Lecture Note 4: Integration

1. Meaning of Integration

Integration is the process of finding a function from its derivative. It is the reverse operation of differentiation. If F′(x) = f(x), then F(x) is called an antiderivative of f(x).

The general antiderivative is written as:

∫ f(x) dx = F(x) + C

where C is an arbitrary constant called the constant of integration. The symbol ∫ is the integral sign, f(x) is the integrand, and dx shows that integration is carried out with respect to x.

Core Theorem

If F(x) is one antiderivative of f(x) on an interval, then every antiderivative of f(x) on that interval has the form:

F(x) + C

Example

Since ^d⁄_dx(x³) = 3x², it follows that:

∫ 3x² dx = x³ + C

Key Takeaways

• Integration reverses differentiation.
• An indefinite integral represents a family of antiderivatives.
• The constant C is required in every indefinite integral.
• An answer to an indefinite integral can be checked by differentiating it.

2. Integration by the Power Rule

For any real number n ≠ -1, the power rule for integration is:

∫ xⁿ dx = ^{xn + 1}⁄_{n + 1} + C

In words: increase the power by one and divide by the new power.

Example 1

Evaluate ∫ ³√x dx.

Solution:

³√x = x^1/3

∫ ³√x dx = ∫ x^1/3 dx = ^x4/3⁄_4/3 + C = ³⁄₄x^4/3 + C

Example 2

Evaluate ∫ ¹⁄_x⁵ dx.

Solution:

¹⁄_x⁵ = x^-5

∫ x^-5 dx = ^x-4⁄_-4 + C = -¹⁄_4x⁴ + C

Exam Points

• Rewrite roots as fractional powers before integrating.
• Rewrite reciprocals as negative powers before integrating.
• The power rule does not apply to x^-1; instead ∫ ¹⁄_x dx = ln|x| + C.
• Always include + C for indefinite integrals.

3. Integration by Substitution

Integration by substitution is the reverse of the chain rule. It is used when an integral contains a composite expression and the derivative of the inner expression is present, either exactly or as a constant multiple.

If u = g(x), then:

du = g′(x) dx

The integral is changed from an expression in x to an expression in u, integrated, and then changed back to x.

Example 1

Evaluate ∫ (4 - 3x)^1/7 dx.

Solution:

Let u = 4 - 3x. Then du = -3 dx, so dx = -¹⁄₃du.

∫ (4 - 3x)^1/7 dx = -¹⁄₃∫ u^1/7 du

= -¹⁄₃ × ^u8/7⁄_8/7 + C = -⁷⁄₂₄u^8/7 + C

= -⁷⁄₂₄(4 - 3x)^8/7 + C

Example 2

Evaluate ∫ cos(3x + 5) dx.

Solution:

Let u = 3x + 5. Then du = 3dx, so dx = ¹⁄₃du.

∫ cos(3x + 5) dx = ¹⁄₃∫ cos u du = ¹⁄₃sin(3x + 5) + C

Example 3

Evaluate ∫ sin(5x - 2) dx.

Solution:

Let u = 5x - 2. Then du = 5dx, so dx = ¹⁄₅du.

∫ sin(5x - 2) dx = ¹⁄₅∫ sin u du = -¹⁄₅cos(5x - 2) + C

Key Takeaways

• Choose u as the inner expression.
• Replace both the expression and dx.
• After integrating, return the answer to the original variable.
• Substitution is especially useful for powers of brackets and trigonometric functions with linear angles.

4. Integration of Trigonometric Expressions

Many trigonometric integrals become simple after applying identities. The most useful identities here are:

sin²x = ^{1 - cos 2x}⁄₂

cos²x = ^{1 + cos 2x}⁄₂

sin²x + cos²x = 1

4.1 Even Powers of Sine and Cosine

When the power is even, use the half-angle identities.

Example 1

Evaluate ∫ sin²x dx.

Solution:

∫ sin²x dx = ∫ ^{1 - cos 2x}⁄₂ dx

= ^x⁄₂ - ¹⁄₄sin 2x + C

Example 2

Evaluate ∫ cos²(3x) dx.

Solution:

cos²(3x) = ^{1 + cos 6x}⁄₂

∫ cos²(3x) dx = ^x⁄₂ + ¹⁄₁₂sin 6x + C

4.2 Odd Powers of Sine and Cosine

When an odd power occurs, separate one sine or cosine factor and convert the remaining even power using sin²x + cos²x = 1.

Example 3

Evaluate ∫ cos³x dx.

Solution:

cos³x = cos x(1 - sin²x)

Let u = sin x, so du = cos x dx.

∫ cos³x dx = ∫ (1 - u²) du = u - ^u3⁄₃ + C

= sin x - ^sin3x⁄₃ + C

Example 4

Evaluate ∫ sin³x cos²x dx.

Solution:

sin³x = sin x(1 - cos²x)

∫ sin³x cos²x dx = ∫ [cos²x sin x - cos⁴x sin x] dx

Using u = cos x, du = -sin x dx, the result is:

-^cos3x⁄₃ + ^cos5x⁄₅ + C

Exam Points

• Use half-angle identities for even powers.
• Separate one sine or cosine factor for odd powers.
• Convert the remaining even power using sin²x = 1 - cos²x or cos²x = 1 - sin²x.
• Choose the substitution that matches the separated differential.

5. Trigonometric Substitution

Trigonometric substitution is used for integrals containing square roots of quadratic expressions. The substitution is chosen from a Pythagorean identity.

Example 1

Evaluate ∫ ^dx⁄_{√(1 - x²)}.

Solution:

Let x = sin θ. Then dx = cos θ dθ and √(1 - x²) = cos θ.

Hence:

∫ ^dx⁄_{√(1 - x²)} = ∫ dθ = θ + C

Since x = sin θ, θ = sin^-1x. Therefore:

∫ ^dx⁄_{√(1 - x²)} = sin^-1x + C

Example 2

Evaluate ∫ ^dx⁄_{x√(x² - 25)}.

Solution:

Let x = 5sec θ. Then dx = 5sec θtan θ dθ and √(x² - 25) = 5tan θ.

Therefore:

∫ ^dx⁄_{x√(x² - 25)} = ∫ ^{5sec θtan θ dθ}⁄_{(5sec θ)(5tan θ)}

= ¹⁄₅∫ dθ = ^θ⁄₅ + C

Since x = 5sec θ, θ = sec^-1(^x⁄₅). Thus:

∫ ^dx⁄_{x√(x² - 25)} = ¹⁄₅sec^-1(^x⁄₅) + C

Key Takeaways

• Use trigonometric substitution for radicals involving quadratic expressions.
• Choose the substitution that matches the structure of the radical.
• Return the final answer to the original variable.
• Constants such as a must be carried carefully throughout the solution.

6. Integration by Parts

Integration by parts is the reverse of the product rule. It is used for integrals involving products of functions.

∫ u dv = uv - ∫ v du

A useful guide for choosing u is LIATE: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential.

Example 1

Evaluate ∫ te^-3t dt.

Solution:

Let u = t, so du = dt. Let dv = e^-3tdt, so v = -¹⁄₃e^-3t.

∫ te^-3tdt = uv - ∫ vdu

= -¹⁄₃te^-3t + ¹⁄₃∫ e^-3tdt

= -¹⁄₃te^-3t - ¹⁄₉e^-3t + C

Example 2

Evaluate ∫ x²sin x dx.

Solution:

Let u = x² and dv = sin x dx. Then du = 2x dx and v = -cos x.

∫ x²sin x dx = -x²cos x + 2∫ x cos x dx

Apply integration by parts again to ∫ x cos x dx:

∫ x cos x dx = x sin x + cos x

Therefore:

∫ x²sin x dx = -x²cos x + 2x sin x + 2cos x + C

Exam Points

• Use integration by parts for products such as x sin x, x e^x, and x ln x.
• Choose u so that it becomes simpler when differentiated.
• Choose dv so that it can be integrated easily.
• Some problems require integration by parts more than once.

7. Integration by Partial Fractions

Partial fractions are used to integrate rational functions, that is, fractions whose numerator and denominator are polynomials. The goal is to split a complicated fraction into simpler fractions.

Common Forms

For distinct linear factors:

^P(x)⁄_{(x + a)(x + b)} = ^A⁄_{x + a} + ^B⁄_{x + b}

For a repeated linear factor:

^P(x)⁄_{(x + a)²(x + b)} = ^A⁄_{x + a} + ^B⁄_{(x + a)²} + ^C⁄_{x + b}

Example 1

Evaluate ∫ ^{x + 2}⁄_{(x + 3)(x + 4)} dx.

Solution:

Write:

^{x + 2}⁄_{(x + 3)(x + 4)} = ^A⁄_{x + 3} + ^B⁄_{x + 4}

Then:

x + 2 = A(x + 4) + B(x + 3)

Put x = -3: A = -1.

Put x = -4: B = 2.

Thus:

∫ ^{x + 2}⁄_{(x + 3)(x + 4)} dx = -∫ ^dx⁄_{x + 3} + 2∫ ^dx⁄_{x + 4}

= -ln|x + 3| + 2ln|x + 4| + C

Example 2

Evaluate ∫ ^dx⁄_{(x - 1)²(x + 2)}.

Solution:

Decompose:

¹⁄_{(x - 1)²(x + 2)} = ^A⁄_{x - 1} + ^B⁄_{(x - 1)²} + ^C⁄_{x + 2}

Solving for the constants gives:

A = -¹⁄₉, B = ¹⁄₃, C = ¹⁄₉

Therefore:

∫ ^dx⁄_{(x - 1)²(x + 2)} = ¹⁄₉ln|^{x + 2}⁄_{x - 1}| - ¹⁄_{3(x - 1)} + C

Key Takeaways

• Use partial fractions for rational functions.
• Factor the denominator first.
• Use repeated terms for repeated factors.
• After decomposition, integrate each simple fraction separately.
• Integrals of the form ∫ ^dx⁄_{x + a} give logarithms.

8. Definite Integrals

A definite integral has limits of integration and gives a numerical value:

∫_a^b f(x) dx

If F′(x) = f(x), then:

∫_a^b f(x) dx = F(b) - F(a)

This result is the evaluation form of the Fundamental Theorem of Calculus.

Substitution in Definite Integrals

When using substitution in a definite integral, either:

• Integrate in terms of the original variable and then apply the original limits.
• Change the limits to the new variable and integrate directly.

Example 1

Evaluate ∫₀² (x + 1)(x² + 2x + 3)⁵ dx.

Solution:

Let u = x² + 2x + 3. Then du = 2(x + 1)dx, so (x + 1)dx = ¹⁄₂du.

When x = 0, u = 3. When x = 2, u = 11.

Therefore:

∫₀² (x + 1)(x² + 2x + 3)⁵ dx = ¹⁄₂∫₃¹¹ u⁵ du

= ¹⁄₁₂(11⁶ - 3⁶)

Example 2

Evaluate ∫₀^{1 dx}⁄_{√(4 - x²)}.

Solution:

Let x = 2sin u. Then dx = 2cos u du and √(4 - x²) = 2cos u.

When x = 0, u = 0. When x = 1, u = ^π⁄₆.

Thus:

∫₀^{1 dx}⁄_{√(4 - x²)} = ∫₀^π/6 du = ^π⁄₆

Exam Points

• A definite integral gives a numerical value.
• Do not add + C after evaluating a definite integral.
• Use F(b) - F(a), not F(a) - F(b).
• When substituting, either change the limits or return to the original variable before applying the limits.

9. Exam-Focused Formula Summary

∫ f(x) dx = F(x) + C, where F′(x) = f(x).

∫ xⁿ dx = ^{xn + 1}⁄_{n + 1} + C, for n ≠ -1.

∫ ¹⁄_x dx = ln|x| + C.

∫ sin x dx = -cos x + C.

∫ cos x dx = sin x + C.

∫ sec²x dx = tan x + C.

∫ csc²x dx = -cot x + C.

∫ u dv = uv - ∫ v du.

∫_a^b f(x) dx = F(b) - F(a).

10. Practice Questions for Revision

A. Power Rule

1. Evaluate ∫ x⁷ dx.
2. Evaluate ∫ ¹⁄_x⁴ dx.
3. Evaluate ∫ √x dx.
4. Evaluate ∫ ³√(x²) dx.

B. Substitution

1. Evaluate ∫ (2x + 1)(x² + x + 3)⁵ dx.
2. Evaluate ∫ sin(4x - 1) dx.
3. Evaluate ∫ cos(7x + 2) dx.
4. Evaluate ∫ x√(x² + 5) dx.

C. Trigonometric Integrals

1. Evaluate ∫ sin⁴x dx.
2. Evaluate ∫ cos³x dx.
3. Evaluate ∫ sin⁵x dx.
4. Evaluate ∫ tan⁴x sec²x dx.

D. Techniques of Integration

1. Evaluate ∫ ^dx⁄_{√(4 - x²)}.
2. Evaluate ∫ x cos x dx.
3. Evaluate ∫ xe^-x dx.
4. Evaluate ∫ x²ln x dx.
5. Evaluate ∫ ^{x + 3}⁄_{x(x + 1)} dx.

E. Definite Integrals

1. Evaluate ∫_-1⁰ (x + 3)⁴ dx.
2. Evaluate ∫₀¹ x²(x³ + 7)^2/3 dx.
3. Evaluate ∫₀^{1 dx}⁄_{√(4 - x²)}.

11. Final Revision Checklist

• Explain integration as the reverse of differentiation.
• Apply the power rule correctly.
• Convert roots and reciprocals into index form.
• Use substitution for composite expressions.
• Use identities to integrate trigonometric powers.
• Use trigonometric substitution for radical expressions.
• Apply integration by parts to products.
• Use partial fractions for rational functions.
• Evaluate definite integrals using F(b) - F(a).
• Check indefinite integrals by differentiating the answer.