01 · Explore
Introduction to Integration
Integration is the inverse process of differentiation, often referred to as anti-differentiation.
Differential Calculus focuses on finding the derivative, which represents the slope of a tangent line. In contrast, Integral Calculus is motivated by the problem of finding the area bounded by the graph of a function. If we are given the derivative of a function, the process of finding the original function is called integration.
The functions that could have a given function as their derivative are called anti-derivatives or primitives. For example, since the derivative of sin x is cos x, we say sin x is an anti-derivative of cos x. However, since the derivative of any constant C is zero, sin x + C is also an anti-derivative for any real number C. This leads to the concept of an indefinite integral, which represents a family of curves differing only by a constant.
The symbol ∫ f(x) dx denotes the indefinite integral of f with respect to x. Here, f(x) is called the integrand, x is the variable of integration, and C is the constant of integration. Two indefinite integrals with the same derivative are equivalent as they represent the same family of curves.
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Question
Why is the constant of integration 'C' added to the result of an indefinite integral?
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02 · Explore
Standard Integral Formulae
Basic integration formulae are derived directly from known differentiation rules.
To integrate effectively, one must memorize standard forms. The power rule for integration states that ∫ xⁿ dx = (xⁿ⁺¹)/(n + 1) + C, provided n ≠ -1. When n = -1, the integral of 1/x is log |x| + C. These rules are the foundations for more complex polynomial integration.
Trigonometric functions have specific anti-derivatives: ∫ cos x dx = sin x + C and ∫ sin x dx = -cos x + C. Note the sign change compared to differentiation. Other standard forms include ∫ sec² x dx = tan x + C and ∫ cosec² x dx = -cot x + C. Exponential functions are unique, as ∫ eˣ dx = eˣ + C.
Inverse trigonometric results are also common. For instance, ∫ 1/√(1 - x²) dx = sin⁻¹ x + C. It is essential to remember that these formulae are valid only within the domains where the functions are defined.
| Function f(x) | Integral ∫ f(x) dx |
|---|---|
| xⁿ (n ≠ -1) | (xⁿ⁺¹)/(n + 1) + C |
| cos x | sin x + C |
| sin x | -cos x + C |
| sec² x | tan x + C |
| eˣ | eˣ + C |
| 1/x | log |x| + C |
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Question
Evaluate the integral of f(x) = x⁴ + sin x.
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Integration by Substitution
The method of substitution reduces a complex integral into a standard form by changing the variable of integration.
Integration by substitution is analogous to the chain rule in differentiation. If an integrand contains a function g(x) and its derivative g'(x), we can substitute u = g(x), which implies du = g'(x) dx. This transforms the integral into a simpler form in terms of u.
Common substitutions involve trigonometric identities. For example, to integrate tan x, we write it as sin x / cos x. By letting t = cos x, then dt = -sin x dx, the integral becomes ∫ -1/t dt = -log |t| + C. This simplifies to log sec x + C. Similar logic applies to cot x, sec x, and cosec x.
When choosing a substitution, look for a part of the integrand whose derivative is also present. If the derivative is not exactly present, it should at most differ by a constant factor.
Substitution Method Example
∫ 2x sin(x² + 1) dx
Let t = x² + 1. Differentiating both sides with respect to x gives dt/dx = 2x, so dt = 2x dx. Substituting these into the integral: ∫ sin(t) dt. The integral of sin(t) is -cos(t) + C. Resubstituting t = x² + 1 gives the final answer: -cos(x² + 1) + C.
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What is the integral of tan x?
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Integration Using Trigonometric Identities
Trigonometric identities are used to simplify integrands involving powers or products of trigonometric functions.
When the integrand involves powers like sin² x or cos² x, we use double angle formulae to reduce the degree. For example, cos² x = (1 + cos 2x)/2 and sin² x = (1 - cos 2x)/2. These linear expressions are much easier to integrate than their squared counterparts.
For products of sines and cosines like sin mx cos nx, we use product-to-sum formulae: 2 sin A cos B = sin(A + B) + sin(A - B). This converts a product into a sum of individual trigonometric terms.
For higher powers like sin³ x, we can either use the identity sin 3x = 3 sin x - 4 sin³ x or split the term into sin² x · sin x and substitute cos x = t.
Integrating cos² x
∫ cos² x dx = ∫ [(1 + cos 2x)/2] dx
Using the identity cos² x = (1 + cos 2x)/2, we split the integral: (1/2) ∫ 1 dx + (1/2) ∫ cos 2x dx. This results in (1/2)x + (1/4)sin 2x + C.
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Question
Which identity is most useful for integrating sin²(2x + 5)?
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Integrals of Particular Functions
Specific algebraic forms have standard logarithmic or inverse trigonometric integral results.
Certain rational and irrational functions follow standard patterns. For example, ∫ dx/(x² - a²) = (1/2a) log |(x - a)/(x + a)| + C. Similarly, ∫ dx/(a² - x²) = (1/2a) log |(a + x)/(a - x)| + C. These are derived using partial fractions or trigonometric substitutions.
Forms involving square roots like ∫ dx/√(x² + a²) result in the logarithmic function log x + √(x² + a²) + C. The form ∫ dx/√(a² - x²) results in sin⁻¹(x/a) + C. Note that in the sin⁻¹ form, there is no 1/a factor outside, unlike the tan⁻¹ form ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C.
To integrate general quadratic expressions in the denominator, such as ax² + bx + c, we first complete the square to transform the expression into one of the standard forms (x ± k)² ± a².
| Integrand Form | Integral Result |
|---|---|
| 1/(x² - a²) | (1/2a) log |(x-a)/(x+a)| + C |
| 1/(x² + a²) | (1/a) tan⁻¹(x/a) + C |
| 1/√(a² - x²) | sin⁻¹(x/a) + C |
| 1/√(x² + a²) | log x + √(x² + a²) + C |
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Question
Evaluate ∫ dx/(x² - 16).
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Integration by Partial Fractions
Rational functions can be integrated by decomposing them into simpler partial fractions.
A rational function P(x)/Q(x) is proper if the degree of P(x) is less than Q(x). If it is improper, we first perform long division. Proper rational functions can be decomposed into partial fractions based on the factors of the denominator Q(x).
If Q(x) has non-repeated linear factors (x - a)(x - b), the decomposition is A/(x - a) + B/(x - b). If there is a repeated linear factor (x - a)², we use A/(x - a) + B/(x - a)². For a non-factorizable quadratic factor (x² + bx + c), the numerator in the partial fraction is of the form (Ax + B).
Once decomposed, the individual terms are integrated using standard logarithmic or inverse trigonometric formulae. This method is highly effective for complex rational integrands.
Partial Fraction Decomposition Process
- 1
Check Degree
Ensure the rational function is proper (degree of numerator < degree of denominator).
- 2
Factorize
Factorize the denominator into linear or irreducible quadratic factors.
- 3
Set Up Fractions
Assign constants (A, B, C) to each factor according to Table 7.2 rules.
- 4
Solve Constants
Equate the numerator and solve for constants by substituting values or comparing coefficients.
- 5
Integrate
Integrate the resulting simpler fractions using standard formulae.
Sequential steps to decompose a rational function for integration.
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How do you handle an improper rational function before integration?
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07 · Explore
Integration by Parts
Integration by parts is used to integrate the product of two functions.
The formula for integration by parts is derived from the product rule of differentiation: ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx. Here, f(x) is the first function and g(x) is the second function. A common mnemonic for choosing the first function is ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, and Exponential.
The choice of the first function is crucial. We choose f(x) such that its derivative f'(x) is simpler, and g(x) such that its integral is easily found. For example, in ∫ x cos x dx, we take f(x) = x (algebraic) and g(x) = cos x (trigonometric).
A special case involves the form ∫ eˣ [f(x) + f'(x)] dx. The result of this integral is simply eˣ f(x) + C. This property is very useful for quickly solving integrals involving exponential functions multiplied by a sum of a function and its derivative.
Integration by Parts: x eˣ
∫ x eˣ dx
Let f(x) = x and g(x) = eˣ. Then f'(x) = 1 and ∫ g(x) dx = eˣ. Applying the formula: x(eˣ) - ∫ 1(eˣ) dx = x eˣ - eˣ + C.
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Question
What is the result of ∫ eˣ (tan⁻¹ x + 1/(1 + x²)) dx?
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Definite Integrals
A definite integral has a unique numerical value and represents the signed area under a curve between two limits.
A definite integral is denoted by ∫ₐᵇ f(x) dx, where 'a' is the lower limit and 'b' is the upper limit. Unlike indefinite integrals, the constant of integration 'C' is not required because it cancels out during subtraction.
The Fundamental Theorem of Calculus connects differentiation and integration. The First Fundamental Theorem states that the area function A(x) = ∫ₐˣ f(t) dt has a derivative A'(x) = f(x). The Second Fundamental Theorem provides the practical tool for evaluation: ∫ₐᵇ f(x) dx = F(b) - F(a), where F is any anti-derivative of f.
When evaluating definite integrals by substitution, it is often easier to change the limits of integration to match the new variable. This avoids the need to resubstitute the original variable at the end.
- 1
Find the indefinite integral F(x) of the integrand f(x).
- 2
Substitute the upper limit 'b' into the anti-derivative to get F(b).
- 3
Substitute the lower limit 'a' into the anti-derivative to get F(a).
- 4
Calculate the difference F(b) - F(a) to find the final value.
The value of the definite integral is F(b) - F(a).
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Question
Evaluate ∫₂³ x² dx.
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Properties of Definite Integrals
Properties of definite integrals simplify the evaluation of complex functions, especially those with symmetry.
Several properties are essential for solving definite integrals efficiently. P₀ states that the variable of integration is a 'dummy' variable: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(t) dt. P₁ states that swapping limits changes the sign: ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx. Additionally, if the limits are the same, the integral is zero: ∫ₐᵃ f(x) dx = 0.
P₂ allows splitting the interval: ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫꜀ᵇ f(x) dx. This is particularly useful for modulus functions. P₄ is a very common property: ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a - x) dx, which often helps in simplifying trigonometric integrands.
P₇ deals with even and odd functions over symmetric limits [-a, a]. If f(x) is odd (f(-x) = -f(x)), the integral is 0. If f(x) is even (f(-x) = f(x)), the integral is 2 ∫₀ᵃ f(x) dx. This property significantly reduces calculation time for symmetric intervals.
| Property Label | Mathematical Statement |
|---|---|
| P₁ | ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx; ∫ₐᵃ f(x) dx = 0 |
| P₂ | ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫꜀ᵇ f(x) dx |
| P₄ | ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a - x) dx |
| P₇ (Odd) | ∫₋ₐᵃ f(x) dx = 0 if f(-x) = -f(x) |
| P₇ (Even) | ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx if f(-x) = f(x) |
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Question
Evaluate ∫₋₁¹ sin⁵ x cos⁴ x dx.
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 49, 50, 51, 52.