Class 12 — Mathematics (NCERT)

Integrals

Class 12

✓Understand the concept of integration as the inverse process of differentiation.
✓Learn and apply various methods for finding indefinite integrals, including substitution, partial fractions, and integration by parts.
✓Evaluate definite integrals using the Fundamental Theorem of Calculus.
✓Utilise properties of definite integrals to simplify and solve complex problems.

Key concepts

Indefinite Integral

Integration is the inverse process of differentiation. If the derivative of a function F(x) is f(x), i.e., d/dx [F(x)] = f(x), then F(x) is called an antiderivative or primitive of f(x). The collection of all antiderivatives of f(x) is called the indefinite integral of f(x) and is denoted by ∫ f(x) dx. The constant 'C' is called the constant of integration.

∫ f(x) dx = F(x) + C, where d/dx [F(x)] = f(x)

Properties of Indefinite Integrals

Indefinite integrals follow certain linearity properties.

∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx\n∫ k f(x) dx = k ∫ f(x) dx, where k is any real number.

Standard Integrals (Basic Formulas)

These are fundamental integral formulas that must be memorised.

∫ x^n dx = (x^(n+1))/(n+1) + C, n ≠ -1\n∫ (1/x) dx = log|x| + C\n∫ e^x dx = e^x + C\n∫ a^x dx = (a^x)/(log a) + C\n∫ sin x dx = -cos x + C\n∫ cos x dx = sin x + C\n∫ sec^2 x dx = tan x + C\n∫ cosec^2 x dx = -cot x + C\n∫ sec x tan x dx = sec x + C\n∫ cosec x cot x dx = -cosec x + C\n∫ (1/√(a^2 - x^2)) dx = sin⁻¹(x/a) + C\n∫ (1/(a^2 + x^2)) dx = (1/a) tan⁻¹(x/a) + C\n∫ (1/(x√(x^2 - a^2))) dx = (1/a) sec⁻¹(x/a) + C\n∫ (1/(x^2 - a^2)) dx = (1/(2a)) log|(x-a)/(x+a)| + C\n∫ (1/(a^2 - x^2)) dx = (1/(2a)) log|(a+x)/(a-x)| + C\n∫ (1/√(x^2 - a^2)) dx = log|x + √(x^2 - a^2)| + C\n∫ (1/√(x^2 + a^2)) dx = log|x + √(x^2 + a^2)| + C\n∫ √(a^2 - x^2) dx = (x/2)√(a^2 - x^2) + (a^2/2) sin⁻¹(x/a) + C\n∫ √(x^2 - a^2) dx = (x/2)√(x^2 - a^2) - (a^2/2) log|x + √(x^2 - a^2)| + C\n∫ √(x^2 + a^2) dx = (x/2)√(x^2 + a^2) + (a^2/2) log|x + √(x^2 + a^2)| + C

Integration by Substitution

This method transforms the given integral into a simpler form by changing the independent variable. If we have ∫ f(g(x)) g'(x) dx, we substitute t = g(x), so dt = g'(x) dx. The integral then becomes ∫ f(t) dt, which can often be integrated using standard formulas.

∫ f(g(x)) g'(x) dx = ∫ f(t) dt, where t = g(x)

Integration by Partial Fractions

This method is used for integrating rational functions P(x)/Q(x), where P(x) and Q(x) are polynomials and the degree of P(x) is less than the degree of Q(x). The rational function is decomposed into simpler rational functions (partial fractions) which can be integrated easily. Different forms exist for non-repeated linear factors, repeated linear factors, and irreducible quadratic factors in the denominator.

Integration by Parts

This method is used to integrate the product of two functions. It is based on the product rule of differentiation. We choose one function as the first function (u) and the other as the second function (v). The choice of u and v is often guided by the "ILATE" rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential) where the function appearing earlier in ILATE is chosen as u.

∫ u v dx = u ∫ v dx - ∫ [ (du/dx) ∫ v dx ] dx

Definite Integral

A definite integral has upper and lower limits of integration. It represents the area under the curve of a function f(x) from x = a to x = b. The Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is F(b) - F(a).

∫_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a), where d/dx [F(x)] = f(x)

Properties of Definite Integrals

Definite integrals possess several useful properties that can simplify their evaluation.

P0: ∫_a^b f(x) dx = ∫_a^b f(t) dt\nP1: ∫_a^b f(x) dx = - ∫_b^a f(x) dx. In particular, ∫_a^a f(x) dx = 0.\nP2: ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx, where a < c < b.\nP3: ∫_a^b f(x) dx = ∫_a^b f(a+b-x) dx\nP4: ∫_0^a f(x) dx = ∫_0^a f(a-x) dx\nP5: ∫_0^(2a) f(x) dx = ∫_0^a f(x) dx + ∫_0^a f(2a-x) dx\nP6: ∫_0^(2a) f(x) dx = 2 ∫_0^a f(x) dx, if f(2a-x) = f(x)\nP7: ∫_0^(2a) f(x) dx = 0, if f(2a-x) = -f(x)\nP8: ∫_(-a)^a f(x) dx = 2 ∫_0^a f(x) dx, if f(x) is an even function (i.e., f(-x) = f(x))\nP9: ∫_(-a)^a f(x) dx = 0, if f(x) is an odd function (i.e., f(-x) = -f(x))

Key facts to remember

1Integration is the reverse process of differentiation; it is also known as finding the antiderivative or primitive.
2An indefinite integral always includes an arbitrary constant 'C', called the constant of integration.
3The three main methods of integration are substitution, partial fractions, and integration by parts.
4The "ILATE" rule helps in choosing the first function (u) for integration by parts.
5A definite integral has upper and lower limits and represents a specific numerical value (e.g., area under a curve), thus it does not include a constant of integration.
6The Fundamental Theorem of Calculus connects differentiation and integration, stating ∫_a^b f(x) dx = F(b) - F(a), where F'(x) = f(x).
7Memorising standard integral formulas is essential for solving problems efficiently.
8Properties of definite integrals are powerful tools for simplifying and evaluating integrals, especially those with specific limits.

Worked examples

Example 1

Evaluate ∫ (x^2 + 1) / (x^4 + 1) dx.

IDivide numerator and denominator by x^2: ∫ (1 + 1/x^2) / (x^2 + 1/x^2) dx

IIRewrite the denominator: x^2 + 1/x^2 = (x - 1/x)^2 + 2

IIISubstitute t = x - 1/x. Then dt = (1 + 1/x^2) dx.

IVThe integral becomes ∫ dt / (t^2 + 2).

VThis is of the form ∫ dx / (x^2 + a^2) where a = √2.

VI∫ dt / (t^2 + (√2)^2) = (1/√2) tan⁻¹(t/√2) + C.

VIISubstitute back t = x - 1/x: (1/√2) tan⁻¹((x - 1/x)/√2) + C

VIIISimplify the argument of tan⁻¹: (1/√2) tan⁻¹((x^2 - 1)/(√2 x)) + C.

Answer

(1/√2) tan⁻¹((x^2 - 1)/(√2 x)) + C

This problem requires a clever substitution after algebraic manipulation.

Example 2

Evaluate ∫ x sin x dx.

IThis is an integral of a product of two functions, so we use integration by parts.

IIAccording to the ILATE rule, 'x' (Algebraic) comes before 'sin x' (Trigonometric).

IIILet u = x and v = sin x.

IVThen du/dx = 1 and ∫ v dx = ∫ sin x dx = -cos x.

VUsing the formula ∫ u v dx = u ∫ v dx - ∫ [ (du/dx) ∫ v dx ] dx:

VI∫ x sin x dx = x(-cos x) - ∫ [ (1)(-cos x) ] dx

VII= -x cos x - ∫ (-cos x) dx

VIII= -x cos x + ∫ cos x dx

9= -x cos x + sin x + C.

Answer

-x cos x + sin x + C

Example 3

Evaluate ∫_0^(π/2) (sin x) / (sin x + cos x) dx.

ILet I = ∫_0^(π/2) (sin x) / (sin x + cos x) dx (Equation 1)

IIApply property P4: ∫_0^a f(x) dx = ∫_0^a f(a-x) dx.

IIIHere a = π/2. So, x becomes π/2 - x.

IVI = ∫_0^(π/2) (sin(π/2 - x)) / (sin(π/2 - x) + cos(π/2 - x)) dx

VI = ∫_0^(π/2) (cos x) / (cos x + sin x) dx (Since sin(π/2 - x) = cos x and cos(π/2 - x) = sin x) (Equation 2)

VIAdd Equation 1 and Equation 2:

VIII + I = ∫_0^(π/2) (sin x) / (sin x + cos x) dx + ∫_0^(π/2) (cos x) / (cos x + sin x) dx

VIII2I = ∫_0^(π/2) (sin x + cos x) / (sin x + cos x) dx

92I = ∫_0^(π/2) 1 dx

102I = [x]_0^(π/2)

112I = (π/2) - 0

122I = π/2

13I = π/4.

Answer

π/4

This is a classic problem demonstrating the power of definite integral properties.

Common mistakes

✗Forgetting to add the constant of integration 'C' in indefinite integrals.
✗Making errors in algebraic manipulation, especially when simplifying expressions before or after integration.
✗Incorrectly applying the substitution method, particularly when changing limits for definite integrals.
✗Wrongly choosing 'u' and 'v' in integration by parts, leading to more complex integrals.
✗Not checking if the degree of the numerator is less than the denominator before applying partial fractions (if not, perform polynomial division first).
✗Errors in applying the properties of definite integrals, such as incorrect signs or limits.

Exam tips

★Thoroughly memorise all standard integral formulas and the formulas for integration by parts and partial fractions.
★Practice a wide variety of problems from the NCERT textbook and exemplars to gain proficiency in applying different methods.
★Always check your indefinite integral by differentiating the result to see if you get the original integrand.
★For definite integrals, pay close attention to the limits of integration and ensure they are correctly applied after finding the antiderivative.
★When using substitution in definite integrals, remember to change the limits of integration according to the new variable.
★Identify the type of integral (e.g., product, rational function, trigonometric) to choose the most appropriate integration method.

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