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
Variable Differential Examples
Example 1:
`0.04`
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 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`.
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 )/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.
Example 2:
`( 1 )/2ln|2x + 3| + c`
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`
Example 3:
` –( 1 )/3cos 3x + c`
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 )`
Example 4:
` x ^3/3–x ^2– 3x + c`
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`
Example 5:
` x - 2 ln (x + 2) + c`
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`
Example 6:
`( 1 )/2x–( 1 )/4 . sin 2x + c`
Explanation:
`∫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`
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.
Substitution Method Integration Examples
Example 1:
`√(4+ x^2 ) + c`
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`
Example 2:
` 2ln|sin √(x )| + c`
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 ∫cosu/sinu . 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`
Example :
`ln(x + 1 +√(2x+x^2 )) + c`
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 )/tanΘ =∫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 `
Integration by Parts Examples
Example 1:
` x ex -ex + c`
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`
Example 2:
` 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 `
Example 3:
`-x^2cos x+ 2x sin x +2cos x+ c`
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
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.
Partial Fraction Integration Examples
Example :
`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.
Definite Integrals Examples
Example 1:
` 48`
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.
Example 2:
`π/12+√(3 )/2– 1`
Explanation:
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`
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`
