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Choose the trigonometric substitution $x=\sin u$ with $-\pi /2\le u\le \pi /2$. The cosine is nonnegative there, so $\sqrt{1-x^2}=\cos u$ and $dx=\cos udu$. The double-angle formula then givesSince $u=\arcsin x$ and $\sin 2u=2x\sqrt{1-x^2}$,The formula is an antiderivative on $(-1,1)$ and has continuous endpoint values at $\pm 1$.

The product in the integrand simplifies to $\sin x{\cos }^{3}x$ below its upper endpoint. With the integration by substitution $t=\cos x$, the limits become $c$ and zero and $dt=-\sin xdx$. ThereforeAlthough the tangent function itself is undefined at the upper endpoint, the product has a zero limit there, so its continuous extension, or its improper integral, gives this same result.

To obtain $J$ directly from a Riemann sum, divide $[0,c]$ into $n$ equal pieces of width $c/n$ and use right endpoints $t_k=kc/n$. The sum isThe integrand is continuous, so the limit of these Riemann sums is its definite integral, namely $c^4/4$, agreeing with the antiderivative calculation.

Use integration by substitution with $u=3-2x$, so $dx=-du/2$ and $x=(3-u)/2$. The indefinite integral becomesThus, with an arbitrary constant of integration,This real antiderivative applies for $x<3/2$ and extends continuously to the stated endpoint. Its left derivative there agrees with the limiting integrand, zero.

A suitable real antiderivative on the disconnected domain isFor $x>0$, $G^{\prime }=1/x$ directly; for $x<0$, $G=\ln (-x)$ and the chain rule again gives $G^{\prime }=1/x$. The graph is even, crosses zero at $x=\pm 1$, tends to $-\infty$ on either side of zero, and tends to $+\infty$ as $|x|\to \infty$. Independent additive constants may be chosen on the two disconnected half-lines; both are set to zero here.

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.

With $y$ held fixed, the inner antiderivative givesThe remaining integral is therefore

The difference and sum of the two antiderivatives in part (i) giveandTheir derivatives also verify the signs immediately.

Write $I=\int e^x\sin xdx$ and $J=\int e^x\cos xdx$. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, giveThus $I+J=e^x\sin x$ and $J-I=e^x\cos x$, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

For $n\ge 2$, write $x^n{e}^{-x^2}=x^{n-1}(x{e}^{-x^2})$ and apply integration by parts, using the antiderivative $-e^{-x^2}/2$ for the second factor:The boundary term is zero: at zero, $n-1>0$, and at infinity Gaussian decay dominates every power. Consequently the Gaussian moment recurrence is

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.

Use the substitution rule with $u=\sqrt{x}$, for which $du=dx/(2\sqrt{x})$ on $x>0$. The integral becomes $\int e^{u}du$, giving the antiderivative

On a nonconstant branch with $\sin y\ne 0$, regard $x$ as a function of $y$. The equation becomes the linear ordinary differential equationAn integrating factor is $\sin y$, givingTaking an antiderivative yields the implicit familyOn a regular branch this can also be written $x=\sin y+C/\sin y$. The latter form loses the constant solutions $y=n\pi$, which must be restored separately.

The implicit relation is a first integral of an ordinary differential equation: along any differentiable solution of the original equation,Conversely, a differentiable graph satisfying this relation solves the original equation wherever $\cos y\ne 0$. The original equation is undefined when $\cos y=0$; points at which $2\sin y-x=0$ also need separate attention before solving explicitly for $y^{\prime }$. This is why dividing immediately by $\tan y$ can lose valid solutions and obscure nonuniqueness.

The function graphs intersect when $2x=4x{e}^{-x}$, so $x=0$ or $x=\ln 2$. For $0<x<\ln 2$, the exponential function satisfies $e^{-x}>1/2$, making $4x{e}^{-x}$ the upper graph. The enclosed area is thereforeThe antiderivative of $x{e}^{-x}$ follows by integration by parts.

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.

For $L=1$, the even function $h(x)=x^2$ has $b_n=0$, because $x^2\sin (n\pi x)$ is an odd function. Its constant coefficient is $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. The remaining Fourier coefficients areTwo applications of integration by parts give the antiderivative, with $k=n\pi$,Its endpoint difference is $2(-1{)}^n/(n\pi {)}^2$, so $a_n=4(-1{)}^n/(n^2{\pi }^2)$. ThereforeThe periodic extension is continuous and piecewise smooth, so pointwise convergence of a piecewise smooth Fourier series gives equality, including at the endpoints.

The partial fraction decomposition isTaking antiderivatives and using the positive endpoints givesHence $\boxed{n=2}$.

Write $I_n=\int {\sin }^{n}xdx$. Apply integration by parts with ${\sin }^{n-1}x$ and $\sin xdx$:Using ${\cos }^{2}x=1-{\sin }^{2}x$ on the right gives $I_n=-\cos x{\sin }^{n-1}x+(n-1)(I_{n-2}-I_n)$. Rearranging provesFor $J_n=\int {\cos }^{n}xdx$, the analogous integration by parts uses ${\cos }^{n-1}x$ and $\cos xdx$, giving $J_n=\sin x{\cos }^{n-1}x+(n-1)(J_{n-2}-J_n)$. HenceAn arbitrary constant of integration absorbs the choices in the lower-order antiderivatives.

Starting from $I_0=J_0=x$, successive use of $n=2,4,6$ givesThese expressions follow directly from the requested recurrence, rather than requiring a separate trigonometric expansion.

With $u=XY$, divide the separated equation on $x\ne 0$, $y\ne 0$ to obtainThus $X^{\prime }/X=\lambda x{e}^{-x}$ and $Y^{\prime }/Y=\lambda /y$. Taking antiderivatives,Choose $\lambda =1$ and the signed branch $Y=y$. A nonconstant solution, smooth for every real $x,y$, isIndeed $u_x=x{e}^{-x}u$ and $u_y=\exp [-(x+1)e^{-x}]$, so $e^x{u}_x-xy{u}_y=0$. This verification includes the axes excluded temporarily during separation.

For integration by parts, take $u=x$ and $dv=\cos xdx$. Then $du=dx$ and $v=\sin x$, giving the antiderivative