Calculus II: Techniques of Integration

Module 1: The Substitution Rule

Theorem: The Substitution Rule. If u=g(x) is a differentiable function whose range is an interval I, and f is continuous on I, then:
∫ f(g(x))g'(x) dx = ∫ f(u) du

Key Concepts:

This technique reverses the Chain Rule for derivatives.
Identify an "inner function" g(x) to define as u.
The derivative of u (i.e., g'(x)dx) must be present in the integral, possibly off by a constant factor.
For definite integrals, convert the limits of integration from x-values to u-values.

Example 1.1: Evaluate ∫ x²√(x³ + 1) dx


1. DEFINE u: Let u be the inner function.
   u = x³ + 1

2. FIND du: Differentiate u with respect to x.
   du = 3x² dx

3. REWRITE INTEGRAL: Adjust for the constant factor.
   (1/3) ∫ √(x³ + 1) * (3x² dx)

4. SUBSTITUTE: Replace all terms with u and du.
   (1/3) ∫ √u du = (1/3) ∫ u¹⁄² du

5. INTEGRATE: Apply the power rule for integration.
   (1/3) * [ (2/3)u³⁄² ] + C = (2/9)u³⁄² + C

6. BACK-SUBSTITUTE: Convert the result back to x.
   (2/9)(x³ + 1)³⁄² + C

Module 2: Integration by Parts

Theorem: Integration by Parts. Derived from the Product Rule for derivatives.
∫ u dv = uv - ∫ v du

Key Concepts:

Used for integrating a product of two functions.
Choose u using the **L.I.P.E.T.** priority list:
**L**ogarithmic, **I**nverse Trig, **P**olynomial, **E**xponential, **T**rigonometric.
The remaining part of the integrand becomes dv.
This process may need to be applied multiple times.

Example 2.1: Evaluate ∫ ln(x) dx


1. CHOOSE u AND dv: Following LIPET, Logarithmic is chosen for u.
   u = ln(x)      dv = dx
   du = (1/x) dx  v = x

2. APPLY FORMULA: uv - ∫ v du
   x*ln(x) - ∫ x * (1/x) dx

3. SIMPLIFY: The new integral is simpler.
   x*ln(x) - ∫ 1 dx

4. INTEGRATE:
   x*ln(x) - x + C

Module 3: Trigonometric Integrals

Key Identities:
sin²x + cos²x = 1
tan²x + 1 = sec²x
sin²x = (1 - cos(2x))/2
cos²x = (1 + cos(2x))/2

Strategy:

For ∫sin^mxcosⁿx dx:
If the power of cos(x) is ODD, save one cos(x) factor and convert the rest to sin(x). Use u=sin(x).
If the power of sin(x) is ODD, save one sin(x) factor and convert the rest to cos(x). Use u=cos(x).
If BOTH powers are EVEN, use the half-angle identities to reduce the powers.

Example 3.1: Evaluate ∫ cos⁴(x) dx


1. STRATEGY: The power is even, so use the half-angle identity.
   ∫ (cos²(x))² dx = ∫ [ (1 + cos(2x))/2 ]² dx

2. EXPAND THE INTEGRAND:
   ∫ (1/4) * (1 + 2cos(2x) + cos²(2x)) dx

3. APPLY IDENTITY AGAIN: Another even power requires another reduction.
   cos²(2x) = (1 + cos(4x))/2

4. SUBSTITUTE AND SIMPLIFY:
   ∫ (1/4) * [1 + 2cos(2x) + (1 + cos(4x))/2] dx
   ∫ (1/4) * [3/2 + 2cos(2x) + (1/2)cos(4x)] dx
   ∫ [3/8 + (1/2)cos(2x) + (1/8)cos(4x)] dx

5. INTEGRATE TERM BY TERM:
   (3/8)x + (1/2)*(1/2)sin(2x) + (1/8)*(1/4)sin(4x) + C
   (3/8)x + (1/4)sin(2x) + (1/32)sin(4x) + C

Module 4: Trigonometric Substitution

Method: For integrands with quadratic expressions in radicals, transform the integral into the trigonometric domain using a reference triangle.

Strategy:

For √(a² - x²), let x = a*sin(θ), so dx = a*cos(θ)dθ.
For √(a² + x²), let x = a*tan(θ), so dx = a*sec²(θ)dθ.
For √(x² - a²), let x = a*sec(θ), so dx = a*sec(θ)tan(θ)dθ.
After integrating, use the reference triangle to convert the result back to x.

Example 4.1: Evaluate ∫ dxx²√(9 - x²)


1. IDENTIFY FORM: The expression matches √(a² - x²) with a=3.
   Let x = 3sin(θ), which means dx = 3cos(θ)dθ.

2. SUBSTITUTE INTO THE INTEGRAL:
   ∫ 3cos(θ)dθ / [ (3sinθ)² * √(9 - 9sin²θ) ]

3. SIMPLIFY THE RADICAL:
   √(9(1-sin²θ)) = √(9cos²θ) = 3cos(θ)

4. SIMPLIFY THE FULL INTEGRAL:
   ∫ 3cos(θ)dθ / [ 9sin²θ * 3cos(θ) ]
   ∫ 1 / (9sin²θ) dθ = (1/9) ∫ csc²(θ) dθ

5. INTEGRATE WITH RESPECT TO θ:
   (1/9) * (-cot(θ)) + C

6. CONVERT BACK TO x: Use a reference triangle.
   From x = 3sin(θ), we have sin(θ) = x/3 (opposite/hypotenuse).
   The adjacent side is √(3²-x²) = √(9-x²).
   Therefore, cot(θ) = adjacent/opposite = √(9-x²)/x.
   
   Final Answer: -(1/9) * [√(9-x²)/x] + C

Module 5: Partial Fraction Decomposition

Theorem: Any rational function P(x)/Q(x) where degree(P) < degree(Q) can be rewritten as a sum of simpler fractions whose denominators are the factors of Q(x).

Strategy:

If degree(P) >= degree(Q), perform polynomial long division first.
Factor the denominator Q(x) completely.
For each linear factor (x-r), create a term A/(x-r).
For repeated linear factors (x-r)ⁿ, create terms for each power: A₁/(x-r) + A₂/(x-r)² + ...
For irreducible quadratic factors (ax²+bx+c), create a term (Ax+B)/(ax²+bx+c).

Example 5.1: Evaluate ∫ x+1x(x-1)² dx


1. SET UP THE DECOMPOSITION:
   (x+1)/[x(x-1)²] = A/x + B/(x-1) + C/(x-1)²

2. CLEAR THE DENOMINATOR by multiplying through:
   x+1 = A(x-1)² + Bx(x-1) + Cx

3. SOLVE FOR COEFFICIENTS (Heaviside Cover-up Method):
   Let x=1:  1+1 = C(1)  => C = 2
   Let x=0:  0+1 = A(-1)² => A = 1

4. SOLVE FOR THE REMAINING COEFFICIENT (B):
   Expand and equate the coefficients of the x² term.
   x+1 = A(x²-2x+1) + B(x²-x) + Cx
   x+1 = (A+B)x² + (-2A-B+C)x + A
   The coefficient of x² on the left is 0.
   0 = A+B => 0 = 1+B => B = -1

5. REWRITE THE INTEGRAL with the simpler fractions:
   ∫ [ 1/x - 1/(x-1) + 2/(x-1)² ] dx

6. INTEGRATE TERM BY TERM:
   ln|x| - ln|x-1| + 2 * (-1/(x-1)) + C
   ln|x| - ln|x-1| - 2/(x-1) + C