P4 6.2 - Integration by substitution | Edexcel A-Level International Mathematics

Use a new variable to turn a complicated integral into a familiar one. You will connect substitution to the chain rule, change every part of the integrand consistently, and evaluate definite integrals with appropriate limits and domains.

Reversing the chain rule

Why does replacing $x^2+4$ by one letter help with $\int2x\sqrt{x^2+4}\,dx$ ? Its derivative $2x$ is already beside it. Together they become a single differential.

For a small change in $x$ , the change in $u=x^2+4$ is approximately $2x$ times as large. In calculus notation, $du/dx=2x$ , so $du=2x\,dx$ . The whole expression $2x\,dx$ is replaced, not just the letter $x$ .

Integration by substitution

Choose $u=g(x)$ and rewrite an integral entirely in terms of $u$ , using $du=g'(x)\,dx$ . Integrate the simpler expression, then return to $x$ for an indefinite integral.

This reverses the chain rule: if $F'(u)=f(u)$ , then differentiating $F(g(x))$ gives $f(g(x))g'(x)$ .

Worked example 1: the derivative is present

Find $\int2x\sqrt{x^2+4}\,dx$ .

Plan. Substitute $u=x^2+4$ , because its derivative supplies the other factor.

u=x^2+4,\qquad du=2x\,dx

\int2x\sqrt{x^2+4}\,dx =\int u^{1/2}\,du =\frac23u^{3/2}+C.

The last step is the power rule in the new variable.

Answer. $\frac23(x^2+4)^{3/2}+C$ .

Check. Differentiation gives $\frac23\frac32(x^2+4)^{1/2}(2x)=2x\sqrt{x^2+4}$ . The square root exists for every real $x$ .

Only the outside coefficient changes in the next example. A constant multiple can be carried through the substitution.

Worked example 2: a missing constant factor

Find $\int5x\sqrt{x^2+4}\,dx$ .

Plan. Use the same substitution and write $5x\,dx=\frac52\,du$ .

\begin{aligned} \int5x\sqrt{x^2+4}\,dx &=\frac52\int u^{1/2}\,du\\ &=\frac52\frac23u^{3/2}+C \end{aligned}

Answer. $\frac53(x^2+4)^{3/2}+C$ .

Check. Its derivative is $\frac53\frac32(x^2+4)^{1/2}(2x)=5x\sqrt{x^2+4}$ .

A substitution makes the reverse-chain pattern explicit and also works when the remaining factors need rearranging.

Watch out

Writing $du=dx$ here loses a factor of $2x$ . Differentiate your chosen substitution before replacing the differential. An indefinite answer in $u$ is unfinished when the question uses $x$ .

Change the outside coefficient once more.

Changing the differential

Sometimes the substitution removes a root but leaves another occurrence of $x$ . Replace that occurrence too before integrating.

For $u=x-2$ , we have $du=dx$ and $x=u+2$ . Consequently $x\sqrt{x-2}\,dx$ becomes $(u+2)\sqrt{u}\,du$ . Expanding this product gives two powers you can integrate. The real domain is $x\ge2$ ; the antiderivative is differentiable in the interior $x>2$ .

Worked example 3: replace the remaining variable

Find $\int x\sqrt{x-2}\,dx$ .

Plan. Let $u=x-2$ , replace $x$ by $u+2$ , then expand.

\begin{aligned} \int x\sqrt{x-2}\,dx &=\int(u+2)u^{1/2}\,du\\ &=\int(u^{3/2}+2u^{1/2})\,du\\ &=\frac25u^{5/2}+\frac43u^{3/2}+C \end{aligned}

Answer. $\frac25(x-2)^{5/2}+\frac43(x-2)^{3/2}+C$ , for $x>2$ .

Check. Differentiating gives $(x-2)^{3/2}+2(x-2)^{1/2}=x\sqrt{x-2}$ , because the common root leaves $(x-2)+2=x$ .

Change only the coefficient of $x$ inside the root. This changes both the rearranged $x$ and the differential.

Worked example 4: a scaled inner expression

Find $\int x\sqrt{2x-2}\,dx$ .

Plan. Let $u=2x-2$ ; then $x=(u+2)/2$ and $dx=du/2$ .

\begin{aligned} \int x\sqrt{2x-2}\,dx &=\frac14\int(u+2)u^{1/2}\,du\\ &=\frac1{10}u^{5/2}+\frac13u^{3/2}+C \end{aligned}

There are two factors of $1/2$ : one from $x$ , the other from $dx$ .

Answer. $\frac1{10}(2x-2)^{5/2}+\frac13(2x-2)^{3/2}+C$ , for $x>1$ .

Check. Its derivative is $\frac12(2x-2)^{3/2}+(2x-2)^{1/2}=x\sqrt{2x-2}$ .

A completely transformed integral has one variable and its matching differential.

Watch out

$\int x\sqrt{u}\,du$ is not ready for a power-rule integration when $x$ depends on $u$ . Use the substitution equation to eliminate it first.

Change only the constant inside the root for your turn.

Changing bounds

The limits name values of the integration variable, so a change of variable changes their values too.

If $u=g(x)$ , the endpoint $x=a$ becomes $u=g(a)$ . Use the corresponding value at each endpoint, even when the new lower value is numerically larger than the new upper value. A decreasing substitution produces a negative differential that compensates for this reversed direction.

Definite substitution

For a differentiable substitution on the interval, with the integrand well defined,

\int_a^b f(g(x))g'(x)\,dx =\int_{g(a)}^{g(b)}f(u)\,du.

Use the transformed limits with an antiderivative in $u$ . Alternatively, return to $x$ before using the original limits.

Not in the formula booklet: learn it.

Compare an increasing and a decreasing choice for exactly the same integral.

Worked example 5: increasing substitution

Use $u=2x+1$ to evaluate $\int_0^1(2x+1)^2\,dx$ .

Plan. Transform the differential and both limits before integrating.

dx=\frac12du,\qquad x=0\Rightarrow u=1,\quad x=1\Rightarrow u=3.

\int_0^1(2x+1)^2\,dx =\frac12\int_1^3u^2\,du =\left[\frac{u^3}{6}\right]_1^3 =\frac{27-1}{6}=\frac{13}{3}.

Answer. $13/3$ .

Check. Expanding in $x$ gives $\int_0^1(4x^2+4x+1)\,dx=4/3+2+1=13/3$ .

Change only the sign of the substitution: choose $u=-(2x+1)$ . The original integral is unchanged, so the answer must agree.

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P4 6.2 - Integration by substitution

Reversing the chain rule

Changing the differential

Changing bounds

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