U-substitution — technique, change of limits for definite integrals: Question bank

Reminder of the players (so no symbol here is unearned):

$u = g (x)$ — the nickname we give an inner function (the thing inside a power/root/exp/log/trig).
$d u = g^{'} (x) d x$ — the differential, the rule that keeps "the amount of stuff" honest when we rename. It comes from $\frac{d u}{d x} = g^{'} (x)$ , the slope of $u$ , multiplied across by $d x$ .
$F$ — an antiderivative of $f$ : a function whose derivative is $f$ , i.e. $F^{'} = f$ . Read " $F^{'} = f$ " as "the slope of $F$ at every point equals $f$ there." (See Antiderivatives & Indefinite Integrals.)
$\int_{a}^{b}$ — a definite integral: " $x$ runs from $a$ to $b$ ." A plain $\int$ with no numbers is indefinite (answer is a function $+ C$ ).

Why questions — the geometric heart

Look at the area-mapping figure above while answering these.

Why does u-substitution keep the area the same even though the interval changes width?

Because

d u = g^{'} (x) d x

rescales the height by the same factor the interval is stretched: a wider strip in

u

carries a proportionally shorter integrand, so width

\times

height (the area) is preserved. This is the visual meaning of the whole theorem.

Why does the derivative of the inner function have to "already be there" for u-sub to work?

Because u-sub reverses the Chain Rule, whose output is always inner-function-times-inner-derivative; if that derivative is absent, there is no chain-rule pattern to undo — geometrically, we'd have no factor to rescale the height with.

Why can constants be pulled through the integral sign but not variables?

Integration is linear:

\int c f = c \int f

holds only when

c

is fixed. A variable changes across the interval, so it interacts with

d x

and cannot be treated as a fixed multiplier.

Why do the limits change for a definite integral but a "

+ C

" appears for an indefinite one instead?

A definite integral asks for a number over a fixed range, so the range must be re-expressed on the new

u

-axis; an indefinite integral asks for a function, so we restore the original variable and add

C

for the lost constant.

Why is treating

\frac{d u}{d x}

as a fraction (to get

d u = g^{'} (x) d x

) actually legitimate here?

It's shorthand justified by the u-sub theorem itself — the theorem proves the swap

f (g (x)) g^{'} (x) d x \to f (u) d u

is valid, so the fraction-manipulation is a correct bookkeeping notation, not a hand-wave.

Why might two students pick different

u

's and both succeed?

Several inner functions can expose a chain-rule pattern; as long as the chosen

u

has its derivative available (up to a constant), any valid choice unwinds to the same antiderivative.

Why is Trigonometric Substitution considered a special case of u-substitution rather than a new idea?

It's still renaming the variable (

x = a sin θ

, etc.) with a matching differential

d x

; it just chooses the substitution to exploit trig identities rather than an inner-function-derivative pairing.

True or false — justify

U-substitution is just a repackaging of the chain rule, run backwards.

True — differentiating the antiderivative

F (g (x))

gives

F^{'} (g (x)) g^{'} (x) = f (g (x)) g^{'} (x)

(using

F^{'} = f

); integrating that pattern returns

F (g (x))

, which is exactly what u-sub recovers. See Chain Rule.

Every integral can be solved by some clever choice of

u

False — u-sub only works when the integrand hides an inner function together with (a constant multiple of) its own derivative. Many integrals need Integration by Parts or other tools instead.

In a definite integral you must either change the limits or back-substitute, but never both.

True — both are complete finishing methods; doing both translates the endpoints twice and corrupts the answer.

Once you write everything in

u

, the

d x

can simply be dropped since it's "just notation."

False —

d x

is part of the integrand and must be converted to

d u

via

d u = g^{'} (x) d x

; a

u

-integral that doesn't end in

d u

is unfinished, not simplified.

u = g (x)

is a constant function, u-substitution still helps.

False — then

g^{'} (x) = 0

, so

d u = 0

and there is no derivative-of-inner appearing; the method has nothing to reverse.

\int \frac{1}{u} d u = ln u + C

is always correct.

False — it's

ln ∣ u ∣ + C

. The domain of

\frac{1}{u}

splits at

u = 0

into positive and negative branches; the absolute value is what covers the negative branch, where

ln u

isn't even defined.

Changing limits can turn a "

0

2

" integral into one where the lower limit exceeds the upper limit.

True — if

g

is decreasing on that interval,

g (a) > g (b)

; that's fine, the FTC handles reversed limits by flipping the sign (see the worked figure below and Definite Integral & Fundamental Theorem of Calculus).

You may pull a stray factor of

\frac{1}{3}

out of the integral to make

d u

match.

True —

\frac{1}{3}

is a constant, and integration is linear over constants, so constant factors move freely in or out.

You may pull a stray factor of

\frac{1}{x}

out of the integral to make

d u

match.

False —

\frac{1}{x}

is a variable, not a constant; it cannot leave the integral. You must solve

u = g (x)

for

x

and substitute it too.

After a correct substitution the new integrand may still contain an

x

False (if done fully) — a completed substitution eliminates every

x

-piece; a lingering

x

signals an incomplete conversion or a bad choice of

u

For a definite integral, after changing limits you should back-substitute at the end to double-check.

False — back-substituting after already changing the limits double-translates the endpoints; the changed-limit answer is already a pure number and needs no back-sub.

\int \frac{1}{g ( x )} g^{'} (x) d x

is always fine as long as

g^{'}

is present.

False — you also need

g (x) > 0

throughout the interval, otherwise

g (x)

(and

1/ g (x)

) leaves the reals; a valid substitution must keep every

u

-value inside the new integrand's domain.

Spot the error — grouped by the misconception

Group A — the $d x \to d u$ swap (the height-rescale) was skipped.

\int 2 x (x^{2} + 1)^{5} d x

: let

u = x^{2} + 1

, so

\int u^{5} d x = \frac{u ^{6}}{6} + C

." Where's the slip?

The integral still says

d x

, not

d u

— the

2 x d x

was never converted, so the height-rescale is missing. Correctly

d u = 2 x d x

, giving

\int u^{5} d u

\int x e^{x^{2}} d x

: let

u = x^{2}

d u = 2 x d x

, so this is

\int e^{u} d u = e^{x^{2}} + C

." Find the missing piece.

We only have

x d x = \frac{1}{2} d u

, not

2 x d x = d u

, so a constant factor

\frac{1}{2}

was dropped; the answer is

\frac{1}{2} e^{x^{2}} + C

Group B — the inner derivative isn't actually present.

\int sin (x^{2}) d x

: let

u = x^{2}

d u = 2 x d x

, so

\int sin u d u

." Why is this wrong?

There is no

2 x d x

in the integrand to supply

d u

; the derivative of the inner function is absent, so u-sub does not apply (this integral has no elementary answer).

\int cos x d x

: let

u = cos x

d u = - sin x d x

." Why is this a poor choice?

The integrand contains no

sin x

to pair with

d u

, and the integral is already elementary; a nickname matching no present derivative only makes things harder.

Group C — limits and orientation on the new $u$ -axis.

\int_{0}^{2} x x^{2} + 1 d x

with

u = x^{2} + 1

gives

\frac{1}{3} u^{3/2}_{0}^{2}

." What went wrong?

The limits weren't changed — the old interval

[0, 2]

was kept on the new

u

-axis. With

u = x^{2} + 1

x = 0 \Rightarrow u = 1

x = 2 \Rightarrow u = 5

, so bounds must be

1

5

\int_{1}^{- 1} \frac{d u}{1 - u}

from a substitution — the lower limit (

1

) exceeds the upper (

- 1

), so I'll just swap them freely." Is that safe?

You may swap only if you also negate:

\int_{1}^{- 1} = - \int_{- 1}^{1}

. Swapping silently, without the sign flip, changes the value (see the orientation figure below).

Group D — domain/absolute-value oversights.

\int_{- 2}^{- 1} \frac{1}{x} d x = ln x_{- 2}^{- 1}

." What breaks?

ln x

is undefined for negative

x

; the antiderivative on this interval is

ln ∣ x ∣

, so the value is

ln ∣ - 1 ∣ - ln ∣ - 2 ∣ = - ln 2

Orientation: why a decreasing $u$ flips the limits

Worked flip:

\int_{0}^{π} sin x cos x d x

with

u = sin x

. What are the new limits, and what is the value?

x = 0 \Rightarrow u = 0

and

x = π \Rightarrow u = sin π = 0

; the limits coincide, so the integral is

\int_{0}^{0} u d u = 0

— the sweep goes up to

u = 1

and back to

0

, and the two halves cancel exactly.

Edge cases

u = g (x)

takes the same value at both endpoints (e.g.

g (a) = g (b)

), what is the definite integral?

Zero — the new limits coincide, and

\int_{c}^{c} f (u) d u = 0

regardless of

f

, because the range has collapsed to a point.

What if the substitution

u = g (x)

is not one-to-one over

[a, b]

(it turns around inside)?

You cannot blindly use

g (a)

and

g (b)

as limits; split the interval at the turning point so

g

is monotonic on each piece, then substitute on each separately. See Definite Integral & Fundamental Theorem of Calculus.

What if

g

maps part of

[a, b]

outside the domain where the integrand makes sense (e.g. into where

g (x)

1/ u

is undefined)?

The substitution is invalid there — u-sub requires

g

to be continuous and to keep every

u

-value inside the new integrand's domain across the whole interval; otherwise you must restrict or split the interval to a region where it does.

Does u-substitution require

g

to be continuous, or just monotonic?

It needs

g

(and

g^{'}

) continuous on

[a, b]

so the differential

d u = g^{'} (x) d x

and the FTC apply; monotonicity is an extra comfort that lets you use

g (a), g (b)

directly, but continuity is the non-negotiable requirement.

If a chosen

u

makes

d u

appear with the wrong sign (e.g.

d u = - sin x d x

but you have

+ sin x d x

), what do you do?

Absorb the sign as a constant

- 1

: write

sin x d x = - d u

; the negative is a constant and rides safely outside the integral.

When integrating

\int \frac{g ^{'} ( x )}{g ( x )} d x

, why must the answer be

ln ∣ g (x) ∣ + C

and not

ln g (x) + C

With

u = g (x)

this is

\int \frac{d u}{u} = ln ∣ u ∣ + C

; the absolute value covers intervals where

g (x) < 0

, on which

ln g (x)

would be undefined.

What happens to the constant of integration

C

when you change limits in a definite integral?

It disappears entirely — evaluating

F (u)

at both limits and subtracting cancels any

+ C

, which is why definite answers are bare numbers.

g^{'} (x) = 0

at an interior point but

g

is still monotonic overall, is u-sub valid?

Yes — a single zero-slope point doesn't break the substitution as long as

g

is still one-to-one on the interval; the change-of-limits formula only needs the endpoint values

g (a), g (b)

Can u-substitution ever increase the difficulty of an integral?

Yes — a poor choice of

u

can leave a leftover

x

that must be re-expressed messily, producing something harder than the original; choosing

u

is a judgement guided by spotting the inner-derivative pair.

Recall One-line survival checklist

Did I convert $d x$ to $d u$ (the height-rescale)? Did I choose $u$ whose derivative is present (up to a constant)? For definite integrals, did I change limits and not back-substitute? For $1/ u$ or $\cdot$ , did I respect absolute values / domain? If a variable won't match $d u$ , did I solve $u = g (x)$ for $x$ instead of pulling it out?

Connections

U-substitution — technique, change of limits for definite integrals
Chain Rule — the rule u-sub inverts.
Antiderivatives & Indefinite Integrals — where the $+ C$ and $ln ∣ u ∣$ live.
Definite Integral & Fundamental Theorem of Calculus — justifies limit-changing and sign flips.
Integration by Parts — contrast: inverts the product rule.
Trigonometric Substitution — a specialized u-sub.

Why questions — the geometric heart

True or false — justify

Spot the error — grouped by the misconception

Orientation: why a decreasing u flips the limits

Edge cases

Connections

Orientation: why a decreasing $u$ flips the limits