Integrals Class 12 NCERT Notes PDF

Summary

This document is an excerpt from class 12th NCERT mathematics textbook focusing on integrals. Topics like integration as inverse of differentiation, standard integration formulas, and example problems are included. Key integral techniques discussed in the text sample include inspection and substitution techniques for different types of functions.

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# INTEGRALS ## TOPIC 1: Integration and Its Properties ### Integration as an Inverse Process of Differentiation - Let F(x) and f(x) be two functions connected together such that _dF(x)/dx_ = f(x), then F(x) is called integral of f(x), or the indefinite integral, or anti-derivative. - _dF(x)/d...

# INTEGRALS ## TOPIC 1: Integration and Its Properties ### Integration as an Inverse Process of Differentiation - Let F(x) and f(x) be two functions connected together such that _dF(x)/dx_ = f(x), then F(x) is called integral of f(x), or the indefinite integral, or anti-derivative. - _dF(x)/dx_ = f(x), then for any arbitrary constant C, _d[F(x)+C]/dx_ = f(x). - Thus, F(x) + C is also an anti-derivative of f(x). Actually, there exist infinitely many anti-derivatives of a function which can be obtained by choosing C arbitrarily from the set of real numbers. Hence, ∫f(x) dx = F(x) + C, where C is an arbitrary constant (also called constant of integration) and symbol '∫' indicates the sign of integration. By varying the parameter C, one gets different anti-derivatives or integrals of the given function. ### Symbols, Terms & Phrases | Symbols/Terms/ Phrases | Meaning | |---|---| | ∫f(x)dx | Integral of f with respect to x | | f(x) in ∫f(x)dx | Integrand | | x in ∫f(x)dx | Variable of integration | | Integrate | Find the integral | | An integral of f | A function F such that F'(x) = f(x) | | Integration | The process of finding the integral | | Constant of integration | Any real number C, considered as constant function | ### Some Standard Formulae | Derivatives | Indefinite Integrals (Anti-derivatives) | |---|---| | _d(x^n)/dx_ = x^(n-1), _n≠-1_ | ∫x^n dx = (x^(n+1))/(n+1) + C, _n≠-1_ | ## CHAPTER CHECKLIST - Integration and Its Properties - Integration by Substitutions - Integration by Partial Fractions - Integration by Parts - Definite Integral - Evaluation of Definite Integral by Substitution ## EXAMPLE 1: Evaluate the following integrals. 1. ∫ x dx 2. ∫ dx 3. ∫ cos x dx 4. ∫ sin x dx 5. ∫ sec²x dx 6. ∫ cosec²x dx 7. ∫ sec x tan x dx 8. ∫ cosec x cot x dx 9. ∫ (1/sin²x) dx 10. ∫ (1/1-x²) dx 11. ∫ (1/1+x²) dx 12. ∫ (1/√1-x²) dx 13. ∫ (1/x²-1) dx 14. ∫ e^x dx 15. ∫ (1/x) dx 16. ∫ log|x| dx 17. ∫ a^x dx (a>0, a≠1) 18. ∫ (ax+b) dx 19. ∫ sin(ax+b) dx 20. ∫ cos(ax+b) dx 21. ∫ tan(ax+b) dx 22. ∫ cot(ax+b) dx 23. ∫ sec(ax+b) dx 24. ∫ cosec(ax+b) dx 25. ∫ sec(ax+b) tan(ax+b) dx 26. ∫ cosec(ax+b) cot(ax+b) dx ## Geometric Interpretation of Indefinite Integral - Geometrically, the statement ∫f(x) dx = ϕ(x) + C = y (say) represents a family of curves. - The different values of C correspond to different members of this family and the graph of these members can be obtained by shifting any one of the curves parallel to itself. - Further, the tangents to the curves at the point of intersection of a line x = a with the curves are parallel. ## EXAMPLE 3: Evaluate ∫ √x dx. ## Properties of Indefinite Integrals (i) The process of differentiation and integration are inverse of each other. - i.e. _d/dx_∫f(x)dx = f(x) and ∫f'(x)dx = f(x) + C, where C is any arbitrary constant. (ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. (iii) ∫(f(x) ± g(x))dx = ∫f(x)dx ± ∫g(x)dx (iv) ∫k f(x)dx = k ∫f(x)dx, where k is any non-zero real number. (v) ∫[k₁f₁(x) + k₂f₂(x) + ... + kₙfₙ(x)]dx = k₁∫f₁(x)dx + k₂∫f₂(x)dx + ... + kₙ∫fₙ(x)dx ## EXAMPLE 2: Evaluate the following integrals. (i) ∫(sin x + cos x) dx (ii) ∫(sin x cot x + 1/sin³x) dx ## Integration by Method of Inspection - We can find an anti-derivative of a given function by searching intuitively a function whose derivative is the given function. - The search for the requisite function for finding an anti-derivative is known as integration by the method of inspection. ## EXAMPLE 5: Write an anti-derivative of 3x² + 4x³ by the method of inspection. ## Comparison between Differentiation and Integration 1. Differentiation and integration both are operations on real valued functions. 2. Differentiation and integration both satisfy the property of linearity. - i.e. _d/dx_ {af(x) ± bg(x)} = a{ _d/dx_ f(x)} ± b{ _d/dx_ g(x)} - and [[af(x) ± bg(x)]dx = a∫f(x)dx + b∫g(x)dx 3. All functions are not differentiable, similarly there are some functions which are not integrable. 4. The derivative of a function, when it exists, is a unique function. But the integral of a function is not unique. However, they are unique upto an additive constant, i.e. any two integrals of a function differ by a constant. - e.g. If _d/dx_ (sinx) = cos x. - Then, ∫cos x dx = sinx + C, C ∈ R 5. When a polynomial function P is differentiated, then the resultant is also a polynomial whose degree is 1 less than the degree of P. ## EXAMPLE 3: If _f'(x)_ = x + (1/x), then the value of f(x) is (a) x² + log x + C (b) (x²/2) + log|x| + C (c) (x²/2) + log x + C (d) None of the above ## Family of curves - Family of curves y = F(x) + C can be represented geometrically by shifting any one of the curves ...A... to itself. Here, A refers to - (a) perpendicular - (b) parallel - (c) Both (a) and (b) - (d) None of these ## Integration by Substitution - In previous topic, we discussed the integrals of those functions which are in standard forms. - But integrals of certain functions cannot be obtained directly, if they are not in one of the standard forms given in previous topic. - For such integrals we use the following method to solve - Integration by substitution - Integration by partial fractions - Integration by parts - The method of reducing a given integral into one of the standard integrals by a proper substitution is called method of substitution. To evaluate an integral of the type ∫f{g(x)}· g'(x) dx, we substitute g(x) = t, so that g'(x) dx = dt. ## EXAMPLE 1: Evaluate ∫(tan⁻¹x)/(1 + tan²x)dx. ## EXAMPLE 2: Evaluate ∫cos 6x √1 + sin 6x dx. ## Method to Find Integration by Substitution - Suppose given integral is I = ∫f(x) dx, where f(x) is not in any standard form. - Then, we use the following steps: - Identify the term for which we use substitution say g(x). - Take suitable substitution to reduce the integral into a standard form, say g(x) = t. - Now, integrate with respect to t and then put t = g(x) to get the required value. ## EXAMPLE 3: Integrate the following function w.r.t. x: (tan √x sec² √x)/√x ## EXAMPLE 4: Evaluate ∫ dx/(1 + tan x) ## Some Important Deductions (i) ∫(ax+b)^n dx = (ax + b)^(n + 1) / a(n + 1) + C, _n≠-1_ and _n_ is a rational number. (ii) ∫sin(ax + b) dx = -cos(ax+b)/a + C (iii) ∫cos(ax + b) dx = sin(ax + b)/a + C (iv) ∫tan(ax + b) dx = -log|cos(ax + b)|/a + C = log|sec(ax + b)|/a + C (v) ∫cot(ax + b) dx = log|sin(ax + b)|/a + C (vi) ∫sec(ax + b) dx = log|sec(ax+b)+tan(ax + b)|/a + C (vii) ∫cosec(ax + b) dx = log|cosec(ax + b)-cot(ax + b)|/a + C (viii) ∫sec(ax + b) tan(ax + b) dx = sec(ax+b)/a + C (ix) ∫cosec(ax + b) cot(ax + b) dx = -cosec(ax+b)/a + C (x) ∫sec²(ax + b) dx = tan(ax + b)/a + C (xi) ∫cosec²(ax + b) dx = -cot(ax + b)/a + C (xii) ∫e^(ax+b) dx = (e^(ax+b))/(a logₐe) + C (xiii) ∫a^(ax+b) dx = (a^(ax+b))/(a logₐa) + C ## Some Standard Formulae - (i) ∫tan x dx = -log|cosx| + C = log|secx| + C - (ii) ∫cotx dx = log|sinx| + C - (iii) ∫sec x dx = log|secx + tanx| + C - (iv) ∫cosec x dx = log|cosecx - cotx| + C ## Integration Using Trigonometric Identities - When the integrand involves some trigonometric functions, then some known trigonometric identities are used to evaluate integral easily. ## EXAMPLE 5: Evaluate the following integrals. (i) ∫√ax + b dx (ii) ∫dx/√x + a + √x + b ## Integration by Partial Fractions - Sometimes, an integral of the form ∫P(x)/Q(x) dx, where P(x) and Q(x) are polynomials in x and Q(x) ≠ 0, also Q(x) has only linear and quadratic factors is given to us, if we cannot integrate it directly or by previous methods, then we use the partial fractions. - **To decompose a fraction into partial fractions** - If the degree of P(x) is greater than the degree of Q(x) then we divide P(x) by Q(x) to get a quotient and a remainder. - The quotient is then integrated directly. - The remainder is decomposed into a sum of partial fractions. - **Standard forms of rational functions** (i) Integral of the types: - ∫(px+q)/((x+a)(x+b)) dx - ∫(px+q)/((x+a)²) dx - ∫(px² + qx + r)/((x+a)(x+b)(x+c)) dx - ∫(px² + qx + r)/((x÷a)(x+b)²) dx - ∫(px² + qx + r)/((x+a)(x+b)) dx - ∫(px² + qx + r)/((x+ a)³) dx - ∫(px² + qx + r)/((x+a)(x²+bx+c)) dx - **Method of Evaluation** 1. **Write the rational function in its partial fraction form.** 2. **Multiply both sides of the equation by the common denominator.** 3. **Solve for the constants by setting the values of the variable x such that some denominators simplify to zero.** 4. **Substitute the values of the constants back into the partial fraction form.** 5. **Integrate each term of the partial fraction.** ## EXAMPLE 1: Resolve (x³-6x²+10x-2)/(x²-5x+6) into partial fractions. ## EXAMPLE 2: Evaluate ∫ x/((x-1)²(x+2)) dx. ## Integration by Parts - Let u and v be two differentiable functions of a single variable x, then the integral of the product of these two functions = Ist function × Integral of the IInd function - Integral of [Derivatives of Ist function × Integral of the IInd function] - i.e. ∫uv dx = u∫v dx - ∫(du/dx) ∫v dx)dx - If in the product, two functions are of different types, then take that function as first function (i.e. u) which comes first in word ILATE, where - I : Inverse trigonometric function - L: Logarithmic function - A: Algebraic function - T: Trigonometric function - E: Exponential function ## Method to Find Integration by Parts - Let the given integration is of the form I = ∫u v dx, where u and v are the functions of x. Then, to evaluate such integrals, we use the following steps: - Firstly, choose the Ist and IInd functions with the help of ILATE, i.e. take that function as Ist function which comes Ist in ILATE and take other function as IInd function. - Now, integrate by using integration by parts, i.e. ∫u·v dx = u∫v dx - (d/dx(u))∫v dx)dx - From step II, we get one of the three possible cases: - If IInd integral is in simple form, then integrate it by using appropriate method. - If IInd integral is the product of two functions, then again use Steps I and II. - If IInd integral is same as the given integral, then put Ist in place of IInd integral. Finally, simplify it and get the required result. ## EXAMPLE 1: Evaluate ∫x cos x dx. ## EXAMPLE 2: Evaluate ∫e^x cos x dx. ## EXAMPLE 3: Evaluate ∫log|1+x²| dx. ## EXAMPLE 4: Evaluate ∫(log x)/(x+1)² dx. ## Some More Special Types of Integrals 1. ∫(x² - a²) dx = x√(x² - a²) / 2 - (a²/2) log|x + √(x² - a²)| + C 2. ∫√(x² + a²) dx = x√(x² + a²) / 2 + (a²/2) log|x + √(x² + a²)| + C 3. ∫√(a² - x²) dx = x√(a² - x²) / 2 + (a²/2) sin⁻¹(x/a) + C ## Method to Evaluate Integrals of the Form √(ax² + bx + c) dx - **Step 1:** Make the coefficient of x² unity. - **Step 2:** Complete the square in the expression under the square root by adding and subtracting (b/2a)² from the integrand inside the square root. - **Step 3:** Substitute x ± b/2a = t and reduce the integral obtained from step II into one of the form ∫√(t² k²)dt or ∫√(k² - t²)dt. Then, apply suitable formula to integrate. ## EXAMPLE 6: Evaluate ∫√(x² - 8x + 7dx. ## EXAMPLE 7: Evaluate ∫√(x² + 4x + 6) dx. ## EXAMPLE 8: Evaluate ∫√(1+3x -x²) dx. ## Integral of the Type ∫e^x[f(x) + f'(x)]dx. - In this type of integral, integrand is the product of two functions. One is in exponential form and the second function is the sum of two functions in which one is derivative of other function. - Then, to evaluate such integrals, we directly use the following formula: - ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C ## EXAMPLE 9: Evaluate ∫e^x(sin x + cos x) dx. ## EXAMPLE 10: Evaluate ∫(x - 3)e^x/(x-1)³ dx. ## Definite Integral - An integral of the form ∫_a^b_ f(x) dx is known as definite integral, where a and b are called the lower and upper limits of a definite integral. The value of the definite integral is given as if it has an anti-derivative F, then its value is the difference between the values of Fat the end points, i.e. F(b) - F(a). - Here, ∫_a^b_ f(x) dx is read as 'the integral of f(x) from a to b. - **Second Fundamental Theorem of Integral Calculus** - Let f be a continuous function defined on the closed interval [a, b] and F be an anti-derivative of f. - Then, ∫_a^b_ f(x)dx = [F(x)]_a^b_ = F(b) - F(a). ## EXAMPLE 1: Evaluate ∫_0^π/2_ cos² x dx. ## EXAMPLE 2: Evaluate ∫_1_² (4x³ - 5x² + 6x + 9) dx. ## EXAMPLE 3: Evaluate ∫_0^π/2_ √(sin ф cos ф) cos ф dф. ## EXAMPLE 4: Evaluate ∫_π/6_^(π/3) √(3sin x + cos x)/√(sin 2x) dx. ## Properties of Definite Integrals - (i) ∫_a^ b_ f(x) dx = ∫_a^b_ f(t) dt. - (ii) ∫_a^b_ f(x) dx = -∫_b^a_ f(x)dx. ## EXAMPLE 5: Evaluate ∫_1^2_ |x-1| dx. ## EXAMPLE 6: Evaluate ∫_2^5_ [|x-2| + |x-3| + |x-5|] dx. ## EXAMPLE 7: Evaluate ∫_0^π/4_ log(1 + tan x) dx. ## EXAMPLE 8: Evaluate ∫_0^π/2_ sin³ (x)/sin³ (x) + cos³ (x) dx. ## EXAMPLE 9: Evaluate ∫_0^π/4_ (sec x + tan x) / (sec x - tan x) dx. ## EXAMPLE 10: Evaluate the integral ∫_0^(2π)/5_ 1/(1 + esin x) dx. ## EXAMPLE 11: Evaluate ∫_0^π/2_ sin² x cos x dx. ## EXAMPLE 12: Evaluate the integral ∫_0^π/2_ 2sin² x dx. ## EXAMPLE 13: Evaluate ∫_0^1_ √x(1 - x) dx. ## EXAMPLE 14: Evaluate ∫_π/4_^π/2_ sin 2x tan⁻¹(sin x) dx. ## EXAMPLE 15: Evaluate ∫_0^1_ (x² + 1)/(x⁴ + 1) dx. ## EXAMPLE 16: Evaluate ∫_0^π/4_ cos³ (x)√(2 sin 2x) dx. ## EXAMPLE 17: Evaluate ∫_0^π/2_ (sin√x - cos√x) / (sin√x + cos√x) dx. ## EXAMPLE 18: Evaluate ∫_0^π/2_ (2log|sin x| - log|sin 2x|) dx. ## EXAMPLE 19: Evaluate ∫_0^π/4_ log(1 + tan x) dx. ## EXAMPLE 20: Evaluate ∫_0^1_ (x² +1)/(x⁴ + 1) dx. ## EXAMPLE 21: Evaluate ∫_0^1_ (x sin x + cos x)² dx. ## EXAMPLE 22: Evaluate ∫_0^π/4_ tan² x dx. ## EXAMPLE 23: Evaluate ∫_0^1_ 2x/(1 -x²)√(1 - x²) dx. ## Evaluation of Definite Integral by Substitution - Change the upper and lower limits corresponding to the new variable. - Integrate the new integral with respect to the new variable. - Find the difference of the values of the answer, obtained in step III, at new upper and lower limits. ## EXAMPLE 1: Evaluate ∫_0^1_ (x tan⁻¹ x)/(1 + x²) dx. ## EXAMPLE 2: Evaluate ∫_1^3_ 1/[x(1 + log x)] dx. ## EXAMPLE 3: Evaluate ∫_0^π/2_ sin⁵ ф cos³ ф dф. ## EXAMPLE 4: Evaluate ∫_0^2_ (4 + 3 sin x)/(4 + 3 cos x) dx. ## EXAMPLE 5: Evaluate ∫_0^1_ |x| dx. ## EXAMPLE 6: Evaluate ∫_0^1_ (x³ + |x| + 1) / (x² +2x + 1) dx. ## EXAMPLE 7: Evaluate ∫_π/4_^π/2_ (sin √x - cos √x) / (sin √x + cos √x) dx. ## EXAMPLE 8: Evaluate ∫_0^π/4_ (cos ax - sin bx)² dx. ## EXAMPLE 9: Evaluate ∫_0^π/2_ (2log|sin x| - log|sin 2x|) dx. ## EXAMPLE 10: Evaluate ∫_0^1_ (1 + x) / (x² + 1) dx. ## EXAMPLE 11: Evaluate ∫_0^π/4_ tan² x dx. ## EXAMPLE 12: Evaluate ∫_0^(π/2)_ cos² x dx. ## EXAMPLE 13: Evaluate ∫_0^1_ (1+2x + 3x²)/(2x²+1) dx. ## EXAMPLE 14: Evaluate ∫_0^1_ x /(x²+1) dx. ## EXAMPLE 15: Evaluate ∫_0^π/4_ (sin² (2x) + cos² (2x)) dx. ## EXAMPLE 16: Evaluate ∫_0^1_ √(1-x²) dx. ## EXAMPLE 17: Evaluate ∫_0^1_ (1+x)e⁻²ˣ dx. ## EXAMPLE 18: Evaluate ∫_0^π/4_ tan³ x sec² x dx. ## EXAMPLE 19: Evaluate ∫_0^1_ x√(1 - x) dx. ## EXAMPLE 20: Evaluate ∫_0^π/4_ sec³ x dx. ## EXAMPLE 21: Evaluate ∫_0^π/4_ (x-1)²(x+2)² dx. ## EXAMPLE 22: Evaluate ∫_0^π/4_ tan²(2x) dx. ## EXAMPLE 23: Evaluate ∫_0^π/4_ (sin⁴x + cos⁴x) dx. ## EXAMPLE 24: Evaluate ∫(1 + sin x)e^x dx. ## EXAMPLE 25: Evaluate ∫_√2/2_^1_ (1/√(1-x²)) dx. ## EXAMPLE 26: Evaluate ∫_0^π/4_ cos³ x √(2sin 2x) dx. ## EXAMPLE 27: Evaluate ∫_0^π/4_ (4 sin x + cos x) / (9 + 16 sin 2x) dx. ## EXAMPLE 28: Evaluate ∫_0^2_ (x - [x]) dx. ## EXAMPLE 29: Evaluate ∫_0^1_ x(1 -x) dx. ## EXAMPLE 30: Evaluate ∫_0^π_ x f(sin x) dx. ## EXAMPLE 31: Evaluate ∫_0^π/2_ x²/(1 + 5* ) dx. ## EXAMPLE 32: ∫_0^π/4_ (sin x + cos x)/(cos² x + 4sin² x) dx. ## EXAMPLE 33: Evaluate ∫_0^π_ (1 + √tan x) / (1 + √tan x) dx. ## EXAMPLE 34: Evaluate ∫_0^π/6_ x/(1 + √tan x) dx. ## EXAMPLE 35: Evaluate ∫_0^π/2_ log(4 + 3 sin x) / (4 + 3 cos x) dx. ## EXAMPLE 36: Evaluate ∫_0^π/4_ (4sinx + cos x) / (16 + 9 sin 2x) dx. ## EXAMPLE 37: Evaluate ∫_0^π/2_ tan⁻¹(1 - x + x²) dx. ## EXAMPLE 38: Evaluate ∫_0^∞_ x cos πx dx. ## EXAMPLE 39: Evaluate ∫_0^π/4_ (a + b cos x)² dx. ## EXAMPLE 40: Evaluate ∫_0^π/2_ sin⁴x cos³x dx ## EXAMPLE 41: Evaluate ∫_0^π/2_ sin⁴x cos³x dx. ## EXAMPLE 42: Evaluate ∫_0^1_ x√(1 - x²) dx. ## EXAMPLE 43: Evaluate ∫_0^1_ (1 + 2x + 3x²)/√(2x² + 1) dx. ## EXAMPLE 44: Evaluate ∫_0^π/4_ x(tan²x) dx. ## EXAMPLE 45: Evaluate ∫_0^2_ (2x+ 1) / (1 - x)(1+ x²) dx. ## EXAMPLE 46: Evaluate ∫_0^1_ (x + 1)²/(x + 2) dx. ## EXAMPLE 47: Evaluate ∫_0^1_ x² /(x⁴ + 1) dx. ## EXAMPLE 48: Evaluate ∫_0^∞_ x(log x)² dx. ## EXAMPLE 49 Evaluate ∫_0^π/4_ (log(1 + tan x))/(1 + tan² x) dx. ## EXAMPLE 50: Evaluate ∫_0^π/2_ (cos ax - sin bx)² dx. ## EXAMPLE 51: Evaluate ∫_0^1_ (2 log |sin x| - log|sin 2x|) dx. ## EXAMPLE 52: Evaluate ∫_0^1_ (1 + sin x)e^x dx. ## EXAMPLE 53: Evaluate ∫_0^π/2_ (x tan²x)/ (1 + x²) dx. ## EXAMPLE 54: Evaluate ∫_0^1_ x/(x² + 1) dx. ## EXAMPLE 55: Evaluate ∫_0^π/4_ x / (1 + x²) dx. ## EXAMPLE 56: Evaluate ∫_0^π/4_ (tan² x + 1) dx. ## EXAMPLE 47: Evaluate ∫_0^1_ (x² + x)/(x³ + 1) dx. ## EXAMPLE 58: Evaluate ∫_0^π/3_ (sin x + √3cos x)/5x dx. ## EXAMPLE 59: Evaluate ∫_0^1_ x²/(x² + 1) dx. ## EXAMPLE 60 Evaluate ∫_0^π/2_ x²/(2 + sin 2x) dx. ## EXAMPLE 61 Evaluate ∫_0^1_ (3x + 1)/(x + 1)²(x + 3) dx. ## EXAMPLE 62 Evaluate ∫_0^1_ (x + 1)²(x + 2) dx. ## EXAMPLE 63: Evaluate ∫_0^π/2_ x²/(x² + 12) dx. ## EXAMPLE 64: Evaluate ∫[sin(log x)+cos(logx)] dx. ## EXAMPLE 65: Evaluate ∫_1^e_ log x / (1 + log x)² dx. ## EXAMPLE 66: Evaluate ∫_1^e_ (sin (log x) / x) dx. ## EXAMPLE 67: Evaluate ∫_0^π/4_ (tan 0 + tan³ 0)/(1 + tan³ 0) d0. ## EXAMPLE 68: Evaluate ∫_0^(π/4)_ sin x/sin 4x dx. ## EXAMPLE 69: Evaluate ∫_0^(π/4)_ tan⁵ x sec² x dx. ## EXAMPLE 70: Evaluate ∫_0^1_ x sin⁻¹ x dx. ## EXAMPLE 71: Evaluate ∫_0^1_ x³ / √(1-x²) dx. ## EXAMPLE 72: Evaluate ∫_0^π/2_ x sin x cos²x dx. ## EXAMPLE 73: Evaluate ∫_0^π/4_ (x sin x) / (1 + cos² x) dx. ## EXAMPLE 74: Evaluate ∫_0^4_ (|x| + |x- 2| + |x -4|) dx. ## EXAMPLE 75: Evaluate ∫_0^π/2_ x sin x dx. ## EXAMPLE 76: Evaluate ∫_0^1_ (x² + 4x + 3) / (x² + 4)(x² + 25) dx. ## EXAMPLE 77: Evaluate ∫_0^π/2_ (x + sin x) / (1+cos x) dx. ## EXAMPLE 78: Evaluate ∫_1^e_ x(log x)² dx. ## EXAMPLE 79: Evaluate ∫_1^e_ (1/ (x(log x)²)) dx. ## EXAMPLE 80: Evaluate ∫_0^1_ (2/x²) / (x- x³)^1/3 dx. ## EXAMPLE 81: Evaluate ∫_0^π/2_ log (sin x) dx. ## EXAMPLE 82: Evaluate ∫_0^π/2_ (x² + x) / √(2x + 1) dx. ## EXAMPLE 83: Evaluate ∫_π/10_^(3π/10) x / (sin x + cos x) dx. ## EXAMPLE 84: Evaluate ∫_0^π/2_ (sin x)/ (sin x + cos x) dx. ## EXAMPLE 85: Evaluate ∫_0^π/2_ (cos x - sin x) dx. ## EXAMPLE 86: Let f(x) = x - [x], for every real number x, where [x] is the integral part of x, then evaluate ∫_0^2_ f(x) dx. ## EXAMPLE 87: Evaluate ∫_0^1_ x sin x dx. ## EXAMPLE 88: Evaluate f(xlogx)²dx ## EXAMPLE 89: Evaluate ∫_1^e_ (3x+5) / (x² + 1) dx ## EXAMPLE 90: Evaluate ∫_0^(π/4)_ log(cos x + sin x) dx. ## EXAMPLE 91: Evaluate ∫_0^1_ [(1+x)⁻²(1 + 2x)⁻¹ ] dx. ## EXAMPLE 92: Prove that ∫_0^π/4_ (1 + cosx sin x)/ sin x dx. = π/4. ## Case-Based Questions - For a function f(x), if f(-x) = f(x), then f(x) is an even function and if f(-x) = -f(x), then f(x) is an odd function. Again, we have: ∫_0^a _ f(x) dx = 2∫_o^a/2_ f(x)dx, if f(x) is even - ∫_0^a_ f(x) dx = 0, if f(x) is odd - **Question 93:** - Based on the above information, answer the following questions. - (i) f(x) = x² sin x is a - (a) even - (b) odd - (c) neither even nor odd - (d) None of the above - (ii) ∫_0^π_ f(x) dx is equal to - (a) π - (b) π/2 - (c) π/4 - (d) 0 - (iii) If g(x) = x sin x, then ∫_0^π_ sin x dx is - (a) π - (b) 2π - (c) 3π - (d) 4π - (iv) ∫_0^π/2_ sin xdx is equal to - (a) 0 - (b) 1 - (c) 2 - (d) 3 - (v) ∫_0^π_ sin x dx is equal to - (a) 0 - (b) 1

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