1.08h - Integration by substitution | OCR A-Level Mathematics A

Turn a complicated-looking integral into a familiar one by choosing a new variable. Recognise reverse-chain forms, rewrite every factor, transform limits and keep logarithm and square-root domains intact.

Recognise a function and its derivative

Must you expand $(x^2+4)^3$ before integrating $x(x^2+4)^3$ ? No: its inner derivative is already waiting outside, apart from a constant factor. Substitution saves you from an entirely avoidable expansion.

You already know

The chain rule gives $\frac{\mathrm d}{\mathrm dx}F(g(x))=F'(g(x))g'(x)$ . Integration reverses that derivative; an indefinite answer has one arbitrary constant.

Set $u=g(x)$ . The notation $\mathrm du=g'(x)\,\mathrm dx$ records the width conversion as well as the new expression.

Reverse-chain power integral

On an interval where the functions are real and defined, for $n\ne-1$ ,

\int g'(x)\bigl(g(x)\bigr)^n\,\mathrm dx =\frac{\bigl(g(x)\bigr)^{n+1}}{n+1}+C.

In the formula sheet.

Check the multiplier outside the inner function rather than simply spotting a bracket.

Worked example 1: the missing factor two

Question. Find $\int x(x^2+4)^3\,\mathrm dx$ .

Plan. Put $u=x^2+4$ , so $x\,\mathrm dx=\mathrm du/2$ .

\int x(x^2+4)^3\,\mathrm dx =\frac12\int u^3\,\mathrm du =\frac{u^4}{8}+C =\frac{(x^2+4)^4}{8}+C.

Why the half? The derivative of the inner expression is $2x$ , twice the factor present.

Answer. $(x^2+4)^4/8+C$ .

Check. Differentiation multiplies by $4(2x)$ , cancelling the denominator $8$ .

Change only the outside coefficient, from $1$ to $3$ . The substitution is unchanged.

Worked example 2: scaling the derivative factor

Question. Find $\int3x(x^2+4)^3\,\mathrm dx$ .

Plan. Use the same inner variable and carry its new constant multiplier.

\begin{aligned} 3x\,\mathrm dx&=\frac32\,\mathrm du &&\text{convert the factor}\\ \int3x(x^2+4)^3\,\mathrm dx&=\frac32\int u^3\,\mathrm du =\frac{3(x^2+4)^4}{8}+C. \end{aligned}

Why keep the coefficient visible? Substitution changes variables, not the size of the integral.

Answer. $3(x^2+4)^4/8+C$ .

Check. Differentiating gives $3x(x^2+4)^3$ .

The same recognition gives $\int(2x+1)^5\,\mathrm dx=(2x+1)^6/12+C$ and $\int x\mathrm e^{x^2}\,\mathrm dx=\mathrm e^{x^2}/2+C$ . Inspection is valid unless a specified substitution or derivation is requested.

Watch out

Wrong: divide only by the new power. Right: account for both the outer power rule and the inner derivative.

Change the bracket's exponent rather than expanding it.

A derivative over its function gives a logarithm

The excluded power $-1$ produces a logarithm. An integrand can contain a cubic denominator without requiring partial fractions if its numerator matches the derivative.

Logarithmic derivative

On an interval where $g(x)\ne0$ ,

\int\frac{k g'(x)}{g(x)}\,\mathrm dx=k\ln|g(x)|+C.

In the formula sheet.

The modulus allows negative values of $g$ ; it does not restore the excluded zeros. Different disconnected intervals can have different constants.

Worked example 3: matching a quadratic denominator

Question. Find $\int(2x+1)/(x^2+x-3)\,\mathrm dx$ .

Plan. Use $u=x^2+x-3$ and $\mathrm du=(2x+1)\,\mathrm dx$ .

\int\frac{2x+1}{x^2+x-3}\,\mathrm dx =\int\frac{\mathrm du}{u}=\ln|x^2+x-3|+C.

Why a logarithm? The transformed power is exactly $u^{-1}$ .

Answer. $\ln|x^2+x-3|+C$ , on intervals excluding $(-1\pm\sqrt{13})/2$ .

Check. The chain rule recovers the numerator over its denominator.

Change only the numerator to $6x+3$ ; the denominator and domain stay fixed.

Worked example 4: a constant multiple

Question. Find $\int(6x+3)/(x^2+x-3)\,\mathrm dx$ .

Plan. Factor the numerator as $3(2x+1)$ .

\int\frac{6x+3}{x^2+x-3}\,\mathrm dx =3\int\frac{\mathrm du}{u}=3\ln|x^2+x-3|+C.

Why retain the factor $3$ ? It is the ratio of the actual numerator to the denominator's derivative.

Answer. $3\ln|x^2+x-3|+C$ on the same valid intervals.

Check. Differentiate: $3(2x+1)/(x^2+x-3)$ .

For a cubic, $\int(x^2+2x)/(x^3+3x^2-5)\,\mathrm dx=\frac13\ln|x^3+3x^2-5|+C$ . For $\tan x=\sin x/\cos x$ , the inner derivative is negative: substitution gives $-\ln|\cos x|+C$ , with radians and $\cos x\ne0$ . These follow from the supplied logarithmic-derivative formula; they are not separately printed results.

Watch out

Wrong: $\int\tan x\,\mathrm dx=\ln|\cos x|+C$ . The derivative of cosine is negative, so the correct coefficient is $-1$ .

Use a denominator that never vanishes and justify its domain.

Rewrite every factor in the new variable

A substitution can help even when no derivative factor is obvious. Rename the difficult expression, then rewrite the numerator and differential too. One substitution followed by algebra must suffice here.

Complete substitution

A transformed integral contains one integration variable throughout its integrand, differential and limits. A mixture such as $\int x/u^3\,\mathrm du$ is unfinished.

A shifted linear denominator turns its numerator into an easy difference of powers.

Worked example 5: rewriting the numerator

Question. Find $\int x/(x+2)^3\,\mathrm dx$ .

Plan. Set $u=x+2$ , hence $x=u-2$ and $\mathrm dx=\mathrm du$ .

\begin{aligned} \int\frac{x}{(x+2)^3}\,\mathrm dx &=\int(u^{-2}-2u^{-3})\,\mathrm du &&\text{rewrite all factors}\\ &=-u^{-1}+u^{-2}+C. \end{aligned}

Why rewrite $x$ ? It is part of the integrand, not a constant during integration in $u$ .

Answer. $-1/(x+2)+1/(x+2)^2+C$ , for $x\ne-2$ .

Check. Its derivative is $((x+2)-2)/(x+2)^3$ .

Change only the numerator to $x+1$ . The rewritten constant changes from $-2$ to $-1$ .

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1.08h - Integration by substitution

Recognise a function and its derivative

A derivative over its function gives a logarithm

Rewrite every factor in the new variable

Continue this lesson.

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