Integration is a technique used to find a function whose derivative is given. It is therefore an inverse process of differentiation. This is why, integration is also termed sometimes as

anti-derivation.Before we proceed further into the study of integration, we first discuss the differentials of Variables that will be helpful in developing methods / techniques used for integration.

Let `f` be a differentiable function in the interval ` < x < b ,` and let it be defined as

`y = f(x)`

then

`δy = f(x+δx) – f(x)`

and

`lim δx → ` 0 ` δy/δx`=`lim δx → 0 (f(x + δx) – f(x))/δx`= ` f ' (x) `

that is,

`dy/dx= f ' (x)`

We know that before the limit `δx → 0` is reached, the expression `δy/δx` differs from `f'(x`)by a very slight amount, let us call it `Ԑ`. Then we can write that

`δy/δx=f '(x) + Ԑ` where Ԑ is very small

Or

`δy= f '(x) δx+ Ԑδx`Here the term `f '(x)δx` is much more important than the term `Ԑδx` and is called the differential of the dependant variable `y` and is denoted by `dy`.

Hence

`dy = f '(x) δx`

And

`dx = (x)' δx= (1)δx = δx`

i.e., the differential of `x` is denoted by `dx` and is defined by the relation `dx = δx`

So the above relation takes the form

`dy = f '(x) dx`

As

`f(x) = x^2` => `f ' (x) = 2x`

`δy = f(x+δx) – f(x) = ( x + δx )^2 – x^2`

= 2x `δx + (δx)^2` = 2x dx + `(dx)^2 ` = ` 2(2) (0.01) + (0.01)^2`

`δy = f(x+δx) – f(x) = ( x + δx )^2 – x^2`

= 2x `δx + (δx)^2` = 2x dx + `(dx)^2 ` = ` 2(2) (0.01) + (0.01)^2`

=>

`δy= 0.04 + 0.0001 = 0.0401`

Also

`dy = f '`(x) dx = 2x dx = ` 2(2) (0.01) = 0.04`

This example shows that `δy` and `dy` differ by a very small amount. (0.0401 – 0.04 = 0.0001 in this example)

1. `∫ (ax + b)^n dx ` =`(ax + b)^(n+1)/(a(n+1))+c ` , `( n ≠ -1) `

2. `∫ sin (ax + b) dx` = ` –( 1 )/acos (ax + b) + c`

3. `∫ cos (ax + b) dx ` = ` ( 1 )/asin (ax + b) + c`

4. `∫ sec2 (ax + b) dx ` = `( 1 )/atan (ax + b) + c`

5. `∫ cosec^2 (ax + b) dx ` = `–( 1 )/acot (ax + b) + c`

6. `∫ sec (ax + b)tan (ax + b) dx` = `( 1 )/asec (ax + b) + c `

7. `∫ cosec (ax + b)cot (ax + b) dx` = `–( 1 )/acosec (ax + b) + c `

8. `∫ e^(λx + µ) dx ` = ` ( 1 )/λe^(λx + µ) + c ` `( λ ≠ 0 ) `

9. `∫ a^(λx + µ) dx ` = ` ( 1 )/(λ lna ) . a^(λx + µ ) + c` `(a > 0, a ≠ 1, λ ≠ 0 ) `

10. `∫1/( (ax + b ))dx` = `∫ (ax + b)^-1 dx ` = `( 1 )/aln|ax + b | + c ` , `(ax + b) ≠ 0 `

11. `∫ tan (ax + b) dx` = ` ( 1 )/aln|sec (ax + b)| + c ` = ` –( 1 )/aln|cos(ax + b)| + c `

12. `∫ cot (ax + b) dx ` = ` ( 1 )/aln|sin(ax + b)| + c `

13. `∫ sec (ax + b) dx` = `( 1 )/aln|sec (ax + b) + tan (ax+b)| + c `

14. `∫ cosec (ax + b) dx ` = `( 1 )/aln| cosec (ax + b)– cot (ax+b)| + c `

2. `∫ sin (ax + b) dx` = ` –( 1 )/acos (ax + b) + c`

3. `∫ cos (ax + b) dx ` = ` ( 1 )/asin (ax + b) + c`

4. `∫ sec2 (ax + b) dx ` = `( 1 )/atan (ax + b) + c`

5. `∫ cosec^2 (ax + b) dx ` = `–( 1 )/acot (ax + b) + c`

6. `∫ sec (ax + b)tan (ax + b) dx` = `( 1 )/asec (ax + b) + c `

7. `∫ cosec (ax + b)cot (ax + b) dx` = `–( 1 )/acosec (ax + b) + c `

8. `∫ e^(λx + µ) dx ` = ` ( 1 )/λe^(λx + µ) + c ` `( λ ≠ 0 ) `

9. `∫ a^(λx + µ) dx ` = ` ( 1 )/(λ lna ) . a^(λx + µ ) + c` `(a > 0, a ≠ 1, λ ≠ 0 ) `

10. `∫1/( (ax + b ))dx` = `∫ (ax + b)^-1 dx ` = `( 1 )/aln|ax + b | + c ` , `(ax + b) ≠ 0 `

11. `∫ tan (ax + b) dx` = ` ( 1 )/aln|sec (ax + b)| + c ` = ` –( 1 )/aln|cos(ax + b)| + c `

12. `∫ cot (ax + b) dx ` = ` ( 1 )/aln|sin(ax + b)| + c `

13. `∫ sec (ax + b) dx` = `( 1 )/aln|sec (ax + b) + tan (ax+b)| + c `

14. `∫ cosec (ax + b) dx ` = `( 1 )/aln| cosec (ax + b)– cot (ax+b)| + c `

All these formulas were presented in their most generalized forms. Their specific case is when `(ax + b ) = x;` that is `a = 1` and `b = 0`.

1. ` ∫ x^n dx ` = `x ^(n+1)/(n+1)+ c ` , ` ( n ≠ -1) `

2. ` ∫ sinx dx ` = ` –cos x + c`

3. ` ∫ cos` x `dx ` = `sin x + c`

4. ` ∫ sec^2` x `dx ` = `tan x + c`

5. ` ∫ cosec^2` x `dx ` = ` –cot x + c`

6. ` ∫ secxtan` x `dx ` = ` sec x + c`

7. ` ∫ cosec ` xcot ` xdx ` = ` –cosec ` x ` + c `

8. ` ∫ e^x dx ` = ` e^x+ c `

9. ` ∫ a^x dx ` = ` ( 1 )/lna . a^x+ c` ` (a > 0, a ≠ 1)`

10. ` ∫1/xdx=ln|x| + c ` , ` x ≠ 0`

11. ` ∫ tan ` x ` dx` = `ln|sec x| + c ` = ` – ln|cosx| + c`

12. ` ∫ cot ` x ` dx ` = `ln|sinx| + c`

13. ` ∫ sec ` x ` dx` =`ln|sec x + tan x| + c `

14. ` ∫ cosec ` x ` dx` = `ln| cosec ` x ` – cot x| + c `

2. ` ∫ sinx dx ` = ` –cos x + c`

3. ` ∫ cos` x `dx ` = `sin x + c`

4. ` ∫ sec^2` x `dx ` = `tan x + c`

5. ` ∫ cosec^2` x `dx ` = ` –cot x + c`

6. ` ∫ secxtan` x `dx ` = ` sec x + c`

7. ` ∫ cosec ` xcot ` xdx ` = ` –cosec ` x ` + c `

8. ` ∫ e^x dx ` = ` e^x+ c `

9. ` ∫ a^x dx ` = ` ( 1 )/lna . a^x+ c` ` (a > 0, a ≠ 1)`

10. ` ∫1/xdx=ln|x| + c ` , ` x ≠ 0`

11. ` ∫ tan ` x ` dx` = `ln|sec x| + c ` = ` – ln|cosx| + c`

12. ` ∫ cot ` x ` dx ` = `ln|sinx| + c`

13. ` ∫ sec ` x ` dx` =`ln|sec x + tan x| + c `

14. ` ∫ cosec ` x ` dx` = `ln| cosec ` x ` – cot x| + c `

We now come to solving some of the examples of integrals that make use of these formulae.

We know that

`∫1/( (ax + b ))dx= ∫ (ax + b)^(-1) dx =( 1 )/aln|ax + b | + c`

=> `∫1/( (2x + 3 ))dx= ∫ (2x + 3)^(-1) dx =( 1 )/2ln|2x + 3| + c`

=> `∫1/( (2x + 3 ))dx= ∫ (2x + 3)^(-1) dx =( 1 )/2ln|2x + 3| + c`

We know that

`∫ sin (ax + b) dx = –( 1 )/acos (ax + b) + c`

=> `∫ sin 3x dx = –( 1 )/3cos 3x + c`

=> `∫ sin 3x dx = –( 1 )/3cos 3x + c`

In all these examples and formulas, ` c ` is the constant of integration, and its value can be evaluated from initial conditions. We shall discuss it later in the coming topics.

The integral of the product of a constant and a function is equal to the product of the constant and the integral of the function.

`∫a f(x) dx = a ∫ f(x) dx ` where a is a constant

The integral of the sum (or difference) of two functions is equal to the sum (or difference) of their integrals.

`∫ [ f1(x) + f2(x) ] dx ` = ` ∫ f1(x) dx +∫ f2(x) dx`

The anti-derivative of the expression of the form

`∫ [ f(x) ]^n f '(x) dx ` = ` [ f(x)] ^(n+1)/(n+1)+ c ( n ≠ -1 )`

The anti-derivative of the expression of the form

`∫ [ f(x) ]^(-1)f '(x) dx ` = ` ln f(x) + c ` `( f(x) > 0 )`

`∫(x + 1)(x – 3) dx = ∫(x^2 – 2x – 3 ) dx = ∫ x^2 dx –∫2x dx – ∫3 dx`

= ` x ^3/3–2.x ^2/2– 3x + c`

= ` x ^3/3–x ^2– 3x + c`

= ` x ^3/3–2.x ^2/2– 3x + c`

= ` x ^3/3–x ^2– 3x + c`

`∫x/( x+2 )dx = ∫( x + 2 - 2 )/( x + 2 )dx = ∫(1-2/( x+2 ))dx`

= `∫1dx-2∫(x + 2 )^(-1)dx `

= ` x - 2 ln (x + 2) + c`

= `∫1dx-2∫(x + 2 )^(-1)dx `

= ` x - 2 ln (x + 2) + c`

`∫sin^2x dx ` = `∫( 1-cos2x)/( 2 )dx`

= ` ∫1/2dx-∫cos2x/2dx `

= ` ( 1 )/2x–( 1 )/2 . ( 1 )/2 . sin 2x+ c`

= `( 1 )/2x–( 1 )/4 . sin 2x + c`

= ` ∫1/2dx-∫cos2x/2dx `

= ` ( 1 )/2x–( 1 )/2 . ( 1 )/2 . sin 2x+ c`

= `( 1 )/2x–( 1 )/4 . sin 2x + c`

We use the technique of ** substitution** whenever it is possible to convert an integral into a standard form or to an easy integral by a suitable change
of a variable. We present some simple examples that will help you develop a good understanding of this technique.

Let

`u =4+x^2` => `du = 2x dx ` => `x dx =( 1 )/2du`=> `∫(x )/√(4+ x^2 )dx ` = `∫( 1 )/( 2√(u ))du ` = `( 1 )/2∫u -1/2du + c`

= `( 1 )/2 u ^(1/2)/(1/2) + c ` = ` u ^(1/2)`

Substituting the value of u gives

`∫(x )/√(4+ x^2 )dx ` = `√(4+ x^2 ) + c`

Let

`u = √(x )` => `du = ( 1 )/( 2√(x ))dx ` => `( 1 )/√(x )dx ` = `2du`=> `∫(cot√(x ))/√(x )dx` = `∫ cot√(x ). ( 1 )/√(x )dx` = `∫cot u . 2du`

=>`∫(cot√(x ))/√(x )dx ` = `∫cotu . 2du`

= `2∫cot u .du` = ` 2 ∫cosu/sinu . du`

= ` 2 ∫(sin u)-1 cos u du`

= `2 ln |sin u| + c`

= ` 2ln|sin √(x )| + c`

We present some suitable substitutions for certain expressions to be integrated.

`√(a^2-x^2 )`

`√(x^2-a^2 )`

`√(a^2+x^2 )`

`√(x+a )(or √(x-a ))`

`√(2ax-x^2 )`

`√(2ax+x^2 )`

`√(x^2-a^2 )`

`√(a^2+x^2 )`

`√(x+a )(or √(x-a ))`

`√(2ax-x^2 )`

`√(2ax+x^2 )`

`x = a sin Θ`

`x = a sec Θ`

`x = a tan Θ`

`√(x+a ) = t ` `( or √(x-a ) = t )`

`x – a = a sin Θ`

`x + a = a sec Θ`

`x = a sec Θ`

`x = a tan Θ`

`√(x+a ) = t ` `( or √(x-a ) = t )`

`x – a = a sin Θ`

`x + a = a sec Θ`

`∫(1 )/√(2x+x^2 )` = `∫dx/√((x+1)^2 - 1 )`

`∫dx/√((x+1)^2 - 1 )`=`∫( secΘ tanΘ dΘ )/√(sec^2 Θ - 1 )`=` ∫( secΘ tanΘ dΘ )/tanΘ =∫secΘ dΘ `

` = ln (sec Θ+tanΘ) + c `

=> `∫(1 )/√(2x+x^2 )` = `ln(x + 1 +√(2x+x^2 )) + c`

Let us substitute

`x+1=sec Θ ` `( 0 <Θ<π/2)`

`dx= secΘ tanΘ dΘ``∫dx/√((x+1)^2 - 1 )`=`∫( secΘ tanΘ dΘ )/√(sec^2 Θ - 1 )`=` ∫( secΘ tanΘ dΘ )/tanΘ =∫secΘ dΘ `

` = ln (sec Θ+tanΘ) + c `

=> `∫(1 )/√(2x+x^2 )` = `ln(x + 1 +√(2x+x^2 )) + c`

Next we present a simple integration technique called **"Integration by Parts".** It involves employing a simple formula that is of great help in evaluating
complex integrals.

**If both u and v are functions of some common variable say x, **

`∫u dv = uv -∫ v du`

If

`u = f(x) and v = g(x)`Then the above formula can be expressed as

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

We make use of the formula for **"Integration by Parts".** According to this formula

`∫ u dv = uv -∫ v du`

If we substitute

`u = x ` and `dv = exdx`

then `du = 1 . dx` and `v = ex`

=>` ∫x ex dx` = `xex -∫ex .1 .dx ` = ` x ex -ex + c`

` x ^6/6(ln x) -x ^6/36+ c `

Making use of the formula for ** "Integration by Parts"**, we have

`∫x^5ln x dx = ∫(ln x) x^5 dx `

= `(ln x) .x ^6/6-∫x ^6/6 . ( 1 )/xdx `

= ` x ^6/6(ln x) -∫x ^6/6 . ( 1 )/xdx `

= ` x ^6/6(ln x) - ( 1 )/6∫x5dx `

= ` x ^6/6(ln x) -( 1 )/6[ x ^6/6+c1 ] `

= ` x ^6/6(ln x) -x ^6/36+ c `

= `(ln x) .x ^6/6-∫x ^6/6 . ( 1 )/xdx `

= ` x ^6/6(ln x) -∫x ^6/6 . ( 1 )/xdx `

= ` x ^6/6(ln x) - ( 1 )/6∫x5dx `

= ` x ^6/6(ln x) -( 1 )/6[ x ^6/6+c1 ] `

= ` x ^6/6(ln x) -x ^6/36+ c `

Making use of the formula for ** "Integration by Parts"**, we have

`∫x^2 sin x = x^2. (-cos x) - ∫2x . (-cos x) dx`

= `-cos x . x^2 + 2 ∫xcos x dx`

= `-cos x . x^2+ 2 [x ( sin x ) -∫sin x dx ]`

= `-cos x . x^2+ 2x sin x -2(-cos x) + c`

= `-x^2cos x+ 2x sin x +2cos x+ c`

= `-cos x . x^2 + 2 ∫xcos x dx`

= `-cos x . x^2+ 2 [x ( sin x ) -∫sin x dx ]`

= `-cos x . x^2+ 2x sin x -2(-cos x) + c`

= `-x^2cos x+ 2x sin x +2cos x+ c`

If`P(x)`and `Q(x)` are polynomial functions and the denominator `Q(x) ≠ 0` in the rational function `(P(x) )/(Q(x))` , can be factorized into linear and quadratic
factors, then the rational function is written as a sum of simpler rational functions, each of which can be integrated by methods already known to us.

In the next section, we present some examples of this technique.

`ln │(x – a)/(x+a)│+ c`

The denominator of the above expression can be factorized as follows:

`x^2-a^2= (x + a) (x – a)`

=> `2a/(x^2 - a^2 )`= `2a/((x + a) (x – a))`=` (A )/((x – a) )+(B )/((x + a) )`

=> `2a/(x^2 - a^2 )`= `2a/((x + a) (x – a))`=` (A )/((x – a) )+(B )/((x + a) )`

Applying the Method of Partial Fractions gives A = 1 and B = -1. Hence

=> `2a/(x^2 - a^2 )`=`(1 )/((x – a) )–1/((x + a) )`

=> `∫2a/(x^2 - a^2 )`=`∫(1 )/((x – a) )dx–∫1/((x + a) ) dx`

= `∫(x – a)^(-1).1dx–∫(x+ a)^(-1) . 1 dx`

= `ln | x – a | – ln | x+a | + c`

= `ln │(x – a)/(x+a)│+ c`

=> `∫2a/(x^2 - a^2 )`=`∫(1 )/((x – a) )dx–∫1/((x + a) ) dx`

= `∫(x – a)^(-1).1dx–∫(x+ a)^(-1) . 1 dx`

= `ln | x – a | – ln | x+a | + c`

= `ln │(x – a)/(x+a)│+ c`

What we have discussed uptil now is **indefinite integral** of the form

`∫f(x) dx = Ω(x) + c `

where c is an arbitrary constant

If `∫f(x) dx = Ω(x) + c`, then the integral of `f(x)`from ato bis denoted by `∫_a^bf(x) dx` and has a definite value `Ω(b) – Ω(a)`, and it is called the
** Definite Integral**.The interval [a, b] is called the range of integration, and the values a and b are known as the

`∫_(-1)^3(x^(3 )+ 3x^2 ) dx ` = `∫_(-1)^3x^(3 ) dx + ∫_(-1)^33x^(2 ) dx `

= ` [ x^(4 )/4]+ x3`

= ` [ x^(4 )/4]+ x3`

Applying the limits from -1 to 3

`∫_(-1)^3(x^(3 )+ 3x^2 ) dx`=`[ (3)^(4 )/4 - (-1)^(4 )/4] + [ (3)3 – (–1)3 ]`

= `[( 81 )/4– ( 1 )/4 ] + [ 27+ 1 ] `

=>`∫_(-1)^3(x^(3 )+ 3x^2 ) dx `=` 20 + 28 ` = `48`

= `[( 81 )/4– ( 1 )/4 ] + [ 27+ 1 ] `

=>`∫_(-1)^3(x^(3 )+ 3x^2 ) dx `=` 20 + 28 ` = `48`

Which is a definite value.Hence the name Definite Integral.

`π/12+√(3 )/2– 1`

Applying the formula of __Integration by Parts__

`∫ x cosxdx = x sin x –∫(sinx) (1) dx `

= `x sin x – [(–cos x) + c1]`

= ` x sin x + cos x + c`

= `x sin x – [(–cos x) + c1]`

= ` x sin x + cos x + c`

Applying the limits of integration

`∫_0^(π/6)x cosx `

`dx = (π/6sinπ/6 + cosπ/6 ) – (0 sin 0 + cos 0)`

= ` π/6.( 1 )/2+√(3 )/2–( 0+ 1 )`

=`π/12+√(3 )/2– 1`

`dx = (π/6sinπ/6 + cosπ/6 ) – (0 sin 0 + cos 0)`

= ` π/6.( 1 )/2+√(3 )/2–( 0+ 1 )`

=`π/12+√(3 )/2– 1`

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