Evaluation of definite integrals by substitution

Ordinary (indefinite) substitution and definite-integral substitution share the first step but differ sharply at the last one. For an indefinite integral, $x = g(t)$ (or $t = h(x)$) simplifies the integrand, but the antiderivative is a function of $x$, so at the end you must substitute back to express the answer in terms of $x$. For a definite integral, the answer is just a number, not a function — so if the limits are converted to the new variable along with the integrand, the final number can be read off entirely in $t$, with no return trip to $x$ required. This shortcut is the whole point of the technique. Formally, suppose $\displaystyle\int_a^b f(x)\,dx$ can be written as $\displaystyle\int_a^b f(g(x))\,g'(x)\,dx$ for some differentiable $g$. Substituting $t = g(x)$, so that $dt = g'(x)\,dx$, and converting the limits by the same rule as the variable,

where: $t = g(x)$ is the new variable, $dt = g'(x)\,dx$ is the matching differential taken from the integrand itself, and $g(a)$, $g(b)$ are the new lower and upper limits obtained by plugging the old limits into the substitution formula — gives

The right side is a complete, self-contained definite integral in $t$: it has its own limits, $g(a)$ and $g(b)$, and evaluating it by the fundamental theorem of calculus gives the final number directly.

Choosing a good substitution. Look for a part of the integrand whose derivative is also sitting in the integrand (up to a constant), often the "inside" of a composite function or the base of a power. A useful check after substituting: the whole integrand, including $dx$, must convert cleanly into an expression in $t$ alone — if an $x$ is left stranded, the substitution was not well matched to the integrand.

Show, for the specific case $\displaystyle\int_0^1 2x\sqrt{1+x^2}\,dx$, that converting the limits to the new variable and integrating there gives the exact same numerical answer as the fundamental theorem would give, while never needing to write the antiderivative back in terms of $x$ at any stage of the calculation.

Step 1. Choose the substitution. The integrand has a factor $2x\,dx$ sitting next to $\sqrt{1+x^2}$, and $\dfrac{d}{dx}(1+x^2) = 2x$ — exactly the factor present. So let $t = 1+x^2$, giving $dt = 2x\,dx$.

Step 2. Rewrite the integrand entirely in $t$. Replacing $2x\,dx$ by $dt$ and $\sqrt{1+x^2}$ by $\sqrt t$,

No trace of $x$ remains — a sign the substitution was well chosen.

Step 3. Convert the limits using the same rule $t = 1+x^2$ that was used for the variable: at $x=0$, $t = 1+0^2 = 1$; at $x=1$, $t = 1+1^2 = 2$. So the new integral runs from $t=1$ to $t=2$.

Step 4. Integrate in $t$ and apply the fundamental theorem directly with the new limits:

The value is $\dfrac{2(2\sqrt2-1)}{3}$, and $x$ never had to reappear after Step 3 — the entire evaluation, limits and all, happened in $t$.

Evaluate $\displaystyle\int_0^{\pi/2}\sin^3 x\cos x\,dx$.

Step 1. Pick the substitution. The factor $\cos x\,dx$ is present, and it is exactly $d(\sin x)$, so let $t = \sin x$, giving $dt = \cos x\,dx$.

Step 2. Rewrite the integrand in $t$: $\sin^3x\cos x\,dx = t^3\,dt$.

Step 3. Convert the limits: at $x=0$, $t=\sin 0 = 0$; at $x = \pi/2$, $t = \sin(\pi/2) = 1$.

Step 4. Integrate and evaluate using the new limits directly:

The answer $\dfrac14$ comes straight from the $t$-limits $0$ and $1$; the trigonometric form never had to be restored.

Competitive stretch

For $\displaystyle\int_0^{\pi/2}\sin^3x\cos x\,dx$, spotting that the power of $\sin x$ is odd is a fast recognition cue: whenever an odd power of one trig function multiplies exactly one factor of its own derivative-partner (here $\cos x\,dx = d(\sin x)$), the substitution $t=\sin x$ (or $t=\cos x$ for an odd power of $\cos$) reduces the whole integral to a pure power of $t$, as above, usually solvable in under a minute.

• Forgetting to change the limits, and evaluating the $t$-antiderivative at the original $x$-limits instead of the converted $t$-limits — this produces a meaningless or wrong number.
• Substituting back to $x$ anyway, then evaluating at the original limits — unnecessary extra algebra for a definite integral, and a common source of arithmetic slips; once the limits are in $t$, stay in $t$.
• Choosing a substitution whose derivative is not actually present in the integrand, leaving a stray $x$ that cannot be written in terms of $t$ — always check the rewritten integrand is purely in $t$.
• Sign or order errors when the new upper limit is numerically smaller than the new lower limit (possible if $g$ is decreasing) — the integral is still (value at new upper limit) minus (value at new lower limit), whichever number that is.

Ask the tutor: Stuck on which substitution to pick for a definite integral, or want to practice converting limits correctly on a few more problems? Ask the AI tutor to work through extra examples with you.