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 :**x^{4}/4 is an antiderivative of x^{3}because (x^{4}/4)’ = x^{3}.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 –**

- ∫x
^{n}dx = (x^{n+1}/(n+1))+C - ∫(1/x)dx = (log
_{e}|x|)+C - ∫e
^{x}dx = (e^{x})+C - ∫a
^{x}dx = ((e^{x})/(log_{e}a))+C - ∫sin(x)dx = -cos(x)+C
- ∫cos(x)dx = sin(x)+C
- ∫sec
^{2}(x)dx = tan(x)+C - ∫cosec
^{2}(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 ∫x^{4}dx.**Solution –**Using the formula, ∫x

^{n}dx = (x^{n+1}/(n+1))+C ∫x^{4}dx = (x^{4+1}/(4+1))+C = (x^{5}/(5))+C**Example 2.**Evaluate ∫2/(1+cos2x)dx.**Solution –**As we know that 1+cos2x = 2cos

^{2}x ∫2/(1+cos2x)dx = ∫(2/(2cos^{2}x))dx = ∫sec^{2}x = tan(x)+C**Example 3.**Evaluate ∫((x^{3}-x^{2}+x-1)/(x-1))dx.**Solution –**∫((x

^{3}-x^{2}+x-1)/(x-1))dx = ∫((x^{2}(x-1)+(x-1))/(x-1))dx = ∫(((x^{2}+1)(x-1))/(x-1))dx = ∫(x^{2}+1)dx = (x^{3}/3)+x+C Using, ∫x^{n}dx = (x^{n+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 ∫e^{2x-3}dx- Solution
Let 2x-3=t => dx=dt/2 ∫e

^{2x-3}dx = (∫e^{t}dx)/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

**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 ∫xsec^{2}xdx- Solution
Taking I= x and II = sec

^{2}x ∫xsin(3x)dx = x(∫sec^{2}xdx)-∫((x)'∫sec^{2}xdx)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-a_{1})(x-a_{2})(x-a_{3})….(x-a_{n}), then we assume thatf(x)/g(x) = (A

_{1}/(x-a_{1}))+(A_{2}/(x-a_{2}))+(A_{3}/(x-a_{3}))+....(A_{n}/(x-a_{n}))__Case 2.__

When g(x) = (x-a)^{k}(x-a_{1})(x-a_{2})(x-a_{3})

….(x-a_{r}),

then we assume thatf(x)/g(x) = (A

_{1}/(x-a)^{1})+(A_{2}/(x-a)^{2})+(A_{3}/(x-a)^{3}) +....(A_{k}/(x-a)^{k})+(B_{1}/(x-a_{1}))+(B_{2}/(x-a_{2}))+(B_{3}/(x-a_{3})) +....(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)∫(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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