# Integrals:Class 12 Maths NCERT Chapter 7

**Key Features of NCERT Material for Class 12 Maths Chapter 7 – Integrals**

In the previous Chapter 6:Application of derivatives we will learn about definition of derivatives, rate of change of quantities and many more.In this Chapter 7:Integrals we will learn about integrals and some properties of indefinite integral.

**Quick revision notes**

Integration is the converse cycle of separation. In the differential calculus, we are given a function and we need to locate the derivative or differential of this function, however in the integral calculus, we are to discover a function whose differential is given. In this way, integration is a cycle which is the reverse of separation.

At that point, ∫f(x) dx = F(x) + C, these integrals are called uncertain integrals or general integrals. C is a self-assertive steady by shifting which one gets distinctive enemies of derivatives of the given function.

Note: Derivative of a function is remarkable yet a function can have infinite enemies of derivatives or integrals.

**Properties of Indefinite Integral**

(i) ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx

(ii) For any real number,say k, ∫k f(x) dx = k∫f(x)dx.

(iii) In general, if f1, f2,………, fn are functions and k1, k2,…, kn are real numbers, then

∫[k1f1(x) + k2 f2(x)+…+ knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫ f2(x) dx+…+ kn ∫fn(x) dx

**Basic Formulae**

**Integration using Trigonometric Identities**

When the integrand includes some geometrical functions, we utilize the accompanying identities to locate the integral:

- 2 sin A . cos B = sin( A + B) + sin( A – B)
- 2 cos A . sin B = sin( A + B) – sin( A – B)
- 2 cos A . cos B = cos (A + B) + cos(A – B)
- 2 sin A . sin B = cos(A – B) – cos (A + B)
- 2 sin A cos A = sin 2A
- cos2 A – sin2 A = cos 2A
- sin2 A = ()
- sin2 A + cos2 A = 1

**Integration by Substitutions**

Substitution strategy is utilized, when a reasonable substitution of variable prompts rearrangements of integral.

On the off chance that I = ∫f(x)dx, at that point by putting x = g(z), we get

I = ∫ f[g(z)] g'(z) dz

Note: Try to substitute the variable whose derivative is available in the first integral and last integral must be written as far as the first factor of integration.

**Integration by Parts**

For a given functions f(x) and q(x), we have

∫[f(x) q(x)] dx = f(x)∫g(x)dx – ∫{f'(x) ∫g(x)dx} dx

Here, we can pick the principal function as per its position in ILATE, where

I = Inverse trigonometric function

L = Logarithmic function

A = Algebraic function

T = Trigonometric function

E = Exponential function

[the function which comes first in ILATE should taken as first function and other as the second]

**Note **

(I) Keep at the top of the priority list, ILATE isn’t a standard as all inquiries of integration by parts is impossible by above technique.

(ii) It merits referencing that integration by parts isn’t appropriate to the result of functions in all cases. For example, the strategy doesn’t work for ∫√x sinx dx. The explanation is that there doesn’t exist any function whose derivative is √x sinx.

(iii) Observe that while finding the integral of the subsequent function, we didn’t include any constant of integration.

**Integration by Partial Fractions**

A rational function is the proportion of two polynomials of the form p(x)/q(x), where p(x) and q(x) are polynomials in x and q(x) ≠ 0. On the off chance that degree of p(x) > degree of q(x), at that point we may divide p(x) by q(x) so that , where t(x) is a polynomial in x which can be integrated easily and degree of p1(x) is less than the degree of q(x) . p1(x)/q(x) can be integrated by expressing p1(x)/q(x) as the sum of partial fractions of the accompanying kind :

where x2 + bx + c can’t be factorised further.

Integrals of the sorts can be transformed into standard form by communicating

Integrals of the types can be transformed into standard form by communicating px + q = A d/dx (ax2 + bx + c) + B = A(2ax + b) + B, where A and B are determined by comparing coefficients on two sides.