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#### Integration

###### Basic Integrals

• Use trigonometrical relationships to facilitate complex trigonometric integrals.
• Integrate by decomposing into partial fractions.
###### Integration by u-Substitution

• Make x equal to something; when differentiated, multiply the substituted form directly.
• Make u equal to something; when differentiated, multiply the substituted form with its reciprocal.
• With definite integrals, change the limits in terms of u.
###### Examples
1. The diagram shows part of curve y = sin3 2x cos3 2x. The shaded region shown is bounded by the curve and the x-axis, and its exact area is denoted by A. Use the substitution u = sin 2x in a suitable integral to find the value of A.
• To find the limit, you are trying to find the points at which y = 0;
• sin x = 0 at x = 0, π, 2π
• cos x = 0 at x = π/2, /4
• Choose the two closest to 0 because the shaded area has gone through y = 0 only twice, therefore 0 and π/2
• Since it is sin 2x and cos 2x, divide both limits by 2, therefore, limits are 0 and π/4
• Integrate by u-substitution, let:
• u = sin 2x
• du/dx = 2 cos 2x
• dx/du = 1/2 cos 2x
• sin3 2x cos3 2x ≡ (sin 2x)(cos 2x)2 cos 2x
• ≡ (sin3 2x × (1 – sin2 2x)) cos 2x
• ≡ (sin3 2x – sin5 2x) cos 2x
• f(x) dx/du = (sin3 2x – sin5 2x) cos 2x × 1/2 cos 2x
• ∴ f(x) dx/du = 1/2(u3 – u5)
• Now integrate: 1/2∫(u3 – u5) = 1/2(u4/4u6/6)
• The limits are x = 0 and x = π/4; in terms of u;
• u = sin 2(0) = 0, and
• u = sin 2(π/4) = 1
• Substitute the limits: 1/2(14/416/6) – 1/2(04/406/6) = 1/24

###### Examples
1. By splitting into partial fractions, show that:

• Write as partial fractions:
• Substitute the limits:
###### Integrating by Parts

• For a definite integral:
• Hierarchical order of selecting u:
• L Logs
• A Algebra
• TTrig
• EExponential (e)
###### Examples
1. Find the exact value of

• Convert to index form: ln x/√x = x-1/2ln x
• Integrate by parts:
• u = ln x ⇒ du/dx = 1/x
• dv/dx = x-1/2 ⇒ v = 2x1/2
• Substitute the limits: ⇒ 4ln 4 – 4
###### Integrating Powers of Sine or Cosine
• To integrate sin x or cos x with a power:
• If the power is odd, pull out a sin x or cos x and use Pythagorean identities and double angle identities.
• If the power is even, use the following identities:
• sin2 x = 1/21/2 cos(2x)
• cos2 x = 1/2 + 1/2 cos(2x)
###### Integrating cosm x sinm x
• If m or n are odd and even, then:
• Factor out one power from the odd trig function
• Use Pythagorean identities to transform the remaining even trig function into the odd trig function
• Let u equal the odd trig function and integrate.
• If m and n are both even, then:
• Replace all even powers using the double angle identities and integrate.
• If m and n are both odd, then:
• Choose one of the trig functions and factor out one power\
• Use Pythagorean identities to transform the remaining even power of the chosen trig function to the other trig function.
• If either m or n or both = 1, then:
• Let u equal the trig function whose power doesn’t equal 1 then integrate.
• If both are 1, then let u equal either.
###### Examples
1. Prove the identity
1. cos 4θ – 4 cos 2θ + 3 ≡ 8 sin4 θ
• Use double angle identities: cos 4θ – 4 cos 2θ + 3 ≡ 1 – 2 sin2 2θ – 4(1 – 2 sin2 θ) + 3
• Open everything and clean:
• ≡ 1 – 2 sin2 2θ – 4 + 8 sin2 θ + 3
• ≡ 1 – 2(sin 2θ)2 – 4 + 8 sin2 θ + 3
• ≡ 1 – 2(2sin θ cos θ)2 – 4 + 8 sin2 θ + 3
• ≡ 1 – 2(4 sin2 θ cos2 θ) – 4 + 8 sin2 θ + 3
• ≡ 1 – 2(4 sin2 θ (1- sin2 θ)) – 4 + 8 sin2 θ + 3
• ≡ 1 – 8 sin2 θ + 8 sin4 θ – 4 + 8 sin2 θ + 3
• ≡ 8 sin4 θ
2. Use the result from (a), to find, in simplified form, the exact value of

• Use identity from (a):
• Substitute the limits: 1/32(2π – √3)
2. Solve the following:
1. By differentiating 1/cos x , show that if y = sec x, then dy/dx = sec x tan x
• Change to index form: 1/cos x = cos-1 x
• Differentiate by chain rule:
• dy/dx = -1(cos x)-2 × (-sin x)
• -1(cos x)-2 × (-sin x) ≡ sin x/cos2 x1/cos x × sin x/cos x
• 1/cos x × sin x/cos x ≡ sec x tan x
2. Show that 1/sec x tan x ≡ sec x + tan x
• Multiply numerator and denominator by sec x + tan x:
• sec x + tan x/(sec x – tan x)(sec x + tan x)sec x + tan x/sec2 x – tan2 x
• sec x + tan x/sec2 x – tan2 x ≡ sec x tan x
3. Deduce that: 1/(sec x tan x)2 ≡ 2 sec2 x – 1 + 2 sec x tan x
• Substitute identity from (b): 1/(sec x – tan x)2 ≡ (sec x + tan x)2
• Open the brackets:
• (sec x + tan x)2 ≡ sec2 x + 2 sec x tan x + tan2 x
• ≡ sec2 x + 2 sec x tan x + sec2 x – 1
• ≡ 2sec2 x – 1 + 2 sec x tan x
4. Hence show that:
• Substitute identity from (c): ∫(2sec2 x – 1 + 2 sec x tan x) dx
• ≡ 2 ∫sec2 x – ∫1 + 2∫sec x tan x
• ≡ 2 tan x – x + 2 sec x
• Substitute boundaries: 1/4(8√2 – π)
###### Trapezium Rule
• Area = Width of any strip × [(1st height + last height) ÷ 2 + sum of all y-values between the second and second-to-last value]
• Width of any strip = b – a/no. of intervals for
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