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Since $d({\sin }^{2}x)/dx=2\sin x\cos x=\sin 2x$, integration by substitution $u={\sin }^{2}x$ immediately givesHere $C$ is an arbitrary constant of integration.

Integrate by horizontal slices. At a fixed height $0\le z\le H$, the parabolic boundary gives $|x|\le L\sqrt{z/H}$; the other inequality gives $|y|\le b(2-z/H)$. Thus the cross-section is a rectangle of areaThe volume integral, followed by integration by substitution $u=z/H$, isThis has units of volume. As a check on the geometry, the width decreases with height while the length of the horizontal slice increases from zero to $2L$.

Use integration by substitution $u=\ln x$, $du=dx/x$. The bounds become $3$ and $4$, and the remaining denominator factorizes as $(u-1)(u-2)$. The partial fraction decomposition isConsequently the definite integral isAll logarithm arguments here are positive on the integration interval.

For the first definite integral, use integration by substitution $t=\tan x$, with $dx=dt/(1+t^2)$, ${\cos }^{2}x=1/(1+t^2)$ and ${\sin }^{2}x=t^2/(1+t^2)$. This givesFor the second calculation, let $D=p{\cos }^{2}x+q{\sin }^{2}x$. Since $p,q>0$, this denominator is bounded away from zero; on compact positive parameter ranges the differentiated integrands have uniform bounds. Thus differentiation under the integral sign is justified and givesAdding the two equations and using the Pythagorean trigonometric identity in the numerators yields $J=-I_p-I_q$, henceFor $p=q$, this becomes $\pi /(2p^2)$, agreeing with direct integration of the constant integrand.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

Multiplication by the integrating factor and the product rule turn the left side into a single derivative:Using integration by substitution $u=\cos x$ gives $y\sin x=-{\cos }^{3}x/3+C$. Therefore the general solution isThere is one arbitrary constant of integration, as expected for a first-order ordinary differential equation.

Use integration by substitution $u=\cos x$, $du=-\sin xdx$. The denominator factorizes as $(u-2)(u-3)$, and partial fraction decomposition givesIntegrating and restoring $x$ yieldsFor real $x$, both $2-\cos x$ and $3-\cos x$ are positive, so no absolute-value signs are needed in this real logarithm. The denominator never vanishes.

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.

The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=n{t}^{n-1}dt$. Direct substitution gives the line integralsThus both values are unchanged. The derivative of $u$ vanishes at the initial endpoint for $n>1$, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.

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.

For $p>0$ and a nonnegative integer $n$, integration by parts gives, when $n>0$,The endpoint term $[s(1-s^p{)}^n{]}_0^1$ vanishes. ThereforeRepeated application yields $J_n(p)={\prod }_{j=1}^{n}pj/(pj+1)$. The integration by substitution $w=s^p$ also identifies $J_n(p)=B(1/p,n+1)/p$ with an Euler beta function. A related weighted integral follows immediately from the same substitution:

A trigonometric substitution is integration by substitution using a trigonometric function to simplify an integrand, especially a square root. For example, $x=a\sin u$, with $a>0$ and $-\pi /2\le u\le \pi /2$, changes $\sqrt{a^2-x^2}$ to $a\cos u$ and $dx$ to $a\cos udu$. The interval restriction fixes the sign of the square root. Similarly $x=a\tan u$ simplifies $\sqrt{a^2+x^2}$. The transformed integral must include the differential and, for a definite integral, the new limits.