IB Mathematics: Analysis and Approaches SL/HL — Topic 5 Calculus

Integration by Substitution

Recognise the inside function and its matching derivative, then change variable completely so the integral becomes standard.

AA SL/HL · SL 5.10

Learning goal

Recognise the SL 5.10 product pattern, define \(u\), form \(du\), rewrite the entire integral in \(u\), then integrate and complete the answer correctly.

Syllabus link

SL 5.10 Integration by substitution and reverse chain rule.

Big idea

Substitution is a change of variable. Integration by inspection is the same mathematical method carried out mentally.

Target pattern

\(\displaystyle \int k\,g'(x)f(g(x))\,dx\).

What kind of integral is this method for?

At AA SL, the integrand is a product of two functions: a function of an inside expression \(g(x)\), and \(g'(x)\) or a constant multiple of it.

\[\int \underbrace{k\,g'(x)}_{\text{multiple of the inner derivative}}\,\underbrace{f(g(x))}_{\text{function of the inner expression}}\,dx.\]

Two factors, a function of the inside expression and a matching derivative, combine to make the substitution pattern

For example, \(\int2x(x^2+1)^4\,dx\) contains \((x^2+1)^4\) and the derivative of its inside expression, since \(\dfrac{d}{dx}(x^2+1)=2x\).

Important precision

Not every product is a direct substitution integral. In \(\int xe^x\,dx\), neither factor is the derivative of an inner expression inside the other factor. Integration by parts, an AA HL method, is appropriate here.

The product may also be hidden:

\[\int\frac{\sin x}{\cos x}\,dx=\int(\sin x)\left(\frac1{\cos x}\right)\,dx.\]

Why “integration by changing the variable” is a better description

We are dealing with \(\int2x(x^2+1)^4\,dx\). The repeated inside expression is \(x^2+1\). Define the new variable clearly:

\[\text{Let }u=x^2+1.\qquad\frac{du}{dx}=2x\quad\Longrightarrow\quad du=2x\,dx.\]

\[\int2x(x^2+1)^4\,dx=\int\underbrace{(x^2+1)^4}_{u^4}\,\underbrace{2x\,dx}_{du}=\int u^4\,du.\]

The variable of integration has genuinely changed: the original integral is written entirely in \(x\), while the new integral is written entirely in \(u\).

An x-integral becomes an integral of u to the fourth power when u equals x squared plus one

The language of the method

We still use the standard names integration by substitution and u-substitution. Remember what they mean: we change the variable of integration from \(x\) to \(u\).

The practical method

The five-step u-substitution method

Define the new variable. Let \(u\) equal the repeated inside expression.
Differentiate. Find \(du/dx\), then write \(du=g'(x)\,dx\).
Rewrite completely. Replace the inside expression and the matching differential factor so that no \(x\) remains. For a definite integral completed in \(u\), also change its limits to \(u\)-values.
Integrate with respect to u. Use the relevant standard integral.
Complete the answer. For an indefinite integral, return to \(x\) and include \(+C\). For a definite integral completed in \(u\), evaluate with the new \(u\)-limits and do not include \(+C\).

Worked example 1: the exact SL 5.10 product pattern

Find \(\int2x(x^2+1)^4\,dx\).

Solution

Step 1: Define the new variable.

\[u=x^2+1\]

Step 2: Differentiate.

\[\frac{du}{dx}=2x\quad\Longrightarrow\quad du=2x\,dx\]

Step 3: Rewrite the integral in terms of the new variable.

Rearrange the factors and identify exactly what each part becomes:

\[\int2x(x^2+1)^4\,dx=\int\underbrace{(x^2+1)^4}_{u^4}\,\underbrace{2x\,dx}_{du}=\int u^4\,du.\]

Step 4: Integrate with respect to u.

\[\int u^4\,du=\frac{u^5}{5}+C.\]

Step 5: Change back to x.

\[\boxed{\frac{(x^2+1)^5}{5}+C}\]

U-substitution Video 1

Use this tutorial after learning to identify the inside expression, form \(du\), and change the integral completely into \(u\).

Watch on YouTube →

When the matching derivative differs by a constant

The derivative factor does not need to match perfectly at first. A constant can be taken outside the integral or absorbed into \(du\).

Worked example 2: adjusting a constant factor

Find \(\int4x\sin(x^2)\,dx\).

Solution

Step 1: Define the new variable.

\[u=x^2\]

Step 2: Differentiate.

\[du=2x\,dx\quad\Longrightarrow\quad4x\,dx=2\,du.\]

Step 3: Rewrite the integral in terms of the new variable.

\[\int4x\sin(x^2)\,dx=\int\underbrace{\sin(x^2)}_{\sin u}\,\underbrace{4x\,dx}_{2\,du}=2\int\sin u\,du.\]

Step 4: Integrate with respect to u.

\[2\int\sin u\,du=-2\cos u+C.\]

Step 5: Change back to x.

\[\boxed{-2\cos(x^2)+C}\]

Worked example 3: an exponential composite

Find \(\int3x^2e^{x^3-2}\,dx\).

Solution

Step 1: Define the new variable.

\[u=x^3-2\]

Step 2: Differentiate.

\[du=3x^2\,dx.\]

Step 3: Rewrite the integral in terms of the new variable.

\[\int3x^2e^{x^3-2}\,dx=\int\underbrace{e^{x^3-2}}_{e^u}\,\underbrace{3x^2\,dx}_{du}=\int e^u\,du.\]

Step 4: Integrate with respect to u.

\[\int e^u\,du=e^u+C.\]

Step 5: Change back to x.

\[\boxed{e^{x^3-2}+C}\]

Do not leave a mixture of x and u

After changing variables, \(\int2xu^4\,du\) is not valid progress: the differential is \(du\), but the integrand still contains \(x\). The whole integral must be written in one variable.

The logarithmic pattern

\[\boxed{\int\frac{g'(x)}{g(x)}\,dx=\ln|g(x)|+C}.\]

Indeed, let \(u=g(x)\). Then \(du=g'(x)\,dx\), so

\[\int\frac{g'(x)}{g(x)}\,dx=\int\frac1u\,du=\ln|u|+C=\ln|g(x)|+C.\]

Work on an interval where \(g(x)\ne0\).

Worked example 4: using the logarithmic shortcut and the five-step method

Find \(\int\frac{\sin x}{\cos x}\,dx\).

Solution

Quick method: use the logarithmic pattern.

Let \(g(x)=\cos x\), so \(g'(x)=-\sin x\). The numerator is \(\sin x=-g'(x)\). Therefore

\[\begin{aligned}\int\frac{\sin x}{\cos x}\,dx&=-\int\frac{g'(x)}{g(x)}\,dx\\&=-\ln|g(x)|+C\\&=-\ln|\cos x|+C.\end{aligned}\]

Now use the full five-step method to obtain the same result.

Step 1: Define the new variable.

\[u=\cos x\]

Step 2: Differentiate.

\[du=-\sin x\,dx\quad\Longrightarrow\quad\sin x\,dx=-du.\]

Step 3: Rewrite the integral in terms of the new variable.

Write the quotient as a product, then replace both factors:

\[\int\frac{\sin x}{\cos x}\,dx=\int\underbrace{\frac1{\cos x}}_{1/u}\,\underbrace{\sin x\,dx}_{-du}=-\int\frac1u\,du.\]

Step 4: Integrate with respect to u.

\[-\int\frac1u\,du=-\ln|u|+C.\]

Step 5: Change back to x.

\[\boxed{-\ln|\cos x|+C}\]

Why the shortcut is useful

The five steps explain why the result works. Once the logarithmic pattern is familiar, identify \(g(x)\), compare the numerator with \(g'(x)\), and adjust any constant or sign.

Worked example 5: a hidden product and a constant adjustment

Find \(\int\frac6{3x+2}\,dx\).

Solution

Write the quotient as \(6\left(\dfrac1{3x+2}\right)\).

Step 1: Define the new variable.

\[u=3x+2\]

Step 2: Differentiate.

\[du=3\,dx\quad\Longrightarrow\quad6\,dx=2\,du.\]

Step 3: Rewrite the integral in terms of the new variable.

\[\int\frac6{3x+2}\,dx=\int\underbrace{\frac1{3x+2}}_{1/u}\,\underbrace{6\,dx}_{2\,du}=2\int\frac1u\,du.\]

Step 4: Integrate with respect to u.

\[2\int\frac1u\,du=2\ln|u|+C.\]

Step 5: Change back to x.

\[\boxed{2\ln|3x+2|+C}\]

Integration by inspection is still u-substitution

Sometimes the pattern is so clear that we can write the answer immediately. This is called integration by inspection or using the reverse chain rule. For example,

\[\int3x^2e^{x^3+1}\,dx=e^{x^3+1}+C.\]

Technically, let \(u=x^3+1\), so \(du=3x^2\,dx\). This change of variable is still the mechanism; inspection carries it out mentally.

The central connection

Reverse-chain-rule inspection is a shortened form of changing the variable.

Integral	New variable	Result by inspection
\(\int2x\cos(x^2)\,dx\)	\(u=x^2\)	\(\sin(x^2)+C\)
\(\int5e^{5x-1}\,dx\)	\(u=5x-1\)	\(e^{5x-1}+C\)
\(\int\frac1{3x+2}\,dx\)	\(u=3x+2\)	\(\frac13\ln\|3x+2\|+C\)

U-substitution Video 2

Further examples, hidden product structures and less immediate reverse-chain-rule patterns.

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Looking ahead: changing limits in a definite integral

Link to SL 5.11. Definite integrals are developed in the next section. When using substitution, the limits may be changed from \(x\)-values to their corresponding \(u\)-values.

Worked example 6: two valid routes for a definite integral

Find \(\int_0^1\frac{2x}{x^2+3}\,dx\).

Solution

Step 1: Define the new variable.

\[u=x^2+3.\]

Step 2: Differentiate.

\[du=2x\,dx.\]

Before choosing either option, identify the two replacements:

\[\underbrace{\frac1{x^2+3}}_{1/u}\,\underbrace{2x\,dx}_{du}.\]

Option 1: change the limits and complete the calculation in u

Step 3: Rewrite the integral and change the limits.

x	u = x² + 3
0	3
1	4

\[\int_0^1\frac{2x}{x^2+3}\,dx=\int_3^4\frac1u\,du.\]

Step 4: Integrate with respect to u.

\[\int_3^4\frac1u\,du=[\ln u]_3^4.\]

Step 5: Evaluate using the new u-limits.

\[\ln4-\ln3=\boxed{\ln\left(\frac43\right)}.\]

Once the limits are changed to \(u\)-values, there is no need to change back to \(x\).

Option 2: return to x, then use the original x-limits

You may instead obtain an antiderivative in \(x\) before applying the original bounds.

Step 3: Rewrite an indefinite integral in u.

Temporarily write the integral without limits:

\[\int\frac{2x}{x^2+3}\,dx=\int\frac1u\,du.\]

Step 4: Integrate with respect to u.

\[\int\frac1u\,du=\ln|u|+C.\]

Step 5: Change back to x, then apply the original x-limits.

Since \(u=x^2+3>0\), the antiderivative is \(\ln(x^2+3)+C\). Therefore

\[\begin{aligned}\int_0^1\frac{2x}{x^2+3}\,dx&=[\ln(x^2+3)]_0^1\\&=\ln4-\ln3\\&=\boxed{\ln\left(\frac43\right)}.\end{aligned}\]

Both options are correct. Option 1 changes the limits and remains entirely in \(u\); Option 2 returns to \(x\) before the original \(x\)-limits are used. There is no \(+C\) in the final answer to a definite integral.

Do not mix the two routes

Use one route consistently. Do not put the original \(x\)-limits on an integral written in \(u\), or use new \(u\)-limits with an antiderivative that you have changed back to \(x\).

Definite integrals with u-substitution

Compare evaluating entirely in \(u\) with returning to an antiderivative in \(x\) before using the original limits.

Watch on YouTube →

How to define the new variable u

A useful definition of \(u\) normally satisfies three tests:

\(u\) is a repeated or clearly nested expression.
The derivative of \(u\) appears elsewhere as another factor, perhaps multiplied by a constant.
After substitution, every \(x\) disappears and the new integral is simpler.

Integral	Let	Reason
\(\int7x(x^2-4)^9\,dx\)	\(u=x^2-4\)	The derivative \(2x\) matches \(7x\) up to a constant.
\(\int\cos x\,e^{\sin x}\,dx\)	\(u=\sin x\)	The derivative of \(\sin x\) is \(\cos x\).
\(\int\frac{x^2}{x^3+5}\,dx\)	\(u=x^3+5\)	The derivative \(3x^2\) is present up to a constant.

Common mistakes and a final check

Common mistakes

Defining \(u\) but failing to replace every occurrence of \(x\).
Forgetting the constant needed to match \(du\).
Treating \(dx\) as decoration.
Forgetting to return to \(x\) and add \(+C\) for an indefinite integral.
Adding \(+C\) to a definite integral.
Writing \(\ln(g(x))\) instead of \(\ln|g(x)|\) when sign is unknown.

The quickest reliable check

Differentiate your final answer. The chain rule should reconstruct the original product:

Differentiation uses the chain rule; integration reverses it by changing the variable.

Practice

A. Identify the change of variable

Complete a statement “Let \(u=\cdots\)” and write \(du\).

\(\int8x(4x^2+1)^7dx\)
\(\int\cos x\,e^{\sin x}dx\)
\(\int\frac{x^2}{x^3+5}dx\)
\(\int\sin(3x)dx\)

B. Evaluate

\(\int8x(4x^2+1)^7dx\)
\(\int\cos x\,e^{\sin x}dx\)
\(\int\frac{x^2}{x^3+5}dx\)
\(\int6x\cos(3x^2)dx\)
\(\int\frac{2x}{(x^2+4)^2}dx\)
\(\int3x^2\sqrt{x^3+1}dx\)
\(\int e^{2x+1}dx\)
\(\int\frac{\sin x}{\cos x}dx\)

C. Explain

Why is \(\int xe^x dx\) not a direct SL 5.10 substitution integral?
A student lets \(u=x^2+1\) in \(\int2x(x^2+1)^4dx\), then writes \(\int2xu^4du\). Identify the error.

Answer key

Show answers 1–14

A

\(u=4x^2+1,\ du=8x dx\)
\(u=\sin x,\ du=\cos x dx\)
\(u=x^3+5,\ du=3x^2dx\)
\(u=3x,\ du=3dx\)
This integral can also be evaluated by inspection.

B

\((4x^2+1)^8/8+C\)
\(e^{\sin x}+C\)
\(\frac13\ln|x^3+5|+C\)
\(\sin(3x^2)+C\)
\(-1/(x^2+4)+C\)
\(\frac23(x^3+1)^{3/2}+C\)
\(\frac12e^{2x+1}+C\)
\(-\ln|\cos x|+C\)

C

Letting \(u=x\) does not simplify \(xe^x\), while letting \(u=e^x\) does not directly remove the remaining \(x\) factor. It is not a direct matching-derivative integral of the SL 5.10 form; integration by parts is needed.
\(du=2x dx\), so the whole factor \(2x dx\) must be replaced: \(\int2x(x^2+1)^4dx=\int u^4du\).

Quick reference and related tutorials

One-minute summary

Look for a function of \(g(x)\) multiplied by a constant multiple of \(g'(x)\).
Let \(u=g(x)\), then replace \(g'(x)\,dx\) by \(du\).
The transformed integral must contain only \(u\) and \(du\).
For an indefinite integral, integrate in \(u\), return to \(x\), and include \(+C\).
For a definite integral, either change the limits and remain in \(u\), or return to an antiderivative in \(x\) before applying the original \(x\)-limits.
Reverse-chain-rule inspection is the same change of variable carried out mentally.

\[\int k\,g'(x)f(g(x))\,dx=k\int f(u)\,du,\qquad u=g(x).\]

U-substitution Video 1 — identifying u, forming du and changing the variable completely.
U-substitution Video 2 — further examples and hidden product structures.
Definite integrals with u-substitution — both valid routes for definite integrals.