1.08PureStretch

Integration by Substitution

Q: Do I have to change the limits?

Either change them to u-values, or substitute back to x before using the original limits. Mixing the two is wrong.

Q: Why does substitution work?

It is the chain rule in reverse: g( f(x)) f^ (x), dx = g(u), du with u = f(x).

Integrals like $\int 2x(x^{2} + 1)^{5}\,\mathrm{d}x$ or $\int \dfrac{x}{\sqrt{2x + 1}}\,\mathrm{d}x$ do not match any standard result. $\textbf{Integration by substitution}$ introduces a new variable that absorbs the awkward part, rewrites the whole integral in that variable, and leaves something standard. It is the reverse of the chain rule, and alongside integration by parts it is one of the two main techniques for non-standard integrals.

50 min Integration

The big picture

Every derivative rule has an integration counterpart. The chain rule gives substitution: if an integrand contains a function together with (a multiple of) its derivative, that inner function is the natural substitution. The method is mechanical once set up — choose $u$ , find $\dfrac{\mathrm{d}u}{\mathrm{d}x}$ , replace $\mathrm{d}x$ , and rewrite every $x$ in terms of $u$ — but three things go wrong repeatedly: leaving stray $x$ terms, forgetting to change the limits in definite integrals, and choosing a substitution that makes things worse. Recognising the pattern $\int \mathrm{f}^{\prime}(x)\,\mathrm{g}(\mathrm{f}(x))\,\mathrm{d}x$ at sight, and knowing when to reach for parts instead, is what this lesson builds.

What you'll be able to do

Recognise integrals of the form $\int \mathrm{f}^{\prime}(x)\,[\mathrm{f}(x)]^{n}\,\mathrm{d}x$ and $\int \dfrac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}\,\mathrm{d}x$
Integrate using a given or chosen substitution
Change the limits when evaluating definite integrals by substitution
Decide between substitution and integration by parts

Integration by inspection

Differentiating $[\mathrm{f}(x)]^{n+1}$ gives $(n+1)\mathrm{f}^{\prime}(x)[\mathrm{f}(x)]^{n}$ . So when an integrand is a bracket to a power multiplied by the derivative of the inside, you can often write the answer directly and adjust the constant.

Similarly, $\dfrac{\mathrm{d}}{\mathrm{d}x}\ln|\mathrm{f}(x)| = \dfrac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}$ , so a fraction whose numerator is the derivative of its denominator integrates to a log.

\int \dfrac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}\,\mathrm{d}x = \ln|\mathrm{f}(x)| + c

Derivative over function.

\int \mathrm{f}^{\prime}(x)[\mathrm{f}(x)]^{n}\,\mathrm{d}x = \dfrac{[\mathrm{f}(x)]^{n+1}}{n+1} + c

Reverse chain rule,

n \neq -1

Example 1Find

\int x^{2}(x^{3} - 4)^{6}\,\mathrm{d}x

and

\int \dfrac{\cos x}{2 + \sin x}\,\mathrm{d}x

1Try

(x^{3} - 4)^{7}

: it differentiates to

21x^{2}(x^{3} - 4)^{6}

, so divide by 21.

2The numerator

\cos x

is the derivative of

2 + \sin x

Answer

\tfrac{1}{21}(x^{3} - 4)^{7} + c

and

\ln(2 + \sin x) + c

Why this works

$\int \tan x\,\mathrm{d}x$ looks non-standard, but $\tan x = \dfrac{\sin x}{\cos x}$ and the numerator is minus the derivative of the denominator, so the integral is $-\ln|\cos x| + c = \ln|\sec x| + c$ .

The method of substitution

Step 1: choose $u$ (usually the awkward inner expression, or given in the question).

Step 2: differentiate to find $\dfrac{\mathrm{d}u}{\mathrm{d}x}$ , and rearrange to replace $\mathrm{d}x$ .

Step 3: rewrite the $\textit{entire}$ integrand in terms of $u$ — no $x$ may remain.

Step 4: integrate with respect to $u$ , then substitute back to $x$ .

Example 1Use the substitution

u = 2x + 1

to find

\int \dfrac{x}{\sqrt{2x + 1}}\,\mathrm{d}x

\dfrac{\mathrm{d}u}{\mathrm{d}x} = 2

, so

\mathrm{d}x = \tfrac12\mathrm{d}u

. Also

x = \tfrac{u - 1}{2}

2Integral becomes

\int \dfrac{(u - 1)/2}{u^{1/2}} \cdot \tfrac12\,\mathrm{d}u = \tfrac14\int (u^{1/2} - u^{-1/2})\,\mathrm{d}u

= \tfrac14\left(\tfrac23u^{3/2} - 2u^{1/2}\right) + c = \tfrac16u^{3/2} - \tfrac12u^{1/2} + c

Answer

\tfrac16(2x + 1)^{3/2} - \tfrac12(2x + 1)^{1/2} + c

Tip — If $x$ appears outside the substituted expression, rearrange $u = \dots$ to get $x$ in terms of $u$ . This step is the one most often skipped.

Definite integrals and changing limits

In a definite integral, change the limits to $u$ -values at the same time as changing the variable. Then evaluate directly in $u$ — no need to substitute back.

Write the new limits next to the old ones as you convert: when $x = a$ , $u = \dots$ ; when $x = b$ , $u = \dots$

Example 1Use

u = 1 + x^{2}

to evaluate

\int_{0}^{\sqrt3} \dfrac{4x}{\sqrt{1 + x^{2}}}\,\mathrm{d}x

\mathrm{d}u = 2x\,\mathrm{d}x

, so

4x\,\mathrm{d}x = 2\,\mathrm{d}u

2Limits:

x = 0 \Rightarrow u = 1

;

x = \sqrt3 \Rightarrow u = 4

\int_{1}^{4} 2u^{-1/2}\,\mathrm{d}u = \big[4u^{1/2}\big]_{1}^{4} = 8 - 4

Answer

4

Substitution or parts?

Substitution suits a composite function multiplied by (roughly) the derivative of its inside. $\textbf{Integration by parts}$ suits a product of two unrelated functions, such as $x\mathrm{e}^{x}$ or $x\cos x$ , and comes from the product rule.

For parts, choose $u$ to be the function that gets simpler when differentiated (usually the polynomial, but $\ln x$ takes priority) and $\dfrac{\mathrm{d}v}{\mathrm{d}x}$ the part you can integrate.

\int u\dfrac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\dfrac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x

Integration by parts.

Example 1Find

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

1No inner function with its derivative, so use parts:

u = x

\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{2x}

\dfrac{\mathrm{d}u}{\mathrm{d}x} = 1

v = \tfrac12\mathrm{e}^{2x}

\tfrac12x\mathrm{e}^{2x} - \int \tfrac12\mathrm{e}^{2x}\,\mathrm{d}x

Answer

\tfrac12x\mathrm{e}^{2x} - \tfrac14\mathrm{e}^{2x} + c

Tip — A quick test: if differentiating one factor would produce the other (up to a constant), substitute. If not, and one factor is a polynomial, try parts.

Think like an examiner

Show $\mathrm{d}x$ being replaced — " $\mathrm{d}x = \tfrac12\mathrm{d}u$ " — as a separate line.
Ensure no $x$ remains after substitution.
Change limits to $u$ -values in definite integrals, and state them.
Include $+\,c$ and return to $x$ for indefinite integrals.

Common misconceptions

Replacing

\mathrm{d}x

with

\mathrm{d}u

without the factor.

Use

\dfrac{\mathrm{d}u}{\mathrm{d}x}

to convert:

\mathrm{d}x = \dfrac{\mathrm{d}u}{\mathrm{d}u/\mathrm{d}x}

Using the original

x

-limits after substituting.

Convert limits to

u

, or substitute back before evaluating.

Leaving a stray

x

inside a

\mathrm{d}u

integral.

Express

x

in terms of

u

from the substitution.

Integrating

\dfrac{x}{x^{2} + 1}

\ln(x^{2} + 1)

The numerator is half the derivative:

\tfrac12\ln(x^{2} + 1) + c

Integration techniques

\int \tfrac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}\,\mathrm{d}x = \ln|\mathrm{f}(x)| + c

Log form.

\mathrm{d}x = \tfrac{\mathrm{d}u}{\mathrm{d}u/\mathrm{d}x}

Changing variable.

\int uv^{\prime}\,\mathrm{d}x = uv - \int vu^{\prime}\,\mathrm{d}x

By parts.

Stretch yourself

Use the substitution $u = \sin x$ to show that $\int_{0}^{\pi/2} \sin^{2}x\cos^{3}x\,\mathrm{d}x = \dfrac{2}{15}$ .

Hint — Write $\cos^{3}x = \cos^{2}x \cdot \cos x = (1 - \sin^{2}x)\cos x$ . The spare $\cos x$ goes with $\mathrm{d}x$ to make $\mathrm{d}u$ .

Reveal solution

SolutionWith

u = \sin x

\mathrm{d}u = \cos x\,\mathrm{d}x

. Limits:

x = 0 \Rightarrow u = 0

and

x = \tfrac{\pi}{2} \Rightarrow u = 1

. Rewrite the integrand:

\sin^{2}x\cos^{3}x = \sin^{2}x(1 - \sin^{2}x)\cos x

. So the integral becomes

\int_{0}^{1} u^{2}(1 - u^{2})\,\mathrm{d}u = \int_{0}^{1} (u^{2} - u^{4})\,\mathrm{d}u = \big[\tfrac13u^{3} - \tfrac15u^{5}\big]_{0}^{1} = \tfrac13 - \tfrac15 = \tfrac{2}{15}

, as required.

\blacksquare

Questions students ask

How do I choose a substitution?

Look for an inner function whose derivative also appears in the integrand, or a root of a linear expression. If one is given, use it.

Do I have to change the limits?

Either change them to

u

-values, or substitute back to

x

before using the original limits. Mixing the two is wrong.

When is integration by parts better?

For products of unrelated functions, like a polynomial times an exponential or trig function, or anything involving

\ln x

multiplied by a power.

Why does substitution work?

It is the chain rule in reverse:

\int \mathrm{g}(\mathrm{f}(x))\mathrm{f}^{\prime}(x)\,\mathrm{d}x = \int \mathrm{g}(u)\,\mathrm{d}u

with

u = \mathrm{f}(x)

Key takeaways

Spot $\mathrm{f}^{\prime}(x)[\mathrm{f}(x)]^{n}$ and $\tfrac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}$ to integrate by inspection.
Substitution: choose $u$ , convert $\mathrm{d}x$ , rewrite everything in $u$ , integrate, return to $x$ .
In definite integrals, change the limits to $u$ -values.
Integration by parts handles products: $\int uv^{\prime} = uv - \int vu^{\prime}$ .
Choose the technique from the structure of the integrand.