8.2.5 Substitution (Reverse Chain Rule)

The substitution method, also known as the Reverse Chain Rule, is a powerful technique for solving integrals where the integrand is a product of a function and its derivative. It simplifies the integration process by making a substitution that transforms the integral into a simpler form.

Steps for Substitution (Reverse Chain Rule):

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Choose a Substitution:

Identify a part of the integrand that can be substituted by a single variable, usually denoted as $u$ .
A common choice is to let $u$ equal an inner function whose derivative is present in the integrand.

Differentiate to Find du:

Differentiate your chosen substitution to express du in terms of dx.
This allows you to rewrite the differential $dx$ in terms of $du$ .

Rewrite the Integral:

Substitute both the function and its differential into the integral, replacing all x-terms with u-terms.
The integral should now be in terms of $u$ and $du$ , often simplifying the process.

Integrate:

Perform the integration with respect to $u$ .

Substitute Back:

After integrating, replace $u$ with the original expression in terms of $x$ to obtain the final answer.

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Example:

Evaluate $\int 2x \cdot e^{x^2} \, dx.$

Choose a Substitution:

Let $u$ = $x^2$ , since the derivative of $x^2$ is $2x$ , which is present in the integrand.

Differentiate to Find du:

$\frac{du}{dx} = 2x, du = 2x \, dx.$

Rewrite the Integral:

Substitute into the integral:

$\int 2x \cdot e^{x^2} \, dx = \int e^u \, du$

Integrate:

The integral of $e^u$ with respect to $u$ is $e^u.$

Substitute Back:

Replace $u$ with $x^2:$

$e^{x^2} + C$

Integration: Which Method to Use

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Is it of the form $f(ax+b)? \Rightarrow$ Reverse Chain Rule Example: $ax + b = (5-3x)$

\int \frac{2}{5-3x} dx = \int 2(5-3x)^{-1} dx = \frac {2 \ln |5-3x|}{-3} + c

\div -3 \ \text{by diff of bracket}

= -\frac{2}{3} \ln |5-3x| + c \quad

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Is it of the form: Integration (i.e., Reverse chain rule)

\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + c

\int f'(x)(f(x))^n dx = \frac{1}{n+1}(f(x))^{n+1} + c

e.g. \int \frac{5x^2}{x^3+6} dx = \frac{5}{3} \int \frac{3x^2}{x^3+6} dx \quad \text {both are equivalent}

= \frac{5}{3} \ln |x^3+6| + c

\text {e.g.} \quad \int 6\cos x (\sin x + 15)^{12} dx = 6\int \cos x (\sin x+15)^{12} \, dx \\ f'(x) = \cos x, f(x)^n = \sin x+15)^{12}

= \frac{1}{13} (\sin x + 15)^{13} + c

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Check for partial fractions (i.e., is the bottom factorised or will it?)

\text {e.g.} \quad \int \frac{1}{x^2-2} dx

= \int \frac{1}{(x-\sqrt{2})(x+\sqrt{2})} dx

Let \quad \frac {1}{(x+\sqrt{2})(x-\sqrt{2})} = \frac {A}{x+\sqrt{2}} + \frac {B}{x-\sqrt{2}}

1 \equiv A(x-\sqrt{2}) + B(x+\sqrt{2})

Let x = \sqrt{2} \Rightarrow 1 = B(2\sqrt{2}) \Rightarrow B = \frac{1}{2\sqrt{2}} = \sqrt{.}

x = -\sqrt{2} \Rightarrow 1 \equiv A(-2\sqrt2) \Rightarrow A = -\frac {\sqrt2}{4}

\Rightarrow \int -\frac {\sqrt2}{4} \left (\frac {1}{x+\sqrt2}\right)+ \frac {\sqrt2}{4} \left (\frac {1}{x-\sqrt2}\right) \, dx

= \frac{\sqrt{2}}{4} \int \left( \frac{-1}{x + \sqrt{2}} + \frac{1}{x - \sqrt{2}} \right) dx

= \frac{\sqrt{2}}{4} \left( -\ln|x + \sqrt{2}| + \ln|x - \sqrt{2}| \right) + c

= \frac{\sqrt{2}}{4} \ln \left| \frac{x - \sqrt{2}}{x + \sqrt{2}} \right| + c

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Check if trig identities are applicable

\sin^2 x + \cos^2 x \equiv 1

\sec^2 x \equiv 1 + \tan^2 x

\cot^2 x + 1 \equiv \csc^2 x

\cos 2x = 2\cos^2 x - 1 \\ = 1 - 2\sin^2 x \\= \cos^2 x - \sin^2 x

\sin 2x = 2\sin x \cos x

Differentiation

$f(x)$	$f'(x)$
$\tan kx$	$k\sec^2 kx$
$\sec x$	$\sec x \tan x$
$\cot x$	$-\csc^2 x$
$\csc x$	$-\csc x \cot x$

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\int f'(x)(f(x))^n dx = \frac{1}{n+1} (f(x))^{n+1} + c

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e.g. \int \tan^2 x \, dx = \int \sec^2 x - 1 \, dx = \tan x - x + c

e.g. \int \sec^6 x \sec x \tan x \, dx\\ = \int \sec^5x \sec x \tan x \, dx\\ \underbrace{}_{f(x)^5} \ \ \ \ \ \ \ \quad \underbrace{}_{f(x)} \\= \frac{1}{6} \sec^6 x + c

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e.g. \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x - x + c

\sin^2 x + \cos^2 x = 1 \quad \Rightarrow \quad 1 + \cot^2 x = \csc^2 x

\int \frac{4 \sin x \cos x}{2 \cos x^2 - 3} \, dx = \int \frac{2 (2\sin x \cos x)}{(2 \cos^2 x - 1)- 2} \, dx

= \int \frac{2 \sin 2x }{\cos 2x - 2}\, dx = = \int \frac{-2 \sin 2x }{\cos 2x - 2}\, dx

Using $\frac{f'(x)}{f(x)} = \ln|f(x)| + c$

= -\ln|\cos 2x - 2| + c

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\int (\tan x + 1)^2 \, dx = \int \tan^2 x + 2 \tan x + 1 \, dx

Let, \quad A = \tan^2 x, \quad B = 2 \tan x

A = \int \tan^2 x \, dx = \int \sec^2 x - 1 \, dx = \tan x - x + c

B = \int 2 \tan x \, dx = \int \frac{2 \sin x}{\cos x}\, dx = -2\int \frac{-\sin x}{\cos x}\, dx =-2 \ln|\cos x| + c

I = \tan x - \cancel x - 2 \ln|\cos x|+\cancel x + c

I = \tan x - 2 \ln|\cos x| + c

5) Substitution or By Parts

By parts is usually used to integrate a function that is a product of two functions that are not related by differentiation. e.g.

\int xe^{x^2} \, dx \quad \checkmark

\int x^2 \sin x \, dx \quad \checkmark

\int x^2 e^{x^3 + 4} \, dx \quad \times \quad (\text{Differential of power is outside, use reverse chain rule or substitution})

Substitution is used for examples like the third one above where there is a differential relationship (i.e. $\frac{d}{dx}(x^3+4) = 3x^2$ , so $x^2$ at front will cancel) or where several terms on a denominator need to be combined to make the integral simpler.

e.g.

\int x \sqrt{6x - 3} \, dx \quad (\text{By substitution})

Let\ u = 6x - 3 \Rightarrow \frac{du}{dx} = 6 \Rightarrow dx = \frac{1}{6} du

\Rightarrow I = \int xu^{\frac{1}{2}} \times \frac{1}{6}du = \frac{1}{6}\int xu^{\frac{1}{2}}du

\text {Since} \quad u = 6x - 3 \Rightarrow u + 3 = 6x \Rightarrow x = \frac{u}{6}+\frac {1}{2}

\Rightarrow I = \frac{1}{6} \int \left (\frac{u}{6}+\frac {1}{2} \right )u^{\frac{1}{2}} du = \frac{1}{6} \int \frac{u^{\frac{3}{2}}}{6}+\frac {1^{\frac{1}{2}}}{2} du

= \frac{1}{6} \left[ \frac{2u^{\frac{5}{2}}}{56} + \frac{2u^{\frac{3}{2}}}{32} \right] + C = \frac{1}{90} u^{\frac{5}{2}} + \frac{1}{18} u^{\frac{3}{2}} + C

= \frac{1}{90} (6x-3)^{\frac{5}{2}} + \frac{1}{18} (6x-3)^{\frac{3}{2}} + C

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Example: $\int x \sqrt{6x - 3} \, dx$ (Same again but by parts)

Integration by parts:

\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx

Let:

$u = x \Rightarrow \frac{du}{dx} = 1$
$v = \dfrac{2(6x-3)^{\frac{3}{2}}}{3 \times 6}\Rightarrow \dfrac{dv}{dx} = (6x-3)^{\frac{1}{2}}$

= \dfrac{1}{9}(6x-3)^{\frac{3}{2}}

Then:

I = x \times \frac{1}{9}(6x-3)^{\frac{3}{2}} - \int \frac{1}{9}(6x-3)^{\frac{3}{2}} \, dx

= \frac{x}{9}(6x-3)^{\frac{3}{2}} - \frac{1}{9} \cdot \frac{2}{5}\frac{(6x-3)^{\frac{5}{2}}}{6} + c

= \frac{x}{9}(6x-3)^{\frac{3}{2}} - \frac{1}{135}(6x-3)^{\frac{5}{2}} + c

However, it looks like we have got two different answers to the same question.

For substitution we got:

\frac{1}{90}(6x-3)^{\frac{5}{2}} - \frac{1}{18}(6x-3)^{\frac{3}{2}} + c

= \frac{1}{90} (6x - 3)^{\frac{3}{2}} \left[ (6x - 3)^{1} + 5 \right] + c

= \frac{1}{90} (6x - 3)^{\frac{3}{2}} (6x + 2) + c

= \frac{2}{90} (6x - 3)^{\frac{3}{2}} (3x + 1) + c

= \frac{1}{45}(6x-3)^{\frac{3}{2}}(3x+1) + c

For by parts we got:

\frac{x}{9}(6x-3)^{\frac{3}{2}} - \frac{1}{135}(6x-3)^{\frac{5}{2}} + c

= \frac{1}{135}(6x-3)^{\frac{3}{2}}[15x - (6x-3)] + c

= \frac{1}{135} (6x - 3)^{\frac{3}{2}} (9x + 3) + c

= \frac{3}{135} (6x - 3)^{\frac{3}{2}} (3x + 1) + c

= \frac{1}{45}(6x-3)^{\frac{3}{2}}(3x+1) + c

Conclusion: Both answers are equivalent.

Substitution (Reverse Chain Rule) Simplified Revision Notes for A-Level AQA Maths Pure

8.2.5 Substitution (Reverse Chain Rule)

Steps for Substitution (Reverse Chain Rule):

Example:

Integration: Which Method to Use

5) Substitution or By Parts

Example: $\int x \sqrt{6x - 3} \, dx$ (Same again but by parts)

500K+ Students Use These Powerful Tools to Master Substitution (Reverse Chain Rule) For their A-Level Exams.

Other Revision Notes related to Substitution (Reverse Chain Rule) you should explore

Substitution (Reverse Chain Rule) Simplified Revision Notes for A-Level AQA Maths Pure

8.2.5 Substitution (Reverse Chain Rule)

Steps for Substitution (Reverse Chain Rule):

Example:

Integration: Which Method to Use

5) Substitution or By Parts

Example: ∫x6x−3 dx\int x \sqrt{6x - 3} \, dx∫x6x−3​dx (Same again but by parts)

500K+ Students Use These Powerful Tools to Master Substitution (Reverse Chain Rule) For their A-Level Exams.

Other Revision Notes related to Substitution (Reverse Chain Rule) you should explore

Example: $\int x \sqrt{6x - 3} \, dx$ (Same again but by parts)