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. 
2. 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, ^3π/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)³ (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(^u4/₄ – ^u6/₆)
   • 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(¹⁴/₄ – ¹⁶/₆) – ¹/₂(⁰⁴/₄ – ⁰⁶/₆) = ¹/₂₄

Integrating ^f´(x)/_f(x)

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
  • T – Trig
  • E – Exponential (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 = 2x^1/₂
   • ∴
   • 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 = ¹/₂ – ¹/₂ cos(2x)
    • cos² x = ¹/₂ + ¹/₂ cos(2x)

Integrating cos^m x sin^m 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 x} ≡ ¹/_{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