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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.

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.

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.

At $t=0$, the initial conditions implyDifferentiate the first relation and solve for the two derivatives:Fix a reference point $x_{\ast }$ and set $V(x)={\int }_{x_{\ast }}^{x}v(s)ds$. Taking antiderivatives givesThe constants must be opposite to preserve the first initial condition; the arbitrary constant of integration $K$ does not change $\psi$. Substitution yields D'Alembert's formula:The fundamental theorem of calculus verifies that its time derivative at zero is $v(x)$, and its value at zero is $u(x)$. The oriented integral also makes the formula valid for negative $t$.

For a partial fraction decomposition, writeMultiplying by $(x-1{)}^2(x+1)$ gives $4=A(x^2-1)+B(x+1)+C(x-1{)}^2$. Setting $x=1,-1$ gives $B=2,C=1$, and comparing the coefficient of $x^2$ gives $A=-1$. ThusIntegrating each term gives the antiderivativeThe constant of integration can be chosen independently on each interval separated by $-1$ and $1$.