Mathematics | Indefinite Integrals Last Updated : 13 May, 2020 Improve Improve Like Article Like Save Share Report Antiderivative – Definition :A function ∅(x) is called the antiderivative (or an integral) of a function f(x) of ∅(x)’ = f(x). Example : x4/4 is an antiderivative of x3 because (x4/4)’ = x3. In general, if ∅(x) is antiderivative of a function f(x) and C is a constant.Then, {∅(x)+C}' = ∅(x) = f(x). Indefinite Integrals – Definition :Let f(x) be a function. Then the family of all ist antiderivatives is called the indefinite integral of a function f(x) and it is denoted by ∫f(x)dx. The symbol ∫f(x)dx is read as the indefinite integral of f(x) with respect to x. Thus ∫f(x)dx= ∅(x) + C. Thus, the process of finding the indefinite integral of a function is called integration of the function. Fundamental Integration Formulas – ∫xndx = (xn+1/(n+1))+C ∫(1/x)dx = (loge|x|)+C ∫exdx = (ex)+C ∫axdx = ((ex)/(logea))+C ∫sin(x)dx = -cos(x)+C ∫cos(x)dx = sin(x)+C ∫sec2(x)dx = tan(x)+C ∫cosec2(x)dx = -cot(x)+C ∫sec(x)tan(x)dx = sec(x)+C ∫cosec(x)cot(x)dx = -cosec(x)+C ∫cot(x)dx = log|sin(x)|+C ∫tan(x)dx = log|sec(x)|+C ∫sec(x)dx = log|sec(x)+tan(x)|+C ∫cosec(x)dx = log|cosec(x)-cot(x)|+C Examples – Example 1.Evaluate ∫x4dx. Solution – Using the formula, ∫xndx = (xn+1/(n+1))+C ∫x4dx = (x4+1/(4+1))+C = (x5/(5))+C Example 2.Evaluate ∫2/(1+cos2x)dx. Solution – As we know that 1+cos2x = 2cos2x ∫2/(1+cos2x)dx = ∫(2/(2cos2x))dx = ∫sec2x = tan(x)+C Example 3.Evaluate ∫((x3-x2+x-1)/(x-1))dx. Solution – ∫((x3-x2+x-1)/(x-1))dx = ∫((x2(x-1)+(x-1))/(x-1))dx = ∫(((x2+1)(x-1))/(x-1))dx = ∫(x2+1)dx = (x3/3)+x+C Using, ∫xndx = (xn+1/(n+1))+C Methods of Integration – Integration by Substitution : Definition –The method of evaluating the integral by reducing it to standard form by proper substitution is called integration by substitution. If f(x) is a continuously differentiable function, then to evaluate the integral of the form ∫g(f(x))f(x)dx we substitute f(x)=t and f(x)’dx=dt. This reduces the integral to the form ∫g(t)dt Examples : Example 1.Evaluate the ∫e2x-3dx Solution Let 2x-3=t => dx=dt/2 ∫e2x-3dx = (∫etdx)/2 = (∫et)/2 = ((e2x-3)/2)+C Example 2.Evaluate the ∫sin(ax+b)cos(ax+b)dx Solution Let ax+b=t => dx=dt/a; ∫sin(ax+b)cos(ax+b)dx = (∫sin(t)cos(t)dt)/a = (∫sin(2t)dt)/2a = -(cos(2t))/4a = (-cos(2ax+2b)/4a)+C Integration by Parts : Theorem :If u and v are two functions of x, then ∫(uv)dx = u(∫vdx)-∫(u'∫vdx)dx where u is a first function of x and v is the second function of x Choosing first function : We can choose first function as the function which comes first in the word ILATE where I – stands for inverse trigonometric functions. L – stands for logarithmic functions. A – stands for algebraic functions. T – stands for trigonometric functions. E – stands for exponential functions. Examples : Example 1.Evaluate the ∫xsin(3x)dx Solution Taking I= x and II = sin(3x) ∫xsin(3x)dx = x(∫sin(3x)dx)-∫((x)'∫sin(3x)dx)dx = x(cos(3x)/(-3))-∫(cos(3x)/(-3))dx = (xcos(3x)/(-3))+(cos(3x)/9)+C Example 2.Evaluate the ∫xsec2xdx Solution Taking I= x and II = sec2x ∫xsin(3x)dx = x(∫sec2xdx)-∫((x)'∫sec2xdx)dx = (xtan(x))-∫(1*tan(x))dx = xtan(x)+log|cos(x)|+C Integration by Partial Fractions : Partial Fractions : If f(x) and g(x) are two polynomial functions, then f(x)/g(x) defines a rational function of x. If degree of f(x) < degree of g(x), then f(x)/g(x) is a proper rational function of x. If degree of f(x) > degree of g(x), then f(x)/g(x) is an improper rational function of x. If f(x)/g(x) is an improper rational function, we divide f(x) by g(x) so that the rational function can be represented as ∅(x) + (h(x)/g(x)).Now h(x)/g(x) is an proper rational function. Any proper rational function can be expressed as the sum of rational functions, each having a simple factor of g(x).Each such fraction is called partial fraction . Cases in Partial Fractions : Case 1. When g(x) = (x-a1)(x-a2)(x-a3)….(x-an), then we assume that f(x)/g(x) = (A1/(x-a1))+(A2/(x-a2))+(A3/(x-a3))+....(An/(x-an)) Case 2. When g(x) = (x-a)k(x-a1)(x-a2)(x-a3) ….(x-ar), then we assume that f(x)/g(x) = (A1/(x-a)1)+(A2/(x-a)2)+(A3/(x-a)3) +....(Ak/(x-a)k)+(B1/(x-a1))+(B2/(x-a2))+(B3/(x-a3)) +....(Br/(x-ar)) Examples : Example 1.∫(x-1)/((x+1)(x-2))dx Solution Let (x-1)/((x+1)(x-2))= (A/(x+1))+(B/(x-2)) => x-1 = A(x-2)+B(x+1) Putting x-2 = 0, we get B = 1/3 Putting x+1 = 0, we get A = 2/3 Substituting the values of A and B, we get (x-1)/((x+1)(x-2))= ((2/3)/(x+1))+((1/3)/(x-2)) ∫((2/3)/(x+1))+((1/3)/(x-2))dx = ((2/3)∫(1/(x+1))dx)+((1/3)∫(1/(x-2))dx) = ((2/3)log|x+1|)+((1/3)log|x-2|)+C Example 2.∫(cos(x))/((2+sin(x))(3+4sin(x)))dx Solution Let I = ∫(cos(x))/((2+sin(x))(3+4sin(x)))dx Putting sin(x) = t and cos(x)dx = dt, we get I = ∫dt/((2+t)(3+4t)) Let 1/((2+t)(3+4t))= (A/(2+t))+(B/(3+4t)) => 1 = A(3+4t)+B(2+t) Putting 3+4t = 0, we get B = 4/5 Putting 2+t = 0, we get A = -1/5 Substituting the values of A and B, we get 1/((2+t)(3+4t)) = ((-1/5)/(2+t))+((4/5)/(3+4t)) I = (∫((-1/5)/(2+t))dt)+(∫((4/5)/(3+4t))dt) = ((-1/5)log|2+t|)+((1/5)log|3+4t|)+C = ((-1/5)log|2+sin(x)|)+((1/5)log|3+4sin(x)|)+C Like Article Suggest improvement Previous Definite Integral Next Application of Derivative - Maxima and Minima | Mathematics Share your thoughts in the comments Add Your Comment Please Login to comment...