INTEGRATION

Suppose your friend gives you a wooden stick. He asks you to break it. Can you do so? Yes, it will be very easy for you to do so. But what will happen if he gives you five to six sticks to break? It will not be that easy to break it. As the number of sticks increases it is difficult to break them. The process of uniting things is an integration of things. Similarly, in mathematics too, we have an integration of two functions. Integration is like drop by drop addition of water in a container. Let us get ourselves familiar with the concepts of integrations.

In differentiation, we studied that if a function f is differentiable in an interval say, I, then we get a set of a family of values of the functions in that interval. Is there any way by which we can get to know about the function if the values of the function within an interval are known?

This process is the reverse of finding a derivative. Integrations are the anti-derivatives. Integrations are the way of adding the parts to find the whole. Integration is the whole pizza and the slices are the differentiable functions which can be integrated

In Calculus, the two important processes are differentiation and integration. We know that differentiation is the process of finding the derivative of a function, whereas integration is the inverse process of differentiation. Here, we are going to discuss the important component of integration called “integrals” here. The integrals are generally classified into two types, namely:

• Definite Integral
• Indefinite Integral

INDEFENITE INTEGRAL

So Differentiation and integration are the reverse processes of each other. Both differentiation and integration are operations which are performed on functions.

IMPORTANT FORMULAS

PROPERTIES OF INDEFENITE INTEGRALS

METHODS OF INTEGRATION

TYPE -1�NORMAL INTEGRATION

TYPE -2� INTEGRATION BY SUBSTITUTION

Hint Put f(x) = u and proceed to obtain the proof.

Different types of Problems

Integrate

Integrate

TYPE -3� INTEGRATION USING TRIGONOMETRIC�SUBSTITUTION

TRIGONOMETRIC IDENTITIES

9.2SinxCosy =Sin(x+y)+Sin(x-y)

10.2CosxSiny = Sin(x+y)-Sin(x-y)

11.2CosxCosy= Cos(x+y)+Cos(x-y)

12.2SinxSiny = Cos(x-y)-Cos(x+y)

TYPE 4�INTEGRATIION USING COMPLETING THE SQUARE

TYPE 5�INTEGRATION OF RATIONAL FUNCTION

Solution:

=

=

= 2 √ (

= 2 √

Problem: Evaluate

TYPE 6�INTEGRATIION OF RATIONAL FUNCTION BY PARTIAL FRACTION

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METHOD OF RESOLVING A GIVEN RATIONAL FUNCTION INTO PARTIAL FRACTION

This method is applicable when the degree of the numerator is less than the degree of the denominator

axⁿ +bx+c can be expressed as the product of linear factors

axⁿ +bx+c

=(x + A)(x + B)(x + C)....... (x + k), then

Compairing co-efficients of x,x² ……,we are getting the values A_1,A₂……A_K

This method is applicable when the degree of the numerator is less than the degree of the denominator

= (x + A)

.

The denominator can be expressed as the product of linear factors in which one or some are repeating.

Compairing co-efficients of constant term, x,x³ ……,we are getting the values A_1,A_2,A₃

This method is applicable when the degree of the numerator is less than the degree of the denominator

• The denominator

cannot be expressed as the product of

distinct linear factors

Then we write

,

.

This method is applicable when the degree of the numerator is greater than the degree of the denominator

Divide the numerator with denominator till the degree of the numerator become less than the degree of denominator and then apply any one of the above methods depending on the type of denominator.

Compairing co-efficients of constant term, x,x³ ……,we are getting the values A_1,A_2,A₃

TYPE 7�INTEGRATION BY PARTS

INTEGRATION BY PARTS

If u and v are any two differentiable variable of a single variable x (say),then, by the product rule of differentiation, we have

+ v

Integrating both sides, we get

uv =

or

...(1)

u = f(x) and

Let

Therefore, expression (1) can be rewritten as

i.e.,

The integral of the product of two functions = (first function) x (integral of the second function)– Integral of [(differential coefficient of the first function) x (integral of the second function)]

Evaluvate

Solution: Put f (x) = x (first function) and g (x) = cos x (second function).

Then integration by parts gives

= x sin x –

Evaluvate

Solution: u = x, dv = sin 3x dx

=

Evaluvate

Solution: let u =

=

Evaluvate :

Solution: Take u = log x, and v = x => dv = 1 dx

.1 dx

= x log x – x + C

= x (log x – 1) + C

Integral of the type :

= f(x)

[Take u = f(x), dv =

Proof :

]

+C

Evaluvate

Solution:

=

=

Integrals of some more types

​

• Let I =

Solution: Taking constant function 1 as the second function

and integrating by parts, we have

I = x

= x

= x

= x

= x

= x

= x

Or 2I = x

I=

Similarly, integrating other two integrals by parts, taking constant function 1 as the second functions, we get

(2)

Hint; Put x = a

(3)

Hint Put x = a

Evaluvate:

Solution: put tan

(dx=

Cos x =

TYPE 11

HOTS INTEGRATION PROBLEMS

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