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Mathematics | Indefinite Integrals

• Difficulty Level : Easy
• 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))-∫(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-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

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