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The reverse chain rule and trigonometric formulae

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What the reverse chain rule does

The chain rule differentiates $f(g(x))$ to $f'(g(x)) \times g'(x)$ — the derivative of the inside appears as a multiplier.
The reverse chain rule spots that multiplier and undoes the process.

$\int f'(g(x))\,g'(x)\,dx = f(g(x)) + c$

In practice you are looking for an integrand that is a function of something, multiplied by the derivative of that something.

Recognising the pattern

Ask: is there an inner function, and is its derivative also present as a factor?
$\displaystyle\int 2x\,e^{x^2}dx$ — the inside is $x^2$ , its derivative is $2x$ , and $2x$ is there. ✓
$\displaystyle\int 3x^2\left(x^3+1\right)^4 dx$ — the inside is $x^3+1$ , its derivative is $3x^2$ , and it is there. ✓
$\displaystyle\int e^{x^2}dx$ — the inside is $x^2$ , its derivative $2x$ is missing. ✗ This cannot be done with the methods in this standard.
The multiplier must be present, or be supplyable as a constant. You may adjust by a constant factor, never by a variable one.

Adjusting by a constant

If the multiplier is present up to a constant, pull the constant out.

$\int x\left(x^2+5\right)^3 dx$

The inside is $x^2 + 5$ , whose derivative is $2x$ . The integrand has $x$ , which is $\frac{1}{2}$ of what is needed.
Multiply inside by 2 and outside by $\frac{1}{2}$ — the two cancel, so nothing changes:

$= \frac{1}{2}\int 2x\left(x^2+5\right)^3 dx = \frac{1}{2} \times \frac{\left(x^2+5\right)^4}{4} + c = \frac{\left(x^2+5\right)^4}{8} + c$

You may not do this with a variable. You cannot "supply" a missing $x$ by putting $\frac{1}{x}$ outside, because $x$ is not a constant and does not pass through the integral sign.

The standard reverse-chain forms

$\int g'(x)\left[g(x)\right]^n dx = \frac{\left[g(x)\right]^{n+1}}{n+1} + c \qquad (n \neq -1)$

$\int g'(x)\,e^{g(x)}dx = e^{g(x)} + c$

$\int \frac{g'(x)}{g(x)}\,dx = \ln\left|g(x)\right| + c$

$\int g'(x)\cos\left(g(x)\right)dx = \sin\left(g(x)\right) + c$

Note that the $\ln$ result from the previous page is just the $n = -1$ case of the first one.

The trigonometric powers problem

$\displaystyle\int \sin^2 x\,dx$ has no reverse-chain structure — there is no $\cos x$ multiplier — so it needs an identity instead.
Rearranging the double-angle formula $\cos 2x = 1 - 2\sin^2 x$ :

$\sin^2 x = \frac{1 - \cos 2x}{2}$

Similarly, from $\cos 2x = 2\cos^2 x - 1$ :

$\cos^2 x = \frac{1 + \cos 2x}{2}$

Then the integration is straightforward:

$\int \sin^2 x\,dx = \int \frac{1 - \cos 2x}{2}dx = \frac{x}{2} - \frac{\sin 2x}{4} + c$

The rule of thumb: an even power of sine or cosine needs a double-angle identity; an odd power can usually be done with the reverse chain rule after splitting off one factor.

Turning products into sums

Products such as $\sin 3x\cos 5x$ cannot be integrated directly, but the product-to-sum formulae convert them:

$\sin A\cos B = \tfrac{1}{2}\left[\sin(A+B) + \sin(A-B)\right]$

$\cos A\cos B = \tfrac{1}{2}\left[\cos(A+B) + \cos(A-B)\right]$

$\sin A\sin B = \tfrac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]$

The 2025 report lists "successfully turned a trigonometric product into a trigonometric sum in order to integrate" as a Merit behaviour.
These formulae are in the Formulae and Tables Booklet — you do not need to memorise them, but you must recognise when to reach for them.

Odd powers of sine and cosine

Split off one factor and use $\sin^2 x + \cos^2 x = 1$ on the rest:

$\int \sin^3 x\,dx = \int \sin^2 x\sin x\,dx = \int \left(1 - \cos^2 x\right)\sin x\,dx$

Now the inside is $\cos x$ and the multiplier $\sin x$ is present (up to sign), so the reverse chain rule works:

$= -\cos x + \frac{\cos^3 x}{3} + c$

Tips and Tricks

Check the multiplier is genuinely there before you start. $\int e^{x^2}dx$ looks similar to $\int xe^{x^2}dx$ but only one of them can be done. Ask "what is the inside, what is its derivative, and is that derivative present?" every time.

You may adjust by a constant, never by a variable. Writing $\frac{1}{2x}$ outside the integral to supply a missing $x$ is invalid, because only constants pass through the integral sign.

Differentiate your answer. The reverse chain rule is where a dropped factor is easiest to make and easiest to catch — one line of differentiation confirms it.

An even power of a trigonometric function is a signal for a double-angle identity; an odd power is a signal to split off one factor and use the Pythagorean identity. Recognising which you have saves a lot of wasted effort.

The Formulae and Tables Booklet has the product-to-sum formulae. Know that they exist and what they are for — the report treats using them as Merit-level behaviour, which means candidates who do not know to look are capped below that.

The report warns generally about "omitted a factor of the integration", which "led to an incorrect integration, so no credit could be given for this question part". On this page in particular, count your factors.

Worked ExampleReverse chain rule with a constant adjustment

Find $\displaystyle\int \frac{x^2}{\sqrt{x^3 + 4}}\,dx$ .

Step 1 — Rewrite the root as a power

Never leave a root in place when integrating:

$\int x^2\left(x^3+4\right)^{-1/2}dx$

Step 2 — Identify the inside function and test for its derivative

The inside function is:

$g(x) = x^3 + 4$

Differentiate it:

$g'(x) = 3x^2$

The integrand contains $x^2$ . That is $\frac{1}{3}$ of what is needed — a constant multiple, so the adjustment is legal.

Step 3 — Adjust by the constant

Multiply inside the integral by 3 and outside by $\frac{1}{3}$ . The two cancel, so the value is unchanged:

$= \frac{1}{3}\int 3x^2\left(x^3+4\right)^{-1/2}dx$

The integrand is now exactly in the form $g'(x)\left[g(x)\right]^n$ .

Step 4 — Apply the reverse chain rule

Treat $\left(x^3+4\right)$ as if it were a single variable: raise the power by one and divide by the new power.

The power is $-\frac{1}{2}$ , so the new power is $\frac{1}{2}$ :

$= \frac{1}{3} \times \frac{\left(x^3+4\right)^{1/2}}{1/2} + c$

Step 5 — Simplify

Dividing by $\frac{1}{2}$ multiplies by 2:

$= \frac{1}{3} \times 2\left(x^3+4\right)^{1/2} + c = \frac{2}{3}\sqrt{x^3+4} + c$

Step 6 — Check by differentiating

Using the chain rule on $\frac{2}{3}\left(x^3+4\right)^{1/2}$ :

$\frac{2}{3} \times \frac{1}{2}\left(x^3+4\right)^{-1/2} \times 3x^2$

$= \frac{2}{3} \times \frac{3x^2}{2\sqrt{x^3+4}} = \frac{x^2}{\sqrt{x^3+4}} \;\checkmark$

This is exactly the original integrand, so the answer is correct.