Indefinite Integrals

Antiderivative –

· Defination :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 –

· Defination :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

· ∫e^2x-3dx = (∫e^tdx)/2

·         = (∫e^t)/2

·         = ((e^2x-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.

· I – stands for algebraic functions.

· I – stands for trigonometric functions.

· I – 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))+....(A_n/(x-a_n))

· 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)

·                   +....(A_k/(x-a)^k)+(B₁/(x-a₁))+(B₂/(x-a₂))+(B₃/(x-a₃))

                  +....(B_r/(x-a_r))

· 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)∫(x+1)dx)+((1/3)∫(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