Methods of Integration: Different Integration Methods

Integration is the inverse operation of differentiation. But it was discovered before. It is used for various applications like to find the area of the surface, to find the area enclosed by the curve, to find the surface, a moment of inertia, the center of gravity, the volume of the solid perimeter of the curve, etc. Sometimes the mathematical expressions cannot be added by the general addition method of mathematics then we use the method of integration to add or sum up the parts to find the whole. Integration is used to find areas on a large scale. Initially, the integration method was used to find the area of large figures by finding the area of smaller parts.

Read More: Inverse Trigonometric Formula

Key Terms: Integration, Integrals, Partial Fraction, Definite Integral, Trigonometric Identities, Indefinite Integral

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What is the Integration?

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The process of finding the differential coefficient of any unknown function is known as integration.

Let f(x) be a function of x and its differential coefficient is f'(x) then

d/dx [f(x)] = f'(x)

Then from the definition of integration f(x) is the integral of F(x)

Symbolically, ∫ f'(x) dx = f(x)

It means that the integration of f'(x) concerning x is f(x).

Read More: Perimeter of Shapes

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Important Terms of Integration 

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Here are a few terms related to integration 

• Integral - The integral of any given function is a function whose differential coefficient is the given function.
• Integrand - The given function f(x) which is to be integrated is called the integrand.
• Integration symbol - The integration is denoted by the symbol ∫ . It is a distracting form of the alphabet S. The symbol ∫ stands for integration and dx is written after function f(x).

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Types of Integration

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Integration is of mainly two types -

 a) Indefinite integral - The integral without upper and lower limit are called an indefinite integral.

Indefinite integral is .. 2 represented as ∫ f(x)dx = F(x) + c

where f(x) is integrand and c is a constant of integration. 

 b) Definite integral - In the definite integral, the upper and lower limit of the integral is given. It lies between the given limit. This type of integration is also called Riemann integral.

The definite integral is represented as ∫_a^xf(x)dx

Here x = upper limit and a = lower limit

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Methods of Integration

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The integration of any function can be solved by various methods or techniques. If the integral is given in the standard form it can be solved using standard formulae of integration, but if the given integral is not in standard form then it must be converted into the solvable form using different methods or techniques. Some important methods generally used to solve the integration of any complex or unsolvable function are as follows :

• Integration by substitution
• Integration by parts
• Integration by partial fraction
• Integration of some special functions
• Integration using trigonometric identities

Here we are discussing all the methods with examples, so you can refer to this article for a better understanding of integration and its methods.

Read More: Statistics

Integration by Substitution Method 

When we have such functions which are not in any standard form or which are not easily convertible into standard forms, then we substitute a proper function of variable t for the variable x. There is no specific rule for applying the substitution method.

Application 

Let I = ∫ f(x) dx

since dI/dx = f(x), (by definition of integration)

let x = φ(t) then dx/dt = φ'(t)

dI/dt = dI/dx.dx/dt

= f(x).φ'(t)

On integrating both sides we get

I = ∫ f {(t). (t) dt

Hence I = ∫ f(x) dx = ∫ f {(t)}.(t) ft

In the substitution method, choose relation x = (t). Such that the integration of new integrand can be done easily. In place of x in f(x) we put (t) and in place of dx we put (t)dt.

Integration by Substitution method

Solved example :

Integration of (ax+b)

Let I = ∫ f(ax+b)dx

put ax+b = t

on differentiating w.r.t. t we get

a dx/dt = dt/dt

a dx/dt = 1

adx = dt

or dx = 1/a dt

I = ∫ f(t). dt/a

I = 1/a ∫ f(t)dt

= 1/a F(t) (where F(t) = ∫ f(t) dt)

∫ f(ax+b)dx = 1/a F(ax+b)

Read More: Mean, Median and Mode

Integration by Special Functions Video Explanation

Integration by Parts

When the two functions are given in the product form then we use the integration by parts method. We say two functions like f₁(x) and f₂(x) be the function of x. Then applying integration by parts formula in both function w.r.t. x.

∫ f₁(x).f₂(x) dx = ∫ f₁(x).d/dx f₂(x)dx + ∫ f₂(x).d/dx f₁(x)dx

To apply this method of integration, proper choice of the first and second functions is essential. If we choose any function as the first or second the answer will be wrong. There is no fixed rule to choosing the first and second function but some helpful tips to choose the first and the second function is as follows:

 i) Take one of the given functions as the first function whose standard result of integration is not known.

 ii) If both given functions have standard results then take the function as first whose differentiation vanishes in some steps.

 iii) If the integral is single, a function then takes 1 as a second function.

 iv) If the differentiation of any function is not zero then take any function as the first function.

For the easiest way to choose the first and second functions, you can use the ILATE rule. The function comes first in the alphabet in ILATE then take it as the first function and the letter comes next then take it as the second function. In the word ILATE, the letters stand for

Integration by Parts

I = Inverse functions.

L = Logarithmic functions.

A = Algebraic functions.

T = Trigonometric functions.

E = Exponential functions

Solved example

Evaluate ∫ xsinx dx

Let I = ∫ xsinx dx

On applying integration by parts formula we get

= x ∫sinx dx - ∫ [d/dx x ∫sinxdx] dx

= -xcosx + ∫cosxdx

I = -xcosx + sinx 

By Parts Method of integrals Video Explanation

Read More: Sin Cos Formula

Integration by Partial Fractions

When the given function is in the form of rational expression p(x)/q(x) then to find the integration, the partial fraction method is to be applied.

Solved example

∫ (x-1) / (x+1) (x-2) dx

Let I = ∫ (x-1)/(x+1) (x-2) dx ….(1)

then (x-1)/ (x+1) (x-2) = A/(x+1) + B/(x-2) ……(2)

(x-1)/(x+1) (x-2) = A(x-2) + B(x+1)/(x+1) (x-2)

x-1 = A(x-2) + B(x+1) …..(3)

put x=2 in eqⁿ (3)

2 - 1 = 0 + B(2+1)

B = 1/3

put x = -1 in equⁿ (3)

(-1) - 1 = A(-1-2) + 0

A = 2/3

on putting values of A and B in eqⁿ (2) we get

(x-1)/(x+1) (x-2) = 2/(x+1)3 + 1/(x-2)3 …..(4)

from eqⁿ (1) and (4)

I = ∫ 2/(x+1)3 + 1/(x-2)3 dx

= 2/3 ∫ dx/(x+1) + 1/3 ∫ dx/(x-2)

I = 2/3 log(x+1) + 2/3 log(x-2)

In this method of integration, four cases arise which are as follows:

• When the denominator contains non-repeated linear factors.
• When the denominator contains repeated linear factors.
• When the denominator contains non-repeated quadratic factors.
• Integration is based on the first substitution than by partial fractions.

Read More: Types of Fraction

Integration by Partial Functions Video Lecture

Integration of Some Particular Functions

Some of the integral can be solved by using a particular substitution. Here are some important substitutions to solve these type of questions.

Solved example

I = ∫ dx/x² + a²

on putting x = at

dx = a.dt

I = a ∫ dt/a²t² + a²

= a/a² ∫ dt/1+t²

= 1/a tan^-¹ t

= 1/a tan^-¹ x/a

∫ dx/x² + a² = 1/a tan^-¹ x/a

Read More: Improper Fractions

Integration Using Trigonometrical Substitution 

In some questions the given function cannot be easily integrated by other methods, then the given integral is transformed by the proper trigonometrical substitution before applying integration by parts. In such cases, the application of formulae of integration by parts will ease the process of integration.

Solved example

Evaluate

∫ sin^-¹ (2x/1+x²) dx

Soln - Let I = ∫ sin^-¹ (2x/1+x²) dx

on putting x = tanθ

dx = sec² θdθ

 I = ∫ sin^-¹ (2tanθ/1+ tan² θ) sec²θ dθ

= ∫ sin^-¹ (sin2θ) sec²θ dθ

= ∫ 2θ sec²θdθ [sin^-¹(sinθ) = θ]. = 2 [θ∫sec²θ dθ - ∫ (d/dθ.θ∫sec²θdθ) dθ]

= 2 [θ.tanθ - ∫1.tanθdθ]

= 2 [θtanθ - logsecθ]

= 2 [θtanθ - log√1+tan²θ]

= 2 [xtan^-¹x - log√1+x²]

Ans = 2 [xtan^-¹x - ½log(1+x²)] 

Read More: Rational Expressions

Reverse Chain Rule

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The reverse chain rule, which is the chain rule's opposite in differentiation, is one of the simplest and most widely used integration techniques. It helps to determine the function's derivative that needs to be integrated. This integration technique is applied when the integration has the form g'(f(x)) f'(x) dx. The integral in this instance is provided by,

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Things to Remember 

• Methods of Integration are employed to solve complex problems. 
• The reverse chain rule is the opposite of the chain rule in differentiation. 
• The definite integral is given by ∫_a^xf(x)dx. 
• An indefinite integral is integral that has neither an upper nor a lower limit.
• A function whose differential coefficient is the provided function is referred to as the integral of any given function.
• Integration is the process of determining any unknown function's differential coefficient.

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Previous Year’s Questions

1. The value of sin⁡² 51^∘+ sin^⁡2 39^∘ is… [KCET – 2020]
2. If cos x = |sin x| then, the general solution is… [KCET – 2019]
3. If 0≤ x< π​/2, then the number of values of x… [JEE Main – 2019]
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5. If tanA+cotA=2, then the value of…  [KCET – 2020]
6. The value of cos²45^∘−sin²15^∘ is… [KCET – 2017]
7. A, B and C are the angles opposite to the corresponding sides of lengths… [JKCET – 2017]
8. The value of tan 8/π​ is equal to… [KCET – 2016]
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