Indefinite Integral

INDEFINITE INTEGRAL

If F'(x) = f(x) in an open interval I, then F is called an anti derivative or primitive of f on I.

If F(x) is an anti derivative of f(x), then F(x) + C, where C is a arbitrary constant, is an anti derivative, called the general anti derivative.

Libnitz used the special symbol ∫ f(x)dx to denote the general anti derivative or general primitive of ƒ and it is called the indefinite integral of f.

f(x) is called the integrand and C is constant of integration.

dx indicates variable of integration is x.

Note:

1. The equation ∫ f(x)dx = F(x) + C is to be considered mearly an alternative way of writing F'(x) = f(x).

2. ∫ f(x)dx represents a family of functions F(x) + C and represents a number.

They have similar notations because the first one is useful in computing the second one.

3. But for the appearance, the symbols ∫ f(x)dx and f(x)dx originate from two different processes differentiation and integration.

From the definition of indefinite integral we have the following results,

Further we have the following list of integrals of standard functions from the derivatives of these functions. This list is important for integration.

There are three main techniques which are used in evaluating integrals.

1. Integration by substitutions

2. Integration by partial fractions

3. Integration by parts

WORKED EXAMPLS

Example1 

Evaluate ∫ (4x3 − 2x + 3) dx

Solution 

Let

Example 2 

Evaluate ∫(x + 3)(x − 2)dx

Solution 

Example 3 

Evaluate

Solution 

Example 4 

Evaluate

Solution 

Example 5 

Evaluate

Solution

Example 6 

Evaluate

Solution

Example 7 

Evaluate

Solution

Example 8 

Evaluate

Solution

Example 9 

Evaluate

Solution

Example 10 

Evaluate

Solution

= tan x - cot x + C

Example 11 

Evaluate ∫(tan x − 2cot x)2dx.

Solution

= tan x + 4 (-cot x) - 9x + C

= tan x - 4 cot x - 9x + C

1. Integration by Substitution

Certain integrals can be reduced to one of the standard forms in the list of integrals by a suitable substitution or transformation of variable. The form of the integrated often suggests the substitutions. If t = g(x) then dt = g'(x) dx.

Then ∫f(x)dx will reduce to standard form ∫F(t)dt = G(t) + C .

Then rewrite in terms of x.

WORKED EXAMPLES

Example 1 

Evaluate

Solution 

Example 2 

Evaluate

Solution 

Example 3 

Evaluate

Solution 

Example 4 

Evaluate

Solution

Example 5 

Evaluate

Solution

Example 6 

Evaluate

Solution

Example 7 

Evaluate

Solution

Note: It can also be worked out by putting t = x + 5. Do it yourself:

Example 8 

Evaluate

Solution

Example 9 

Evaluate

Solution

Example 10 

Evaluate

Solution

Example 11 

Evaluate

Solution

Example 12 

Solution

You can do this problem by putting t = sin2x. Do it yourself!

Example 13 

Evaluate

Solution

Example 14 

Evaluate

Solution

It can also be worked out by putting t = 1 + x2. Do it yourself!

Example 15 

Find

Solution

Example 16 

Find

Solution

Example 17 

Find ∫ sint(sec2 (cost)dt.

Solution

Let I = ∫ sint(sec2 (cost)dt

Put  y = cos t ⸫ dy = -sin t dt ⇒ -dy = sin t dt

= - tan y + C = - tan (cos t) + C

Example 18 

Evaluate

Solution

Example 19 

Evaluate

Solution

Example 20 

Evaluate

Solution

Example 21 

Evaluate

Solution

Example 22 

Find

Solution

Example 23 

Evaluate

Solution

Example 24 

Evaluate

Solution

Example 25 

Find ∫cos x°dx.

Solution

Example 26 

Find

Solution

Example 27 

Evaluate

Solution 

EXERCISE

Evaluate the following indefinite integrals

ANSWERS TO EXERCISE

2. Special Type: Reciprocal Form

Certain integrals ∫ƒ(x)dx can be evaluated by rewriting in terms of 1/x. Observe the next few examples

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3 

Evaluate

Solution

Example 4 

Find

Solution

Example 5 

Evaluate

Solution

3. Integration of Trigonometric Functions of Products and Powers

WORKED EXAMPLES

Example 1 

Evaluate ∫sin 3x⸱sin bx dx.

Solution 

I = ∫sin 3x⸱sin bx dx

We convert into a sum or difference.

We know

cos (A + B) - cos(A + B) = -2 sinA ⸱ sinB

Example 2 

Evaluate

Solution

Example 3 

Find

Solution

We know that

Example 4

Find ∫ cos2 x · sin3 x dx.

Solution 

Let

Example 5 

Find

Solution

Let I = ∫ sin3 x ⋅ cos2x dx [power of sine is odd]

Put t = cos x. ⸫ dt = -sin x dx ⇒ - dt = sin x dx

Example 6 

Find

Solution

Example 7 

Solution

Example 8 

Evaluate

Solution

Example 9

Solution

Example 10 

Evaluate

Solution

Example 11 

Find

Solution 

Let

Example 12 

Evaluate

Solution

Example 13 

Find

Solution

Example 14 

Find

Solution

Type 1: The Integrals of the Form

where numerator and denominator are functions of the first degree in sinx or cosx or both.

To evaluate it, put l sin x + m cos x = A d/dx (a sin x + b cos x) + B(a sin x + b cos x)

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Equating coefficients of sin x on both sides, we get

-4A + 5B = 3   (1)

Equating coefficients of cos x on both sides, we get 

5A + 5B = 2     (2)

Solve (1) and (2) to find A, B

Substituting in (1), we get

Example 2 

Evaluate

Solution

Type 2

In the following problems we choose the substitution t in terms of sin x and cos x, so that dt is the numerator.

WORKED EXAMPLES

Example 1 

Evaluate

Solution 

Example 2 

Evaluate

Solution

Example 3 

Evaluate

Solution

Type 3

WORKED EXAMPLES

Example 1

Solution

Example 2

Solution

Example 3 

Find

Solution

Example 4 

Find

Solution

EXERCISE

ANSWERS TO EXERCISE 

3. Integration of Irrational Functions by Trigonometric Substitutions

When the integrand contains expressions of the form

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Solution

Example 3 

Evaluate

Solution

Example 4 

Evaluate

Solution 

Example 5 

Evaluate

Solution 

EXERCISE

Evaluate the following integrals

ANSWERS TO EXERCISE 

4. Integration of Rational Algebraic Functions

If the degree of the numerator is greater than or equal to the degree of the denominator, We divide till the remainder degree is less than the degree of denominator.

WORKED EXAMPLES

Example 1 

Find

Solution

Example 2 

Find

Solution

Example 3

Show that

Solution

Being rational function divide the Numerator by the Denominator

Example 4 

Find

Solution

5. Integration by Partial Fractions

If the integrand is a proper rational function and the denominator is factorisable into rational factors of first degree or second degree, then the integrand can be expressed into partial fractions. Then we integrate.

WORKED EXAMPLES

Example 1 

Evaluate 

Solution

Since degree of numerator is less then the degree of denominator

i.e., the integrand is a proper fraction, we use partial fraction method.

Example 2 

Evaluate

Solution

Since the degree of numerator is less than the degree of denominator, it is a proper fraction.

We use partial function method.

Multiplying both sides by x(2x − 1)(x + 2), we get

Example 3 

Evaluate

Solution

Since degree of numerator is less than the degree of denominator. We use partial fraction method.

Here the denominator contains repeated factor (x − 1)2

Note: For repeated factor (x − 1)2 there will be 2 terms as written in the Example 3. For a second degree factor assume numerator as first degree ax + b.

Example 4 

Evaluate

Solution

Example 5

Solution

Example 6 

Evaluate

Solution

The numerator degree is less than the denominator degree.

So, it is a proper fraction with denominator containing repeated factor and second degree factor.

Equating co efficient of x, we get

Equating coefficients of x, we get

Example 7 

Evaluate

Solution

Multiplying by (x - 1) (x - 2), we get

7x – 6 = A(x - 2) + B(x - 1)

Put x = 1, then 7 - 6 = A(1−2) ⇒ -A = 1 ⇒  A = -1

Put x = 2, then 14 - 6 = B(2 − 1) ⇒ B = 8

EXERCISE

Evaluate the following integrals:

ANSWERS TO EXERCISE 

Special Cases

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Put t = 0, then I = A

Example 2 

Evaluate

Solution 

Let

Put x + 1 = tan θ

⸫ dx = sec2 θ dθ

Example 3 

Evaluate

Solution

Since there is no odd powers of x and the denominator factors are of the form x2 ± a2, the partial fraction can be written treating x2 as variable y.

Example 4 

Evaluate 

Solution

Example 5 

Evaluate

Solution

Example 6 

Evaluate

Solution

Type 1: Integrals of the Form

where ax2 + bx + c cannot be factorized into real factors.

In the first integral, take out a and completing squares it will reduce to one of the forms

Equating the coefficients find A and B. Then integrate w. r. to x. 

One integral will be of type (1) and the other is of the form

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3 

Prove that

Solution

Example 4

Solution

Equating constant terms, we get

Example 5 

Evaluate

Solution

Equating constants terms,

Example 6 

Evaluate

Solution

Type 2: Integrals of the form (1) and (2)

(1) is evaluated by completion of squares method and using the formulae

(2) is evaluated by putting and proceeding as above.

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3 

Evaluate

Solution

Example 4 

Evaluate

Solution

Example 5 

Evaluate

Solution

Example 6 

Evaluate

Solution

Type 3

by taking out A and completing squares.

To evaluate it, we use any one of the formulae given below.

WORKED EXAMPLES

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3

Evaluate

Solution

Type 4: Integrals of the form

In (i) take out a and complete the squares and it will reduce to the form

To evaluate it, we use any one of the formulae given below.

WORKED EXAMPLE

Example 1 

Evaluate

Solution

Example 2 

Evaluate

Solution

Example 3 

Evaluate

Solution

Example 4 

Solution

Example 5 

Evaluate

Solution

EXERCISE 

Evaluate the following integrals

ANSWERS TO EXERCISE