Mathematics | Indefinite Integrals

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 = (log_e|x|)+C
3. ∫e^xdx = (e^x)+C
4. ∫a^xdx = ((e^x)/(log_ea))+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 ∫e^2x-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))-∫(1*tan(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-a_n), 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-a_r),
       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 
       

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