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### Indefinite Integrals

If you already know how to find the derivative of a function, understanding indefinite integrals becomes easier. An indefinite integral is defined as the parent function of a derivative. For example, if you have the function
f(x) = 3x^2 + 4x + 5
its derivative is
f'(x) = 6x + 4
Then we say that f(x) is the integral of f'(x). Hence, integration is opposite of differentiation.

Now suppose we have the function
f(x) = 3x^2 + 4x + 10
its derivative is
f'(x) = 6x + 4
Notice that the derivative of f(x) = 3x^2 + 4x + 10 is same as that of the derivative of f(x) = 3x^2 + 4x + 5. In other words, the integral of f'(x) = 6x + 4 can be either f(x) = 3x^2 + 4x + 10 or f(x) = 3x^2 + 4x + 5. This implies that one function can have more than one integrals.

In fact, one function can have an infinite number of integrals. Each of these integrals differs from the others only by a constant. As seen above, the two functions f(x) = 3x^2 + 4x + 5 and f(x) = 3x^2 + 4x + 10 differ only by a constant; the former has 5 and the latter has 10 instead.

The above fact is easily understood by the help of differentiation. We know that the derivative of a constant is zero. So in both the functions above, which differ only by a constant, the derivative of the constant does not matter since it is zero. Hence both have the same derivative. In opposite sense, both the functions are integrals of the same function, their derivative.

This is an important property of indefinite integrals. There can be an infinite number of indefinite integrals of a given function. Hence they are given the name 'indefinite'. This is also the reason that the proper way to represent an indefinite integral is to always put a variable '+ C' after the function in place of any constants present in the function. For example, as from the above examples, the integral of the function
f'(x) = 6x + 4
is the function
f(x) = 3x^2 + 4x + C
Notice that we put a '+ C' at the end instead of a + 5 or a + 10. This is the correct representation of an indefinite integral; it always has a trailing + C in it in place of any constants. This is a rule, which, if not followed, can get you incorrect answers.

The + C we put in the above integrals is called the constant of integration.

Now we should become familiar with some terms in integration:

• Indefinite Integrals are also called antiderivatives because they are the result of a mathematical operation which is opposite that of finding the derivative of a function
• Indefinite Integrals are also called primitive functions because in any process of differentiation, they are the original functions whose derivative is obtained.
• The process of finding the integral of a function is called integration
• Integrand  is the function being integrated. It is the expression between the \int and dx sign.

#### Representation of Integration:

The symbol for integration is \int. It is put before the function whose integral you have to find. Note that in integration, you need to put another symbol dx at the end of the integral which denotes the termination of the function being integrated. For example, integration of x^2 + 4x + 1 is represented as
\int x^2 + 4x + 1 dx = 2x + 4 + c
Again note the three remarkable things about the above equation:

1. It starts with the integration symbol \int
2. There is a dx denoting the end of the function which you are integrating
3. There is a trailing '+ c' at the end of the evaluated indefinite integral
Also note that the function between the \int and dx sign is the integrand. These three rules are essential in writing any integration statement.

#### A Note on dx

The symbol dx is always written after the integrand. It represents a sort of a full stop. When a sentence is over you put a full stop after it as a punctuation mark. Similarly in integration you put a dx to represent that the integrand is only till here.

More importantly, this is not the only use of dx. It is mathematically very significant. The variable which comes after d, which, here, is x, is called the variable of integration. It is the variable with respect to which the integral is being evaluated. For example, the following integration statement:
\int x^2 + 2x dx = 2x + 2 + c
is read as "Integral of x^2 + 2x with respect to x is equal to ...". This is quite important. If suppose there is 'y' in place of 'x' in it, that is,
\int x^2 + 2x dy
the above expression is a totally different from the previous one in which we had dx. Since here have a dy, it means the integration is being done with respect to 'y', and any other variable, like 'x', will be treated like a constant. That is, in the process of integrating the above expression, we will treat 'x' as a number not a variable because the expression is being integrated with respect to 'y'. Hence the integral will beome:
\int x^2 + 2x dy = (x^2 + 2x)y + c
Thus, the dx has a very important significance is specifying the variable of integration.

#### Why 'Indefinite Integral'? Why not just 'Integral'?

There are two types of integrals, definite and indefinite. Before learning definite integrals you should be acquainted with the indefinite ones. (definite integrals are nothing but a numerical extension of indefinite integrals). In this introductory post we are clearly specifying 'indefinite integral'. However, saying just 'integral' generally means the indefinite integral.

### Rules and formulas to do integration

Since the process of integration is exactly opposite to that of differentiation, most of the formulas for derivatives, if reversed, can give you formulas for integration. Some of them are:

#### Integrating simple powers of x

The first one is opposite to the power rule in differentiation. The power rule of differentiation says that d/(dx) x^n = nx^(n-1). Its opposite in integration is:
\int x^n dx = x^(n+1)/(n+1) + C
(In differentiation you take the power, multiply it with x and then decrease the power by 1; In integration, you take the power, increase it by 1, then divide x by it as well.)

Exception: This rule can be used for all values of n except -1. If you want to integrate x^-1, you can't apply this rule. The rule discussed after this one will help you integrate that.

Notice the + C at the end of the integral. It denotes a constant and is called the constant of integration. Any indefinite integral must always be written with a trailing + C otherwise it is considered incomplete.

Example usages:
1. Whole number powers:
1. Simple without coefficients:
1. \int x^2 dx = x^(2 + 1)/(2 + 1) = x^3/3 + C
2. \int x^3 dx = x^(3 + 1)/(3 + 1) = x^4/4 + C
2. With coefficients:
1. \int 3x^2 dx = 3x^(2 + 1)/(2 + 1) = 3x^3/3 = x^3 + C
2. \int x^4/5 dx = x^(4 + 1)/(5*(4 + 1)) = x^5/25 + C
3. \int -5x^2 dx = -5x^(2 + 1)/(2 + 1) = -5x^3/3 = -5x^3/3 + C
2. Negative whole number powers:
1. Simple without coefficients:
1. \int x^-2 dx = x^(-2 + 1)/(-2 + 1) = x^-1/(-1) = -1/x + C
2. \int x^-3 dx = x^(-3 + 1)/(-3 + 1) = x^-2/(-2) = -1/(2x^2) + C
2. With coefficients:
1. \int 3x^-2 dx = 3x^(-2 + 1)/(-2 + 1) = 3x^-1/(-1) = -3/x + C
2. \int -4x^-3 dx = -4x^(-3 + 1)/(-3 + 1) = -4x^-2/(-2) = 2/(x^2) + C
3.  \int x^-4/5 dx = x^(-4 + 1)/(5*(-4 + 1)) = -x^-3/15 + C
4. \int -(1/5)x^-4 dx = (-1/5)x^(-4 + 1)/(-4 + 1) = (1/15)x^3 + C
3. Fractional/real number powers:
1. Simple without coefficients:
1. \int x^(1/2) dx = x^(1/2 + 1)/(1/2 + 1) = x^(3/2)/(3/2) = 2/3 x^(3/2) + C
2. \int x^(4/3) dx = x^(4/3 + 1)/(4/3 + 1) = x^(7/3)/(7/3) = 3/7 x^(7/3) + C
2. With coefficients:
1. \int 2x^(1/2) dx = 2x^(1/2 + 1)/(1/2 + 1) = 2x^(3/2)/(3/2) = 4/3 x^(3/2) + C
2. \int -1/2 x^(4/3) dx = -1/2 x^(4/3 + 1)/(4/3 + 1) = -1/2 x^(7/3)/(7/3) = -3/14 x^(7/3) + C

#### Integrating x^-1

The rule for integrating x^-1 is:
\int x^-1 dx = log|x|
Notice that there are absolute value bars in the logarithm. Further, the logarithm denoted has a base of 'e', that is, it is the natural logarithm. All integration rules are stated in terms of natural logarithms.

Example usage:
\int x^-1 dx = log|x| + C

#### Integrals of Exponential functions

As the derivative e^x is e^x, so the integral of e^x is e^x. Thus,
\int e^x dx = e^x + C
where e is a constant.

Further, as the derivative of a^x is a^x log(a), so we can say that the integral of a^x log(a) is a^x. That is, \int a^x log(a) dx = a^x/log(a). Since we want to get the integral of a^x, divide by log(a) on both sides of this equation to get \int a^x dx = a^x/log(a) + C. Thus the integration formula for the exponential function a^x, where a can be any number greater than zero, is
\int a^x dx = a^x/log(a) + C
Example usages:
1. Exponential function with base 'e'
1. \int e^x dx = e^x + C
2. \int 5e^x dx = 5e^x + C
3. \int -3e^x dx = -3e^x + C
4. \int 1/3 e^x dx = 1/3 e^x + C
2. Exponential function with base 'a'
1. \int a^x dx = a^x/log(a)
2. \int 2^x dx = 2^x/log(2)
3. \int (1/2)^x dx = (1/2)^x/log(1/2)
Note that the exponential function with any numbers as its base can only have positive numbers as the base.

#### Integrals of Trigonometric functions

Have a look at the differentiation formulas for trigonometric functions. If you take the opposite of each formula you get the integration formulas for trigonometric functions. For example, d/(dx) sin(x) = cos(x) means that \int cos(x) dx = sin(x) + C. Similarly for other trigonometric functions:

 Derivative formulas for trigonometric functions Integration formulas for trigonometric functions d/(dx) sin(x) = cos(x) \int cos(x) dx = sin(x) + C d/(dx) cos(x) = -sin(x) \int sin(x) dx = -cos(x) + C d/(dx) tan(x) = sec^2(x) \int sec^2(x) dx = tan(x) + C d/(dx) cot(x) = -cosec^2(x) \int cosec^2(x) dx = -cot(x) + C d/(dx) sec(x) = sec(x) tan(x) \int sec(x) tan(x)(x) dx = sec(x) + C d/(dx) cosec(x) = -cosec(x) cot(x) \int cosec(x) cot(x) dx = -cosec(x) + C
Example problems are discussed in a separate post.

#### Integration formulas from derivatives of inverse trigonometric functions

Let us take an example to understand this section. The derivative of sin^-1(x) is 1/sqrt(1 - x^2). Thus, the integral of 1/sqrt(1 - x^2) is sin^-1(x) + C. Similarly some integration formulas are derived from differentiation formulas of inverse trigonometric functions. These are as follows:

 Derivative formulas for inverse trigonometric functions Integration formulas derived from them d/(dx) sin^-1(x) = 1/sqrt(1 - x^2) \int 1/sqrt(1 - x^2) dx = sin^-1(x) + C d/(dx) cos^-1(x) = -1/sqrt(1 - x^2) \int -1/sqrt(1 - x^2) dx = cos^-1(x) + C d/(dx) tan^-1(x) = 1/(1 + x^2) \int 1/(1 + x^2) dx = tan^-1(x) + C d/(dx) cot^-1(x) = -1/(1 + x^2) \int -1/(1 + x^2) dx = cot^-1(x) + C d/(dx) sec^-1(x) = 1/(x*sqrt(x^2 - 1)) \int 1/(x*sqrt(x^2 - 1)) dx = sec^-1(x) + C d/(dx) cosec^-1(x) = -1/(x*sqrt(x^2 - 1)) \int -1/(x*sqrt(x^2 - 1)) dx = cosec^-1(x) + C

#### Integration formulas for tan(x), cot(x), sec(x) and cosec(x)

If you look at all the differentiation formulas for trigonometric functions, there is no one formula in which any of these functions is the derivative itself. For example, the derivative of tan(x) is sec^2(x) but there is no trigonometric function whose derivative is tan(x). Thus we are unable to simply get the formula for integrating these functions. Their formulas are derived by means of the use of substitution, an advanced technique on integration. Thus these formulas are discussed in a separate post after substitution.

#### Advanced techniques of integration

Integration involves the use of advanced algebraic techniques in addition to the above formulas, unlike differentiation. These techniques are:
1. Substitution
2. The use of Partial Fractions
3. Integration by Parts