# Integration Calculus

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.

## Differentials of Variables

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

## Find δy and dy of the function defined as f(x) = x^2 ; when  x = 2 and dx = 0.01

#### Explanation:

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= 0.04 + 0.0001 = 0.0401

Also

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

#### Conclusion:

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

## Some Standard Formulae for Anti-Derivatives in General Form

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 ln⁡a ) . 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.

## Some Standard Formulae for Anti-Derivatives In Simple Form

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 )/ln⁡a . 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.

## Find the integral of 1/( (2x + 3))

#### Explanation:

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

## Find the integral of sin3x

#### Explanation:

We know that

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

## Theorems of Integrals

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 )

## Evaluate the integral of (x + 1)(x – 3)

#### Explanation:

∫(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

## Find the integral of x/( x+2 )

#### Explanation:

∫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

## Evaluate the integral of sin^2x

#### Explanation:

∫sin^2x dx  = ∫( 1-cos⁡2x)/( 2 )dx

=  ∫1/2dx-∫cos⁡2x/2dx

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

= ( 1 )/2x–( 1 )/4 . sin 2x + c

## Integration by Method of Substitution

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.

## Evaluate the integral of  (x )/√(4+ x^2 )

#### Explanation:

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

## Evaluate the integral of (cot√(x ))/√(x )

#### Explanation:

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 ∫cos⁡u/sin⁡u . du

=  2 ∫(sin u)-1 cos u du

= 2 ln |sin u| + c

=  2ln|sin √(x )| + c

## Some Useful Substitutions

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

## Expression Involving

√(a^2-x^2 )
√(x^2-a^2 )
√(a^2+x^2 )
√(x+a )(or √(x-a ))
√(2ax-x^2 )
√(2ax+x^2 )

## Suitable Substitution

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

## Evaluate the integral of (1 )/√(2x+x^2 )

#### Explanation:

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

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 )/tan⁡Θ =∫secΘ dΘ

 = ln (sec Θ+tanΘ) + c

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

## Integration by Parts

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.

## Formula for Integration by Parts:

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

## Find the integral of x e^x

#### Explanation:

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

## Evaluate the integral of x^5ln x

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

#### Explanation:

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

## Evaluate the integral of x^2 sin x

#### Explanation:

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

## Integration involving Partial Fractions

IfP(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.

## Evaluate the integral of 2a/(x^2 - a^2 )

ln │(x – a)/(x+a)│+ c

#### Explanation:

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) )

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

## Definite Integrals

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 lower and upper limits respectively.

## Evaluate∫_(-1)^3(x^(3 )+ 3x^2 ) dx

#### Explanation:

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

=  [ 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

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

## Evaluate∫_0^(π/6)x cos⁡x dx

π/12+√(3 )/2– 1

#### Explanation:

Applying the formula of Integration by Parts

∫ x cos⁡xdx = x sin x –∫(sinx) (1) dx

= x sin x – [(–cos x) + c1]

=  x sin x + cos x + c

Applying the limits of integration

∫_0^(π/6)x cos⁡x

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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