Integrating Using Substitution

AP Calculus · Integration and Accumulation of Change

Integrating Using Substitution

The chain rule in reverse: find the inner function whose derivative is already there.

key idea$\int f(g(x))g'(x)\,dx=\int f(u)\,du,\quad u=g(x)$

The procedure

Substitution is the chain rule read backwards. You are looking for a composition whose inner function's derivative is also sitting in the integrand — up to a constant. The five stages below never change; only the choice of $u$ does.

Choose $u$ = the inner function

Look for something in parentheses, under a root, in an exponent, or in a denominator. That is almost always $u$. If two choices look plausible, pick the one whose derivative you can find elsewhere in the integrand.

in the integrand out $u=g(x)$

Differentiate to get $du$

Compute $\dfrac{du}{dx}=g'(x)$, then write $du=g'(x)\,dx$. Solve for $dx$ if that is more convenient: $dx=\dfrac{du}{g'(x)}$.

in $u$ out $du$

Replace everything

Every $x$ must disappear — including the one hidden in $dx$. If an $x$ survives that you cannot express in terms of $u$, the substitution has failed and you should choose differently.

in an integral in $x$ out an integral in $u$ alone

Integrate in $u$

The new integral should be one of the basic forms. If it is not, back up — substitution was either the wrong tool or you picked the wrong $u$.

in $\int f(u)\,du$ out $F(u)+C$

Undo the substitution — or change the limits

For an indefinite integral, write $u$ back in terms of $x$. For a definite integral, either convert back and use the original limits, or convert the limits to $u$-values at stage 2 and never return to $x$.

out an answer in $x$, or a number

If you see…	Try $u=$	Because $du$ supplies…
$\left(3x^{2}+1\right)^{5}\cdot x\,dx$	$3x^{2}+1$	$6x\,dx$ — the stray $x$
$\dfrac{\ln x}{x}\,dx$	$\ln x$	$\tfrac1x\,dx$ — the denominator
$\sqrt{4x+7}\,dx$	$4x+7$	a constant $4\,dx$; adjust with $\tfrac14$
$\dfrac{f'(x)}{f(x)}\,dx$	$f(x)$	exactly the numerator → gives $\ln\lvert u\rvert$
$\sin^{3}x\cos x\,dx$	$\sin x$	$\cos x\,dx$
$e^{\,\tan x}\sec^{2}x\,dx$	$\tan x$	$\sec^{2}x\,dx$

In every row the derivative of the chosen $u$ already appears in the integrand, give or take a constant. That is the signal that substitution will work.

all four must hold

The integrand contains a composition — something inside something else.
The inner function's derivative is also present, up to a constant multiple.
After substituting, no $x$ remains anywhere, including in $dx$.
The resulting $u$-integral is one you already know.

Worked examples

EX 1

The clean case

$u=3x^{2}+1$

Evaluate $\displaystyle\int x\left(3x^{2}+1\right)^{5}dx$.

Choose $u$

The inner function is the bracket: $u=3x^{2}+1$.

Get $du$

$\dfrac{du}{dx}=6x$, so $du=6x\,dx$, hence $x\,dx=\tfrac16\,du$.

Replace everything

$\displaystyle\int x\left(3x^{2}+1\right)^{5}dx=\int u^{5}\cdot\tfrac16\,du$. No $x$ remains.

Integrate in $u$

$\tfrac16\cdot\dfrac{u^{6}}{6}=\dfrac{u^{6}}{36}$.

Undo the substitution

$\displaystyle\frac{\left(3x^{2}+1\right)^{6}}{36}+C$.

EX 2

Only a constant is missing

$u=4x+7$

Evaluate $\displaystyle\int \sqrt{4x+7}\,dx$.

Choose $u$

$u=4x+7$, the expression under the root.

Get $du$

$du=4\,dx$, so $dx=\tfrac14\,du$. Notice the integrand contains no $x$ factor — that is fine, because $g'=4$ is a constant we can supply ourselves.

Replace everything

$\displaystyle\int u^{1/2}\cdot\tfrac14\,du$.

Integrate in $u$

$\tfrac14\cdot\dfrac{u^{3/2}}{3/2}=\dfrac{u^{3/2}}{6}$.

Undo

$\displaystyle\frac{(4x+7)^{3/2}}{6}+C$. A missing constant can always be fixed; a missing variable factor cannot.

EX 3

Definite integral — change the limits

$u=x^{2}+1$

Evaluate $\displaystyle\int_{0}^{2} \frac{x}{x^{2}+1}\,dx$.

Choose $u$

$u=x^{2}+1$, the denominator.

Get $du$ and convert the limits

$du=2x\,dx$, so $x\,dx=\tfrac12\,du$. When $x=0$, $u=1$; when $x=2$, $u=5$.

Replace everything, limits included

$\displaystyle\int_{1}^{5}\frac{1}{u}\cdot\tfrac12\,du=\tfrac12\int_{1}^{5}\frac{du}{u}$.

Integrate

$\tfrac12\Bigl[\ln\lvert u\rvert\Bigr]_{1}^{5}=\tfrac12(\ln 5-\ln 1)=\tfrac12\ln 5$.

Never convert back

Because the limits are now $u$-values, the answer is final. Substituting $x$ back and keeping $1$ and $5$ would be a genuine error.

EX 4

When a stray $x$ survives

back-substitute

Evaluate $\displaystyle\int x\sqrt{x-3}\,dx$.

Choose $u$

$u=x-3$, so $du=dx$.

Spot the problem

Replacing gives $\displaystyle\int x\sqrt{u}\,du$ — an $x$ has survived, so stage 3 is not complete.

Express the stray $x$ in terms of $u$

From $u=x-3$ we get $x=u+3$. Now everything is in $u$: $\displaystyle\int (u+3)\sqrt{u}\,du$.

Expand and integrate

$\displaystyle\int\left(u^{3/2}+3u^{1/2}\right)du=\frac{2}{5}u^{5/2}+2u^{3/2}$.

Undo

$\displaystyle\frac{2}{5}(x-3)^{5/2}+2(x-3)^{3/2}+C$. A surviving $x$ does not always mean failure — sometimes you can solve for it.

Where it goes wrong

Forgetting $dx$ is part of the substitution

$dx$ must be replaced too. Writing $\int u^{5}dx$ leaves a mixed integral that means nothing.

Adjusting for a variable instead of a constant

You may pull a constant like $\tfrac16$ outside the integral. You may never pull an $x$ outside — if the missing factor involves the variable, substitution is the wrong choice.

Definite limits left in $x$

After converting to $u$, the limits must be $u$-values. Mixing $u$-integrand with $x$-limits is the most common definite-integral error on this topic.

Choosing the outer function as $u$

$u$ is the inside. For $\sqrt{4x+7}$ take $u=4x+7$, not $u=\sqrt{4x+7}$; the latter leaves you worse off than you started.

Course map. Topic 6.9 · LO FUN-6.D — calculate integrals using substitution (EK FUN-6.D.1–FUN-6.D.2).