## Integral Formulas

**Integral Formulas – **Integration can be considered the reverse process of differentiation or called Inverse Differentiation. Integration is the process of finding a function with its derivative. Basic integration formulas on different functions are mentioned here. Apart from the basic integration formulas, classification of integral formulas and a few sample questions are also given here, which you can practice based on the integration formulas mentioned in this article. When we speak about integration by parts, it is about integrating the product of two functions, say y = uv. More integral calculus concepts are given, so keep learning integral formulas to solve problems accurately. Also, watch the video given below to clear your concept.

## List of Integral Formulas

The list of basic integral formulas are

- ∫ 1 dx = x + C
- ∫ a dx = ax+ C
- ∫ x
^{n }dx = ((x^{n+1})/(n+1))+C ; n≠1 - ∫ sin x dx = – cos x + C
- ∫ cos x dx = sin x + C
- ∫ sec
^{2}x dx = tan x + C - ∫ csc
^{2}x dx = -cot x + C - ∫ sec x (tan x) dx = sec x + C
- ∫ csc x ( cot x) dx = – csc x + C
- ∫ (1/x) dx = ln |x| + C
- ∫ e
^{x }dx = e^{x}+ C - ∫ a
^{x }dx = (a^{x}/ln a) + C ; a>0, a≠1

These integral formulas are equally important as differentiation formulas. Some other important integration formulas are:

## Classification of Integral Formulas

The above listed integral formulas are classified based on following functions.

- Rational functions
- Irrational functions
- Trigonometric functions
- Inverse trigonometric functions
- Hyperbolic functions
- Inverse hyperbolic functions
- Exponential functions
- Logarithmic functions
- Gaussian functions

As we have already gone through integral formulas for exponential functions, logarithmic functions, trigonometric functions and some basic functions. Let’s have a look at the additional integration formulas, i.e. the integral formulas for some special functions listed below:

## Integration Formulas

Integration formulas can be applied for the integration of algebraic expressions, trigonometric ratios, inverse trigonometric functions, logarithmic and exponential functions. The integration of functions results in the original functions for which the derivatives were obtained. These integration formulas are used to find the antiderivative of a function. If we differentiate a function f in an interval I, then we get a family of functions in I. If the values of functions are known in I, then we can determine the function f. This inverse process of differentiation is called integration. Let’s move further and learn about integration formulas used in the integration techniques.

## What Are Integration Formulas?

The integration formulas have been broadly presented as the following six sets of formulas. Basically, integration is a way of uniting the part to find a whole. The formulas include basic integration formulas, integration of trigonometric ratios, inverse trigonometric functions, the product of functions, and some advanced set of integration formulas. Integration is the inverse operation of differentiation. Thus the basic integration formula is ∫ f'(x).dx = f(x) + C

### Basic Integration Formulas

Using the fundamental theorems of integrals, there are generalized results obtained which are remembered as integration formulas in indefinite integration.

- ∫ x
^{n}.dx = x^{(n + 1)}/(n + 1)+ C - ∫ 1.dx = x + C
- ∫ e
^{x}.dx = e^{x}+ C - ∫1/x.dx = log|x| + C
- ∫ a
^{x}.dx = a^{x }/loga+ C - ∫ e
^{x}[f(x) + f'(x)].dx = e^{x}.f(x) + C

## Integration Formulas of Trigonometric functions

The process of finding the integral is integration. Here are a few important integration formulas remembered for instant and speedy calculations. When it comes to trigonometric functions, we simplify them and rewrite them as functions that are integrable. Here is a list of trigonometric and inverse trigonometric functions.

- ∫ cosx.dx = sinx + C
- ∫ sinx.dx = -cosx + C
- ∫ sec
^{2}x.dx = tanx + C - ∫ cosec
^{2}x.dx = -cotx + C - ∫ secx.tanx.dx = secx + C
- ∫ cosecx.cotx.dx = -cosecx + C
- ∫ tanx.dx =log|secx| + C
- ∫ cotx.dx = log|sinx| + C
- ∫ secx.dx = log|secx + tanx| + C
- ∫ cosecx.dx = log|cosecx – cotx| + C

### Integration Formulas of Inverse Trigonometric Functions:

- ∫1/√(1 – x
^{2}).dx = sin^{-1}x + C - ∫ /1(1 – x
^{2}).dx = -cos^{-1}x + C - ∫1/(1 + x
^{2}).dx = tan^{-1}x + C - ∫ 1/(1 +x
^{2}).dx = -cot^{-1}x + C - ∫ 1/x√(x
^{2}– 1).dx = sec^{-1}x + C - ∫ 1/x√(x
^{2}– 1).dx = -cosec^{-1 }x + C

## Advanced Integration Formulas

- ∫1/(x
^{2}– a^{2}).dx = 1/2a.log|(x – a)(x + a| + C - ∫ 1/(a
^{2}– x^{2}).dx =1/2a.log|(a + x)(a – x)| + C - ∫1/(x
^{2}+ a^{2}).dx = 1/a.tan^{-1}x/a + C - ∫1/√(x
^{2}– a^{2})dx = log|x +√(x^{2}– a^{2})| + C - ∫ √(x
^{2}– a^{2}).dx =1/2.x.√(x^{2}– a^{2})-a^{2}/2 log|x + √(x^{2}– a^{2})| + C - ∫1/√(a
^{2}– x^{2}).dx = sin^{-1 }x/a + C - ∫√(a
^{2}– x^{2}).dx = 1/2.x.√(a^{2}– x^{2}).dx + a^{2}/2.sin-1 x/a + C - ∫1/√(x
^{2}+ a^{2 }).dx = log|x + √(x^{2}+ a^{2})| + C - ∫ √(x
^{2}+ a^{2 }).dx =1/2.x.√(x^{2}+ a^{2 })+ a^{2}/2 . log|x + √(x^{2}+ a^{2 })| + C

## Different Integration Formulas

There are 3 types of integration methods and each method is applied with its own unique techniques involved in finding the integrals. They are the standardized results. They can be remembered as integration formulas.

**Integration by parts formula:**

When the given function is a product of two functions, we apply this integration by parts formula or partial integration and evaluate the integral. The integration formula while using partial integration is given as:

∫ f(x).g(x) = f(x).∫g(x).dx -∫(∫g(x).dx.f'(x)).dx + c

For example: ∫ xe^{x }dx is of the form ∫ f(x).g(x). Thus we apply the appropriate integration formula and evaluate the integral.

f(x) = x and g(x) = e^{x}

Thus ∫ xe^{x }dx = x∫e^{x }.dx – ∫( ∫e^{x }.dx. x). dx+ c

= xe^{x }– e^{x }+ c

**Integration by substitution formula:**

When a function is a function of another function, then we apply the integration formula for substitution. If I = ∫ f(x) dx, where x = g(t) so that dx/dt = g'(t), then we write dx = g'(t)

We can write I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt

For example: Consider ∫ (3x +2)^{4 }dx

We can use the integration formula of substitution here. Let u = (3x+2). ⇒ du = 3. dx

Thus ∫ (3x +2)^{4 }dx =1/3. ∫(u)^{4}. du

= 1/3. u^{5 }/5 = u^{5 }/15

= (3x+2)^{5 }/15

**Integration by partial fractions formula:**

If we need to find the integral of P(x)/Q(x) that is an improper fraction, wherein the degree of P(x) < that of Q(x), then we use integration by partial fractions. We split the fraction using partial fraction decomposition as P(x)/Q(x) = T(x) + P1 (x)/ Q(x), where T(x) is a polynomial in x and P1 (x)/ Q(x) is a proper rational function. If A B and C are the real numbers, then we have the following types of simpler partial fractions that are associated with various types of rational functions.

Form of Rational Fractions | Form of Partial Fractions |
---|---|

(px + q)/(x-a)(x – b) | A/(x – a) + B/ (x-b) |

(px + q)/(x-a)^{n} | A_{1}/(x-a) + A_{2}/(x-a)^{2} + ………. A_{n}/(x-a)^{n} |

(px^{2} + qx + r)/(ax^{2} + bx + c) | (Ax + B)/(ax^{2} + bx + c) |

(px^{2} + qx + r)/(ax^{2} + bx + c)^{n} | (A_{1}x + B_{1})/(ax^{2} + bx + c) + (A_{2}x + B_{2})/(ax^{2} + bx + c)^{2} + …(A_{n}x + B_{n})/(ax^{2} + bx + c)^{n} |

(px^{2} + qx + r)/(x-a)(x-b)(x-c) | A/(x – a) + B/ (x-b) + C/ (x-c) |

(px^{2} + qx + r)/(x^{2 }+bx +c) | A/(x-a) +(Bx+C)/(x^{2 }+bx +c) |

For example: ∫ 3x+7/ x^{2 }-3x + 2

Resolving it into partial fractions, we get

3x+7/ x^{2 }-3x + 2 = A/(x-2) + B/ (x-1)

= A(x-1) + B(x-2)/ (x-2)(x-1)

Equating the numerators, we get 3x +7 = A(x-1)+B(x-2)

Find B by giving x = 1⇒ 10 = B

Find A by giving x = 2⇒ 13 = A

Thus 3x+7/ x^{2 }-3x + 2 = 13/(x-2) + 10(x-1)

Applying the integration formulas, we get

∫ (3x+7/ x^{2 }-3x + 2) = ∫ 13/(x-2) + ∫ 10(x-1)

∫ (3x+7/ x^{2 }-3x + 2) = 13 log |x-2| – 10 log |x-1| + C

## Application of Integration Formulas

In general, there are two types of integrals. They are definite and indefinite integrals.

### Definite Integration Formula

These are the integrations that have a pre-existing value of limits; thus making the final value of integral definite.

We apply the integration formulas discussed so far, in approximating the area bounded by the curves, in evaluating the average distance, velocity and acceleration oriented problems, in finding the average value of a function, to approximate the volume and the surface area of the solids, in finding the center of mass and work, in estimating the arc length, in finding the kinetic energy of a moving object using improper integrals.

Let us compute the distance traveled by an object using integration formulas. We know that distance is the definite integral of velocity.

Given: the velocity of an object = v(t)= -t^{2 }+ 5t. Let us find the displacement and the total distance traveled on (1,3).

The initial and the final positions of the object are 1 and 3 respectively.

= -t^{3}/3 + 5t^{2 }/2

= [(-27/3-1/3)+(45/2 – 5/2)]

= -28/3 +40/3 =12/3

Displacement =4

The distance between two points is the absolute value of its displacement. The total distance taken between positions 1 and 3 and in the reverse is 4 + 4 = 8

Thus the displacement of the object is 4 and the distance traveled by the object is 4 + 4 = 8

Let us see how to use the indefinite integration formulas in the following solved examples.

## Integration Formulas Examples

**Example 1:Find the value of **∫** (9x+ 25)/ (x+3) ^{2 }.dxSolution:**

(9x+ 25)/ (x+3)^{2 }is a rational function.

Using the partial fraction decomposition, we have (9x+ 25)/(x+3)^{2 }= A/(x+3) + B/(x+3)^{2 }

Taking LCD, we get

(9x+ 25)/(x+3)^{2 }= [A(x+3) +B]/(x+3)^{2 }

Equating the numerator, we get

9x+ 25 = A(x+3) +B

Solving for B when x = -3, we get B = -2

Solving for A when x = 0, we get A = 9

Thus the partial fraction is decomposed as 9/(x+3) -2/(x+3)^{2}

As stated in the integration formulas above, find the integral of 9/(x+3) -2/(x+3)^{2 }.

∫[9/(x+3)]dx – ∫ -2(x+3)^{2 }.dx = 9 ln(x+3) – 2 /(x+3) +C

**Answer: **∫**(9x+ 25)/ (x+3) ^{2 }.dx = 9 ln(x+3) – 2 /(x+3) +C**

**Example 2: Simplify and find the value of ∫ Cosecx(Cosecx – Cotx).dx using the integration formula.****Solution: **

∫Cosecx(Cosecx – Cotx).dx

= ∫Cosec^{2}x.dx – ∫ Cotx.Cosecx.dx

Using the basic trigonometric integration formulas,

= -Cotx – (-Cosecx)

= -Cotx + Cosecx

= Cosecx – Cotx + C

**Answer: **∫**Cosecx(Cosecx – Cotx).dx = Cosecx – Cotx + C **

### How Do You Integrate Using Integration Formulas?

We can use the below steps to integrate:

- Firstly define a small part of an object in certain dimensions which on adding repetitively makes the whole object.
- Use integration formulae over that small part along the varying dimensions.

### What Is the Use of Integration Formulas?

The integration is used to find the area of any objects. Real-life examples are to find the center of mass of an object, center of gravity, and mass moment of Inertia for a sports utility vehicle. It is also used for calculating the velocity and trajectory of an object, predict the alignment of planets, and in electromagnetism. Use integration formulas in all these cases.

### What Are Integration Techniques Involving Integration Formulas?

Substitution, integration by parts, reverse chain rule, and partial fraction expansion are a few integration techniques.

### What is The Integration Formula of Integral UV?

The formula for integral UV is used to integrate the product of two functions. The integration formula of UV form is given as

u.v.du = u.dv- v.du.

### What Are The Integration Formulas For Trigonometric Functions?

The trigonometric functions are simplified into integrable functions and then their integrals are evaluated. The basic integration formulas for trigonometric functions are as follows.

- ∫ cosx.dx = sinx + C
- ∫ sinx.dx = -cosx + C
- ∫ sec
^{2}x.dx = tanx + C - ∫ cosec
^{2}x.dx = -cotx + C - ∫ secx.tanx.dx = secx + C
- ∫ cosecx.cotx.dx = -cosecx + C
- ∫ tanx.dx =log|secx| + C
- ∫ cotx.dx = log|sinx| + C
- ∫ secx.dx = log|secx + tanx| + C
- ∫ cosecx.dx = log|cosecx – cotx| + C

### How Do You apply Integration formulas to Find The Integrate LN?

∫ ln(x) dx is of the UV form. Apply integration formula by parts rule using the following steps to integrate LN.

1) Identify uv: Take u= ln(x) and dv = 1 . dx ⇒ du = 1/x and v = x

2) Apply formula: ∫ uv dx = uv -∫ vdu

= x. ln(x) – ∫ x. 1/x

= x ln(x) – x + C

3) Simplify and evaluate the integral.

### How do You Use Integration Formulas?

We can use the below steps to integrate:

1) Firstly define a small part of an object in certain dimensions which on adding repetitively makes the whole object.

2) Use integration formulas over that small part along the varying dimensions.

### What is The Use of Integration Formulas?

The indefinite integration formulas are used to find the area, volume, or displacement of any objects. Real-life examples are to find the center of mass of an object, the volume of a cylinder, the area under the curve or between the curves, and so on.

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