Pure 8.5 - Integration by substitution and by parts | Edexcel A-Level Mathematics

Some difficult integrals become familiar when you change the variable; others become easier when you undo the product rule. You will choose and carry out substitution and integration by parts, keep limits and logarithm domains consistent, and check an answer by differentiating it.

Choosing the integration method

Why does changing a letter solve one integral, while splitting a product solves another? Compare $\int 10x/(5x^2+1)\,\mathrm{d}x$ with $\int xe^{2x}\,\mathrm{d}x$ . In the first, the numerator is the derivative of the denominator. In the second, differentiating the factor $x$ would simplify the product. Those are different clues.

You already know

The chain rule differentiates a function of another function. The product rule gives $(uv)'=u'v+uv'$ . Basic antiderivatives include $\int x^n\,\mathrm{d}x=x^{n+1}/(n+1)+C$ for $n\ne-1$ and $\int x^{-1}\,\mathrm{d}x=\ln|x|+C$ , on an interval avoiding zero.

Read the structure before calculating. The useful question is what differentiation rule could have produced it.

Structure	First method to consider	Why it helps
$(2x+1)(x^2+x)^4$	Substitution $u=x^2+x$	The derivative of the inside is already a factor
$10x/(5x^2+1)$	Logarithmic recognition	It is the derivative of the denominator divided by the denominator
$x^2\sin x$	Parts, twice	Differentiating the polynomial eventually removes it
$\ln x$	Parts, writing $1\cdot\ln x$	Integrating $1$ supplies an $x$ that cancels the derivative $1/x$

These are clues rather than exclusive rules. A supplied substitution is an instruction to follow, even when another route exists. A useful surprise is that $\int_1^e\ln x\,\mathrm{d}x=1$ exactly: the two terms involving $e$ cancel at the upper endpoint. You will see why in the last part.

Substitution and integration by parts

Substitution rewrites an integral completely in a new variable, reversing the chain rule. Integration by parts replaces an integral of a product using the product rule in reverse. A product alone does not force parts: $2x(x^2+1)^5$ has a much simpler substitution.

The whole lesson develops the distinction in that last sentence.

Key idea

Choose the method that turns the remaining integral into something simpler.

Try the decision before doing any integration.

Substitution as the reverse chain rule

In $\int 2x(x^2+1)^5\,\mathrm{d}x$ , the expression $x^2+1$ occurs inside a power and its derivative $2x$ sits beside it. Name the inside and the whole integral becomes a power integral.

Substitution

If $u=g(x)$ , then $\mathrm{d}u=g'(x)\,\mathrm{d}x$ . Thus, on an interval where the substitution is valid,

\int f(g(x))g'(x)\,\mathrm{d}x=\int f(u)\,\mathrm{d}u.

Return to the original variable in an indefinite answer. Not in the formula booklet: learn it.

The differential records the chain factor; it is part of the substitution.

Worked example 1: the inside derivative is present

Question. Find $\int 2x(x^2+1)^5\,\mathrm{d}x$ .

Plan. Put $u=x^2+1$ , integrate the resulting power, then substitute back.

\begin{aligned} u&=x^2+1,\quad \mathrm{d}u=2x\,\mathrm{d}x && \text{differentiate the inside}\\ \int 2x(x^2+1)^5\,\mathrm{d}x&=\int u^5\,\mathrm{d}u && \text{replace both factors}\\ &=\frac{u^6}{6}+C && \text{power rule} \end{aligned}

Why this substitution? It replaces the entire composite expression and the differential factor together.

Answer. $(x^2+1)^6/6+C$ .

Check. Its derivative is $6(x^2+1)^5(2x)/6=2x(x^2+1)^5$ .

Now change just the coefficient outside the power. The same inside remains useful, but a constant must be carried through.

Worked example 2: a missing constant factor

Question. Find $\int 5x(x^2+1)^5\,\mathrm{d}x$ .

Plan. Use the same substitution and write $5x\,\mathrm{d}x=(5/2)\,\mathrm{d}u$ .

\begin{aligned} \int 5x(x^2+1)^5\,\mathrm{d}x&=\frac52\int u^5\,\mathrm{d}u && u=x^2+1\\ &=\frac{5u^6}{12}+C && \text{integrate the power} \end{aligned}

Why the factor $5/2$ ? The derivative of the inside is $2x$ , not $5x$ .

Answer. $5(x^2+1)^6/12+C$ .

Check. Differentiating gives $(5/12)6(x^2+1)^5(2x)=5x(x^2+1)^5$ .

You may write the same working without introducing $u$ , once the chain factor is secure.

Watch out

Wrong: $\int(5x-2)^4\,\mathrm{d}x=(5x-2)^5/5+C$ . Differentiation gives five times the integrand. Right: $(5x-2)^5/25+C$ . Divide by the inner derivative as well as by the new power.

Change the outside coefficient once more, keeping the inside and power fixed.

Carrying out a given substitution

A substitution does not need an obvious derivative factor already sitting in the integral. If $u=\sqrt{x+1}$ , you can solve for $x$ , obtaining $x=u^2-1$ . That lets you replace an otherwise awkward numerator as well as the root.

Complete substitution

A complete substitution expresses every integrand factor and the differential in the new variable. A definite integral also needs new bounds. An expression such as $\int x u^{-1}\,\mathrm{d}u$ is unfinished because $x$ still remains.

Make a short substitution list before combining anything: the original variable, the root or bracket, the differential, and any bounds.

Worked example 3: solving for the original variable

Question. Use $u=\sqrt{x+1}$ to find $\int x/\sqrt{x+1}\,\mathrm{d}x$ , for $x>-1$ .

Plan. Rewrite $x$ , the root and $\mathrm{d}x$ in terms of $u$ , then integrate a polynomial.

\begin{aligned} x&=u^2-1,\quad \sqrt{x+1}=u,\quad \mathrm{d}x=2u\,\mathrm{d}u && \text{rearrange and differentiate}\\ \int\frac{x}{\sqrt{x+1}}\,\mathrm{d}x&=\int\frac{u^2-1}{u}\,2u\,\mathrm{d}u && \text{replace every factor}\\ &=2\int(u^2-1)\,\mathrm{d}u && u>0\\ &=\frac23u^3-2u+C && \text{power rule} \end{aligned}

Why solve for $x$ ? Without $x=u^2-1$ , the numerator cannot be rewritten and the substitution is incomplete.

Answer. $\frac23(x+1)^{3/2}-2\sqrt{x+1}+C$ .

Check. Its derivative is $\sqrt{x+1}-1/\sqrt{x+1}=x/\sqrt{x+1}$ .

Change only the task from indefinite integration to evaluation between two bounds. The substitutions themselves stay exactly the same.

Worked example 4: changing bounds as well

Question. Use $u=\sqrt{x+1}$ to evaluate $\int_0^3 x/\sqrt{x+1}\,\mathrm{d}x$ .

Plan. Use the complete replacement above and convert the endpoints before integrating.

\begin{aligned} x=0&\Rightarrow u=1,\quad x=3\Rightarrow u=2 && \text{new bounds}\\ \int_0^3\frac{x}{\sqrt{x+1}}\,\mathrm{d}x&=2\int_1^2(u^2-1)\,\mathrm{d}u && \text{all in }u\\ &=\left[\frac23u^3-2u\right]_1^2 && \text{integrate}\\ &=\frac43-\left(-\frac43\right)=\frac83 && \text{upper minus lower} \end{aligned}

Why change bounds now? Once $x$ has disappeared, the bounds must refer to $u$ too.

Answer. $8/3$ .

Check. Evaluating the original-variable antiderivative from Worked example 3 at $3$ and $0$ gives the same difference.

A trigonometric substitution can remove a root in a different way. Recall $1-\sin^2u=\cos^2u$ , which is not printed separately in the formula booklet. Choose the stated interval carefully: $\sqrt{\cos^2u}=|\cos u|$ , so replacing it by $\cos u$ needs a non-negative cosine.

Worked example 5: a supplied trigonometric substitution

Question. Use $x=2\sin u$ , with $0\le u<\pi/2$ , to evaluate $\int_0^1 1/\sqrt{4-x^2}\,\mathrm{d}x$ .

Plan. Replace the root and differential using the same substitution and change both bounds.

\begin{aligned} \mathrm{d}x&=2\cos u\,\mathrm{d}u,\quad \sqrt{4-x^2}=2\cos u && \cos u>0\\ x=0&\Rightarrow u=0,\quad x=1\Rightarrow u=\pi/6 && \text{use the stated branch}\\ \int_0^1\frac{\mathrm{d}x}{\sqrt{4-x^2}}&=\int_0^{\pi/6}\frac{2\cos u}{2\cos u}\,\mathrm{d}u && \text{complete replacement}\\ &=[u]_0^{\pi/6}=\frac\pi6 && \text{integrate }1 \end{aligned}

Why that branch? It makes the root positive and chooses one angle for each $x$ .

Answer. $\pi/6$ .

Check. On $[0,1]$ , the integrand lies between $1/2$ and $1/\sqrt3$ , and $\pi/6\approx0.524$ lies between those bounds.

Change just the upper endpoint. The algebraic cancellation is unchanged; only its corresponding angle changes.

Worked example 6: a new endpoint in the same substitution

Question. Use $x=2\sin u$ to evaluate $\int_0^{\sqrt2}1/\sqrt{4-x^2}\,\mathrm{d}x$ .

Plan. Use the branch $0\le u<\pi/2$ and replace the new upper bound.

\begin{aligned} x=\sqrt2&\Rightarrow\sin u=1/\sqrt2\Rightarrow u=\pi/4 && \text{new endpoint}\\ \int_0^{\sqrt2}\frac{\mathrm{d}x}{\sqrt{4-x^2}}&=\int_0^{\pi/4}\frac{2\cos u}{2\cos u}\,\mathrm{d}u && \text{root and differential}\\ &=[u]_0^{\pi/4}=\frac\pi4 && \text{evaluate} \end{aligned}

Why keep the original substitution? Its cancellation does not depend on the upper endpoint.

Answer. $\pi/4$ .

Check. The interval is longer than in Worked example 5 and the positive integrand adds area; $\pi/4>\pi/6$ , as expected.

One useful substitution is enough for the cases studied here. The skill is completing it, rather than trying a succession of unrelated changes.

Watch out

Wrong: $\sqrt{4-4\sin^2u}=2\cos u$ for every real $u$ . Right: $2|\cos u|$ . Use the chosen interval to justify removing the modulus.

Use a third upper endpoint, showing all replacements rather than just quoting an answer.

Logarithms from f prime over f

For $\int6x/(3x^2+5)\,\mathrm{d}x$ , substitution gives $u=3x^2+5$ and $\mathrm{d}u=6x\,\mathrm{d}x$ . What remains is $\int u^{-1}\,\mathrm{d}u$ , the one power for which the ordinary power formula fails.

Logarithmic substitution

On an interval where $f$ is differentiable and non-zero,

\int\frac{f'(x)}{f(x)}\,\mathrm{d}x=\ln|f(x)|+C.

If $f(x)>0$ , write $\ln f(x)$ . Not in the formula booklet: learn it.

The modulus allows either sign of the denominator; it does not allow a zero denominator.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 8.5 - Integration by substitution and by parts

Choosing the integration method

Substitution as the reverse chain rule

Carrying out a given substitution

Logarithms from f prime over f

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