Substitution (Reverse Chain Rule) (AQA A-Level Mathematics): Revision Notes

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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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  1. Choose a Substitution:
  • Identify a part of the integrand that can be substituted by a single variable, usually denoted as uu.
  • A common choice is to let uu equal an inner function whose derivative is present in the integrand.
  1. Differentiate to Find du:
  • Differentiate your chosen substitution to express du in terms of dx.
  • This allows you to rewrite the differential dxdx in terms of dudu.
  1. 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 uu and du du, often simplifying the process.
  1. Integrate:
  • Perform the integration with respect to u u.
  1. Substitute Back:
  • After integrating, replace uu with the original expression in terms of xx to obtain the final answer.
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Example:

Evaluate ∫2x⋅ex2 dx.\int 2x \cdot e^{x^2} \, dx.

  1. Choose a Substitution:
  • Let uu =x2 x^2, since the derivative of x2x^2 is 2x2x, which is present in the integrand.
  1. Differentiate to Find du:
  • dudx=2x,du=2x dx.\frac{du}{dx} = 2x, du = 2x \, dx.
  1. Rewrite the Integral:
  • Substitute into the integral:

∫2x⋅ex2 dx=∫eu du\int 2x \cdot e^{x^2} \, dx = \int e^u \, du

  1. Integrate:
  • The integral of eue^u with respect to uu is eu.e^u.
  1. Substitute Back:
  • Replace uu with x2:x^2:

ex2+Ce^{x^2} + C

Integration: Which Method to Use

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  1. Is it of the form f(ax+b)?⇒f(ax+b)? \Rightarrow Reverse Chain Rule Example: ax+b=(5−3x)ax + b = (5-3x)
∫25−3xdx=∫2(5−3x)−1dx=2ln⁡∣5−3x∣−3+c\int \frac{2}{5-3x} dx = \int 2(5-3x)^{-1} dx = \frac {2 \ln |5-3x|}{-3} + c÷−3 by diff of bracket\div -3 \ \text{by diff of bracket}=−23ln⁡∣5−3x∣+c= -\frac{2}{3} \ln |5-3x| + c \quad
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  1. Is it of the form: Integration (i.e., Reverse chain rule)
∫f′(x)f(x)dx=ln⁡∣f(x)∣+c\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + c∫f′(x)(f(x))ndx=1n+1(f(x))n+1+c\int f'(x)(f(x))^n dx = \frac{1}{n+1}(f(x))^{n+1} + ce.g.∫5x2x3+6dx=53∫3x2x3+6dxboth are equivalente.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}=53ln⁡∣x3+6∣+c= \frac{5}{3} \ln |x^3+6| + ce.g.∫6cos⁡x(sin⁡x+15)12dx=6∫cos⁡x(sin⁡x+15)12 dxf′(x)=cos⁡x,f(x)n=sin⁡x+15)12\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}=113(sin⁡x+15)13+c= \frac{1}{13} (\sin x + 15)^{13} + c
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  1. Check for partial fractions (i.e., is the bottom factorised or will it?)
e.g.∫1x2−2dx\text {e.g.} \quad \int \frac{1}{x^2-2} dx=∫1(x−2)(x+2)dx= \int \frac{1}{(x-\sqrt{2})(x+\sqrt{2})} dxLet1(x+2)(x−2)=Ax+2+Bx−2Let \quad \frac {1}{(x+\sqrt{2})(x-\sqrt{2})} = \frac {A}{x+\sqrt{2}} + \frac {B}{x-\sqrt{2}}1≡A(x−2)+B(x+2)1 \equiv A(x-\sqrt{2}) + B(x+\sqrt{2})Letx=2⇒1=B(22)⇒B=122=.Let x = \sqrt{2} \Rightarrow 1 = B(2\sqrt{2}) \Rightarrow B = \frac{1}{2\sqrt{2}} = \sqrt{.}x=−2⇒1≡A(−22)⇒A=−24x = -\sqrt{2} \Rightarrow 1 \equiv A(-2\sqrt2) \Rightarrow A = -\frac {\sqrt2}{4}⇒∫−24(1x+2)+24(1x−2) dx\Rightarrow \int -\frac {\sqrt2}{4} \left (\frac {1}{x+\sqrt2}\right)+ \frac {\sqrt2}{4} \left (\frac {1}{x-\sqrt2}\right) \, dx=24∫(−1x+2+1x−2)dx= \frac{\sqrt{2}}{4} \int \left( \frac{-1}{x + \sqrt{2}} + \frac{1}{x - \sqrt{2}} \right) dx=24(−ln⁡∣x+2∣+ln⁡∣x−2∣)+c= \frac{\sqrt{2}}{4} \left( -\ln|x + \sqrt{2}| + \ln|x - \sqrt{2}| \right) + c=24ln⁡∣x−2x+2∣+c= \frac{\sqrt{2}}{4} \ln \left| \frac{x - \sqrt{2}}{x + \sqrt{2}} \right| + c
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  1. Check if trig identities are applicable
sin⁡2x+cos⁡2x≡1\sin^2 x + \cos^2 x \equiv 1sec⁡2x≡1+tan⁡2x\sec^2 x \equiv 1 + \tan^2 xcot⁡2x+1≡csc⁡2x\cot^2 x + 1 \equiv \csc^2 xcos⁡2x=2cos⁡2x−1=1−2sin⁡2x=cos⁡2x−sin⁡2x\cos 2x = 2\cos^2 x - 1 \\ = 1 - 2\sin^2 x \\= \cos^2 x - \sin^2 xsin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x

Differentiation

f(x)f(x)f′(x)f'(x)
tan⁡kx\tan kxksec⁡2kxk\sec^2 kx
sec⁡x\sec xsec⁡xtan⁡x\sec x \tan x
cot⁡x\cot x−csc⁡2x-\csc^2 x
csc⁡x\csc x−csc⁡xcot⁡x-\csc x \cot x
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∫f′(x)(f(x))ndx=1n+1(f(x))n+1+c\int f'(x)(f(x))^n dx = \frac{1}{n+1} (f(x))^{n+1} + c
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e.g.∫tan⁡2x dx=∫sec⁡2x−1 dx=tan⁡x−x+ce.g. \int \tan^2 x \, dx = \int \sec^2 x - 1 \, dx = \tan x - x + ce.g.∫sec⁡6xsec⁡xtan⁡x dx=∫sec⁡5xsec⁡xtan⁡x dx⏟f(x)5       ⏟f(x)=16sec⁡6x+ce.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.∫cot⁡2x dx=∫(csc⁡2x−1) dx=−cot⁡x−x+ce.g. \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x - x + csin⁡2x+cos⁡2x=1⇒1+cot⁡2x=csc⁡2x\sin^2 x + \cos^2 x = 1 \quad \Rightarrow \quad 1 + \cot^2 x = \csc^2 x∫4sin⁡xcos⁡x2cos⁡x2−3 dx=∫2(2sin⁡xcos⁡x)(2cos⁡2x−1)−2 dx\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=∫2sin⁡2xcos⁡2x−2 dx==∫−2sin⁡2xcos⁡2x−2 dx= \int \frac{2 \sin 2x }{\cos 2x - 2}\, dx = = \int \frac{-2 \sin 2x }{\cos 2x - 2}\, dx

Using f′(x)f(x)=ln⁡∣f(x)∣+c\frac{f'(x)}{f(x)} = \ln|f(x)| + c

=−ln⁡∣cos⁡2x−2∣+c= -\ln|\cos 2x - 2| + c
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∫(tan⁡x+1)2 dx=∫tan⁡2x+2tan⁡x+1 dx\int (\tan x + 1)^2 \, dx = \int \tan^2 x + 2 \tan x + 1 \, dxLet,A=tan⁡2x,B=2tan⁡xLet, \quad A = \tan^2 x, \quad B = 2 \tan xA=∫tan⁡2x dx=∫sec⁡2x−1 dx=tan⁡x−x+cA = \int \tan^2 x \, dx = \int \sec^2 x - 1 \, dx = \tan x - x + cB=∫2tan⁡x dx=∫2sin⁡xcos⁡x dx=−2∫−sin⁡xcos⁡x dx=−2ln⁡∣cos⁡x∣+cB = \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| + cI=tan⁡x−x−2ln⁡∣cos⁡x∣+x+cI = \tan x - \cancel x - 2 \ln|\cos x|+\cancel x + cI=tan⁡x−2ln⁡∣cos⁡x∣+cI = \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.
∫xex2 dx✓\int xe^{x^2} \, dx \quad \checkmark ∫x2sin⁡x dx✓\int x^2 \sin x \, dx \quad \checkmark ∫x2ex3+4 dx×(Differential of power is outside, use reverse chain rule or substitution)\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. ddx(x3+4)=3x2 \frac{d}{dx}(x^3+4) = 3x^2, so x2x^2 at front will cancel) or where several terms on a denominator need to be combined to make the integral simpler.

e.g.

∫x6x−3 dx(By substitution)\int x \sqrt{6x - 3} \, dx \quad (\text{By substitution}) Let u=6x−3⇒dudx=6⇒dx=16duLet\ u = 6x - 3 \Rightarrow \frac{du}{dx} = 6 \Rightarrow dx = \frac{1}{6} du ⇒I=∫xu12×16du=16∫xu12du\Rightarrow I = \int xu^{\frac{1}{2}} \times \frac{1}{6}du = \frac{1}{6}\int xu^{\frac{1}{2}}du Sinceu=6x−3⇒u+3=6x⇒x=u6+12\text {Since} \quad u = 6x - 3 \Rightarrow u + 3 = 6x \Rightarrow x = \frac{u}{6}+\frac {1}{2} ⇒I=16∫(u6+12)u12du=16∫u326+1122du\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 =16[2u5256+2u3232]+C=190u52+118u32+C= \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 =190(6x−3)52+118(6x−3)32+C= \frac{1}{90} (6x-3)^{\frac{5}{2}} + \frac{1}{18} (6x-3)^{\frac{3}{2}} + C
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Example: ∫x6x−3 dx\int x \sqrt{6x - 3} \, dx (Same again but by parts)

Integration by parts:

∫udvdx dx=uv−∫vdudx dx\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx

Let:

  • u=x⇒dudx=1u = x \Rightarrow \frac{du}{dx} = 1
  • v=2(6x−3)323×6⇒dvdx=(6x−3)12v = \dfrac{2(6x-3)^{\frac{3}{2}}}{3 \times 6}\Rightarrow \dfrac{dv}{dx} = (6x-3)^{\frac{1}{2}}
=19(6x−3)32= \dfrac{1}{9}(6x-3)^{\frac{3}{2}}

Then:

I=x×19(6x−3)32−∫19(6x−3)32 dxI = x \times \frac{1}{9}(6x-3)^{\frac{3}{2}} - \int \frac{1}{9}(6x-3)^{\frac{3}{2}} \, dx=x9(6x−3)32−19⋅25(6x−3)526+c= \frac{x}{9}(6x-3)^{\frac{3}{2}} - \frac{1}{9} \cdot \frac{2}{5}\frac{(6x-3)^{\frac{5}{2}}}{6} + c=x9(6x−3)32−1135(6x−3)52+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:

190(6x−3)52−118(6x−3)32+c\frac{1}{90}(6x-3)^{\frac{5}{2}} - \frac{1}{18}(6x-3)^{\frac{3}{2}} + c=190(6x−3)32[(6x−3)1+5]+c= \frac{1}{90} (6x - 3)^{\frac{3}{2}} \left[ (6x - 3)^{1} + 5 \right] + c=190(6x−3)32(6x+2)+c= \frac{1}{90} (6x - 3)^{\frac{3}{2}} (6x + 2) + c=290(6x−3)32(3x+1)+c= \frac{2}{90} (6x - 3)^{\frac{3}{2}} (3x + 1) + c=145(6x−3)32(3x+1)+c= \frac{1}{45}(6x-3)^{\frac{3}{2}}(3x+1) + c

For by parts we got:

x9(6x−3)32−1135(6x−3)52+c\frac{x}{9}(6x-3)^{\frac{3}{2}} - \frac{1}{135}(6x-3)^{\frac{5}{2}} + c=1135(6x−3)32[15x−(6x−3)]+c= \frac{1}{135}(6x-3)^{\frac{3}{2}}[15x - (6x-3)] + c=1135(6x−3)32(9x+3)+c= \frac{1}{135} (6x - 3)^{\frac{3}{2}} (9x + 3) + c=3135(6x−3)32(3x+1)+c= \frac{3}{135} (6x - 3)^{\frac{3}{2}} (3x + 1) + c=145(6x−3)32(3x+1)+c= \frac{1}{45}(6x-3)^{\frac{3}{2}}(3x+1) + c

Conclusion: Both answers are equivalent.

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