Mathematics | Indefinite Integrals (original) (raw)

Last Updated : 13 May, 2020

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 -

1. ∫xndx = (xn+1/(n+1))+C
2. ∫(1/x)dx = (loge|x|)+C
3. ∫exdx = (ex)+C
4. ∫axdx = ((ex)/(logea))+C
5. ∫sin(x)dx = -cos(x)+C
6. ∫cos(x)dx = sin(x)+C
7. ∫sec2(x)dx = tan(x)+C
8. ∫cosec2(x)dx = -cot(x)+C
9. ∫sec(x)tan(x)dx = sec(x)+C
10. ∫cosec(x)cot(x)dx = -cosec(x)+C
11. ∫cot(x)dx = log|sin(x)|+C
12. ∫tan(x)dx = log|sec(x)|+C
13. ∫sec(x)dx = log|sec(x)+tan(x)|+C
14. ∫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 -

1. 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
2. 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))-∫(1tan(x))dx
     = xtan(x)+log|cos(x)|+C
3. 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