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

Become a member today!

Register (it’s Free)