominator. Using the chain rule in reverse gives the antiderivativeThe absolute value makes this real on every interval where the denominator is nonzero. The integration constant may be chosen independently on the intervals separated by $x=(-3\pm \sqrt{17})/2$.

The integrating factor is $e^{kx}$, giving $(e^{kx}y{)}^{\prime }=a{e}^{kx}\sin (mx)$. The required antiderivative isAfter dividing by $e^{kx}$, the general solution is the oscillatory term plus $C{e}^{-kx}$. The initial condition fixes $C=1+am/(k^2+m^2)$, soThe real, nonzero parameters ensure $k^2+m^2>0$; direct substitution verifies both the equation and $y(0)=1$.

For period $2$, use the Fourier series conventionThe Fourier coefficients have prefactor $1$ on $[-1,1]$. First, $a_0={\int }_{-1}^1{e}^{x}dx=2\sinh 1$. For $q=n\pi$ the useful antiderivatives areAt the endpoints $\sin (n\pi )=0$ and $\cos (n\pi )=(-1{)}^n$, soConsequently, on $-1<x<1$,The Dirichlet-Jordan convergence theorem justifies equality at every point of continuity of the periodic extension. At odd integers the extension has a jump discontinuity between $e$ and $e^{-1}$, so the series converges to their average $\cosh 1$, regardless of any assigned endpoint value. This is the $\kappa =L=1$ case of the Fourier series of a periodically extended exponential.

The chain rule identifies an antiderivative directly:The upper limit in the PDF is $\sqrt{\pi /2}$. By the fundamental theorem of calculus, the required integral is

First write the forcing as $\sin x\cos x$. An integrating factor for the linear ordinary differential equation is $e^{-\sin x}$, givingSet $u=\sin x$ inside the antiderivative. ThenThe initial value forces $C=1$, henceThe resulting expression is explicit and valid for all real $x$; the substitution does not impose a global monotonicity condition on $\sin x$ when verifying an antiderivative.

Use the real antiderivative $\ln |x|$, valid on either half-line away from zero:The negative answer is consistent with a negative integrand and increasing integration limits.

The fundamental theorem of calculus states that integrating a continuous function from a fixed lower endpoint produces an antiderivative. More precisely, if $f$ is continuous at the interior point $x$, thenIndeed, the difference quotient is the average of $f$ over the short interval from $x$ to $x+h$, which tends to $f(x)$ by continuity. The accompanying evaluation rule is ${\int }_a^{b}f=F(b)-F(a)$ whenever $F^{\prime }=f$ on the interval.

The integrand and both limits depend on $x$, so use the Leibniz integral rule. The upper and lower endpoint contributions and the interior derivative giveFor $x\ne 0$, an antiderivative in $t$ is $t\sin (xt)/x+\cos (xt)/x^2$. Evaluating it and collecting terms givesAt $x=0$ the original Leibniz expression has zero endpoint terms and an integral over a zero-length interval, giving $\boxed{F^{\prime }(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.

For $b>0$, split the interval into $n$ equal widths $\Delta x=b/n$. Since $x^2$ increases on this interval, the left and right rectangles give lower and upper Riemann sums:Using the finite sum of squares,The rectangle bounds differ by $U_n-L_n=b^3/n\to 0$. The squeeze theorem therefore gives, directly from elementary areas,For $b=0$ the result is zero. If negative $b$ is allowed, reverse the integration limits and use the reflection $x=-u$; the same signed formula holds. No antiderivative is needed for this first-principles calculation.

Complete the square in the denominator and split the numerator into a multiple of its derivative plus a constant:ThusThe natural logarithm needs no absolute-value correction because its quadratic argument is strictly positive for every real $x$. The second term is an inverse tangent antiderivative.

For the indefinite integral, substitute $u=x^2$, so $x^{3}dx=udu/2$. An integration by parts then givesFor the definite integral, the original PDF's upper limit is $\sqrt{\pi /2}$, so the corresponding upper value of $u$ is $\pi /2$. ThereforeBoth requested integrals are thus evaluated with the same antiderivative.

The derivative of $\cot u$ is $-{\csc }^{2}u$. The chain rule therefore givesThe antiderivative applies on any interval containing no zero of $\sin (3x+1)$.

A probability density function must be nonnegative. The printed condition $s\ne 0$ is therefore too weak: for $s<0$ the given prefactor makes the function negative everywhere. The location-scale form of the logistic distribution requires $s>0$, which is assumed in the remaining parts.

Set $t=(x-\mu )/(2s)$, so $dx=2sdt$. The normalization of a probability distribution follows from the antiderivative of the squared hyperbolic secant:For completeness, the same expression with $s<0$ integrates to $-1$, since the transformed limits reverse. If a signed scale were desired, replacing its prefactor by $1/(4|s|)$ would give a valid density, but that is a correction to the printed formula.

Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.

On the negative half-line, an antiderivative of $1/x$ is $\ln |x|$. The fundamental theorem of calculus therefore givesThe negative sign agrees with the negative integrand throughout the interval.

Reverse the order in the nonnegative double integral, using $0\le y<\infty$ and $y\le x\le 2y$. For $y>0$, an antiderivative in $x$ is $-(x/y+1/y^2)e^{-xy}$. ThusLet $J$ denote the requested improper integral. Its numerator is $y^2+O(y^4)$ near zero, so its apparent singularity is removable; it also decays at infinity. We may therefore separate the preceding expression into the convergent integralsIt follows thatThe two terms divided separately by $x^2$ would each have a divergent integral at zero; the cancellation in their difference is essential.

Use integration by parts with $u=\ln x$ and $dv=x^{2}dx$. This gives the antiderivativeEvaluating at the two bounds, with $\ln 1=0$, yields

Insert the two densities into the convolution of probability densities:Using the antiderivative $-\arctan (z-y)$ with respect to $y$ givesThis is the convolution of a Cauchy distribution and a uniform distribution with half-width $1/2$.

For period two, the Fourier series convention isSince $f$ is an even function, every $b_n$ is zero. Also $a_0={\int }_{-1}^1\cosh xdx=2\sinh 1$. For $n\ge 1$, an antiderivative useful for the cosine coefficient isEvaluation at $\pm 1$ gives $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. ThusThe periodic extension is continuous at $x=1$, so its Fourier series converges to $\cosh 1$ there. Since $\cos (n\pi )=(-1{)}^n$,Therefore $\boxed{p=2,q=\pi }$, with $q=-\pi$ giving the same expression because only $q^2$ occurs. This is the $\beta =1$ case of Fourier series of periodic hyperbolic sine and cosine.

On an interval excluding $t=0$, an integrating factor for this first-order linear differential equation is $t^4$. Multiplying through givesAn antiderivative then gives $t^{4}y=3t^5/5+C$, soThe constant is arbitrary on each connected interval of the domain; the original differential equation is undefined at zero.