Mathematics 9709 · Pure Mathematics 3

Integration by substitution (when they give you the substitution)

Substitution is the chain rule run backwards.
In CAIE they almost always give you the substitution (“use the substitution $u=\dots$ ”), so you never have to think one up.
Your whole job is to do the swap cleanly.

The method: rewrite everything in u

Write the given $u=g(x)$ (or $x=h(u)$ ).
Differentiate to get $\dfrac{du}{dx}$ , then rearrange for $dx$ : $dx = \dfrac{du}{\,du/dx\,}$ .
Replace every $x$ and the $dx$ so the integral is entirely in $u$ .
Integrate in $u$ .
For an indefinite integral, substitute back to $x$ . For a definite integral, convert the limits and do not substitute back.

The habit you must keep: rewrite everything in $u$ . If even one $x$ is left in the $u$ -integral, it is wrong and the marks are gone.

Common mistake

The

dx

step trips people up:

dx=\dfrac{du}{\,du/dx\,}

is a reciprocal. If

u=\sqrt{x}

then

\dfrac{du}{dx}=\dfrac{1}{2\sqrt{x}}

, so

dx = 2\sqrt{x}\,du = 2u\,du

. Flip it; do not just copy

du/dx

Definite integrals: convert the limits

This is where students lose the most marks in the topic.
When the integral has limits, those limits are $x$ -values. Once you switch to $u$ , the limits must become $u$ -values too.

Route	What you do at the end
Convert the limits	Change each limit to its $u$ -value, evaluate in $u$ . Do not substitute back.
Substitute back	Return to $x$ , then use the original $x$ -limits.

Common mistake

Pick one route and stick to it. The mistake examiners look for: staying in

u

and still plugging the original

x

-limits into a

u

-expression. That mixes the two routes, so the value is wrong. Either convert the limits and stay in

u

, or substitute back and use

x

-limits. Never both.

Why converting the limits is the safer route

Substituting back to $x$ at the end means turning a tidy $u$ -answer like $\tfrac43 u^{3/2}$ back into $x$ . Then you work that out and hope your algebra is right. Converting the limits instead lets you finish entirely in $u$ : fewer symbols, no swapping back, and the new limits are usually nicer numbers. In the worked example, $x=\pi$ and $x=\tfrac{3\pi}{2}$ become a clean $u=1$ and $u=2$ . For a definite integral, convert the limits by default.

Worked example

9709/31 Jun 2024 Q8: $\int_{\pi}^{3\pi/2} \frac{\sin 2x}{\sqrt{1-\sin x}}\,dx$ , $u = 1-\sin x$

$u=1-\sin x \Rightarrow \dfrac{du}{dx}=-\cos x$ , so $\cos x\,dx=-\,du$ .
Use $\sin 2x = 2\sin x\cos x$ and $\sin x = 1-u$ . The bottom is $\sqrt{1-\sin x}=\sqrt{u}$ , so the integrand becomes $\dfrac{2(1-u)\cos x}{\sqrt{u}}\,dx=\dfrac{2(1-u)}{\sqrt{u}}(-\,du)=\dfrac{2(u-1)}{\sqrt{u}}\,du$ .
Convert the limits: $x=\pi\Rightarrow\sin x=0\Rightarrow u=1$ , and $x=\tfrac{3\pi}{2}\Rightarrow\sin x=-1\Rightarrow u=2$ .
Expand: $\dfrac{2(u-1)}{\sqrt{u}}=2u^{1/2}-2u^{-1/2}$ , which integrates to $\tfrac43 u^{3/2}-4u^{1/2}=\left[\tfrac43 u^{3/2}-4u^{1/2}\right]_1^2$ .
At $u=2$ : $\tfrac{8\sqrt2}{3}-4\sqrt2=-\tfrac{4\sqrt2}{3}$ ; at $u=1$ : $\tfrac43-4=-\tfrac83$ . Subtract.

Answer

\int_{\pi}^{3\pi/2} \frac{\sin 2x}{\sqrt{1-\sin x}}\,dx = \tfrac83 - \tfrac43\sqrt{2}

(9709/31 Jun 2024 Q8). The answer is in the required $a+b\sqrt{2}$ form with $a=\tfrac83$ , $b=-\tfrac43$ . It tests two skills: a trig identity to set up the swap, and converting the limits.

The exam pattern

Because the substitution is given, the job stays small: you do not need to choose one yourself (that is a 90+ skill, off-syllabus for this layer).
Common given forms: $u=\sqrt{x}$ , $u=1-\sin x$ , and the trig form $x=a\tan\theta$ (which links back to the arctan family in §8.2).

Your turn— tap to reveal the worked answer (9709/31 Nov 2021 Q4)

“Using the substitution $u=\sqrt{x}$ , find the exact value of a definite integral of $\dfrac{x+1}{x}$ type.” With $x=u^2$ and $dx=2u\,du$ , rewrite everything in $u$ , then convert the limits before you work it out. (9709/31 Nov 2021 Q4)

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Integration by substitution (when they give you the substitution)

Substitution is the chain rule run backwards.
In CAIE they almost always give you the substitution (“use the substitution $u=\dots$ ”), so you never have to think one up.
Your whole job is to do the swap cleanly.

The method: rewrite everything in u

Write the given $u=g(x)$ (or $x=h(u)$ ).
Differentiate to get $\dfrac{du}{dx}$ , then rearrange for $dx$ : $dx = \dfrac{du}{\,du/dx\,}$ .
Replace every $x$ and the $dx$ so the integral is entirely in $u$ .
Integrate in $u$ .
For an indefinite integral, substitute back to $x$ . For a definite integral, convert the limits and do not substitute back.

The habit you must keep: rewrite everything in $u$ . If even one $x$ is left in the $u$ -integral, it is wrong and the marks are gone.

Common mistake

The

dx

step trips people up:

dx=\dfrac{du}{\,du/dx\,}

is a reciprocal. If

u=\sqrt{x}

then

\dfrac{du}{dx}=\dfrac{1}{2\sqrt{x}}

, so

dx = 2\sqrt{x}\,du = 2u\,du

. Flip it; do not just copy

du/dx

Definite integrals: convert the limits

This is where students lose the most marks in the topic.
When the integral has limits, those limits are $x$ -values. Once you switch to $u$ , the limits must become $u$ -values too.

Route	What you do at the end
Convert the limits	Change each limit to its $u$ -value, evaluate in $u$ . Do not substitute back.
Substitute back	Return to $x$ , then use the original $x$ -limits.

Common mistake

Pick one route and stick to it. The mistake examiners look for: staying in

u

and still plugging the original

x

-limits into a

u

-expression. That mixes the two routes, so the value is wrong. Either convert the limits and stay in

u

, or substitute back and use

x

-limits. Never both.

Why converting the limits is the safer route

Worked example

9709/31 Jun 2024 Q8: $\int_{\pi}^{3\pi/2} \frac{\sin 2x}{\sqrt{1-\sin x}}\,dx$ , $u = 1-\sin x$

$u=1-\sin x \Rightarrow \dfrac{du}{dx}=-\cos x$ , so $\cos x\,dx=-\,du$ .
Use $\sin 2x = 2\sin x\cos x$ and $\sin x = 1-u$ . The bottom is $\sqrt{1-\sin x}=\sqrt{u}$ , so the integrand becomes $\dfrac{2(1-u)\cos x}{\sqrt{u}}\,dx=\dfrac{2(1-u)}{\sqrt{u}}(-\,du)=\dfrac{2(u-1)}{\sqrt{u}}\,du$ .
Convert the limits: $x=\pi\Rightarrow\sin x=0\Rightarrow u=1$ , and $x=\tfrac{3\pi}{2}\Rightarrow\sin x=-1\Rightarrow u=2$ .
Expand: $\dfrac{2(u-1)}{\sqrt{u}}=2u^{1/2}-2u^{-1/2}$ , which integrates to $\tfrac43 u^{3/2}-4u^{1/2}=\left[\tfrac43 u^{3/2}-4u^{1/2}\right]_1^2$ .
At $u=2$ : $\tfrac{8\sqrt2}{3}-4\sqrt2=-\tfrac{4\sqrt2}{3}$ ; at $u=1$ : $\tfrac43-4=-\tfrac83$ . Subtract.

Answer

\int_{\pi}^{3\pi/2} \frac{\sin 2x}{\sqrt{1-\sin x}}\,dx = \tfrac83 - \tfrac43\sqrt{2}

(9709/31 Jun 2024 Q8). The answer is in the required $a+b\sqrt{2}$ form with $a=\tfrac83$ , $b=-\tfrac43$ . It tests two skills: a trig identity to set up the swap, and converting the limits.

The exam pattern

Because the substitution is given, the job stays small: you do not need to choose one yourself (that is a 90+ skill, off-syllabus for this layer).
Common given forms: $u=\sqrt{x}$ , $u=1-\sin x$ , and the trig form $x=a\tan\theta$ (which links back to the arctan family in §8.2).

Your turn— tap to reveal the worked answer (9709/31 Nov 2021 Q4)