Methods of Integration

Integration can be defined as the summation of values when the number of terms tends to infinity. It is used to unite a part of the whole. Integration is just the reverse of differentiation and has various applications in all spheres such as physics, chemistry, space, engineering, etc.

Integration of a function or a curve can be used to find useful information such as the area under the curve or volume of the curve, etc. Integration may be of 2 types which are definite and indefinite integration depending upon whether the limits of integration are mentioned or not.

We know that integration is represented using the symbol ∫ over a function f(x) as follows:

∫f(x)dx 

Various methods of integration are as follows:

• Integration By Substitution
• Integration Using Trigonometric Identities
• Integration Using Trigonometric Substitution
• Integration by Parts
• Integration by Partial Fraction
• Integration of Some Special functions

Integration By Substitution

This method is used when we find it difficult to integrate a function as it is. In this method, a certain term in the function is substituted as a new variable and the whole function is changed to a new function of a new variable. This means that:

\int f(x)dx = \int f(u)du

Let us understand it with an example. 

Example: Solve ∫(x - 4) dx.

Solution: 

Given I = \int (x-4) dx

Substitute (x-4) = u

Differentiate both sides w.r.t x to get 

dx = d

Thus I = \int u~du

I = \frac{u^2}{2} + c

Again putting u = (x-4)

I = \frac{(x-4)^2}{2}+c

Integration using Trigonometric Identities

This method involves integrating the given function by transforming it using trigonometric identities. The value of the given function is substituted using some other function that is derived by using trigonometric identities. To know more about trigonometric identities, please refer to Trigonometric Identities.

Let us understand this method with an example.

Example: Solve \int 2cos^2x~dx

Solution:

Given I = \int 2cos^2x~dx

We know that cos(2x) = 2cos²x -1

2cos²x = cos(2x)+1

Substituting the value of 2cos²x, we get 

I = \int [cos(2x)-1]dx \\ I = \frac{sin(2x)}{2} - x + c

Integration by Parts

This method is used in cases where the function to be integrated is a product of two or more functions. Let f(x) = g(x)h(x), then f(x) can be integrated by parts by using the below formula:

∫g(x) · h(x) · dx = g(x) · ∫h(x) dx – ∫(g′(x) · ∫h(x) dx) dx

The sequence of h(x) and g(x) should be decided using the ILATE rule which tells the priority of functions and stands for Inverse trigonometric, logarithmic, algebraic, trigonometric, and exponential functions. This means that inverse trigonometric function should be written before logarithmic, logarithmic should be written before algebraic and so on.

Example: Solve \bold{f(x) = \int x~\log(x) dx}

Solution:

Given I  = \int x~log(x) dx

From ILATE rule, we know that logarithmic function has higher priority than algebraic function. Hence,

I  = \int log(x)~x~ dx

Using integration by parts,

I = log(x)\int xdx - \int (\frac{d}{dx}(log~x)\int x~dx)dx \\ I = \frac{x^2log(x)}{2}-\int (\frac{1}{x}\times \frac{x^2}{2})dx \\ I = \frac{x^2log(x)}{2}-\int \frac{1}{2}(x)dx \\ I = \frac{x^2log(x)}{2}-\frac{x^2}{4} + c

Integration by Partial Fraction

If the function to be integrated is of the form f(x) = g(x)/h(x) where g(x) and h(x) are polynomials, then we use the method of partial fraction. There are multiple cases in partial fractions depending upon the type of f(x).

In all these cases, we need to take the LCM of the partial fractions to make the denominator the same. After that, we compare the numerator on the LHS and RHS. Then substitute the suitable value of x in order to make any one part of the numerator zero and determine the value of A, B and C.

Let us understand partial fractions with an example.

Example: Integrate the function f(x) = x/(x-2)(x+3).

Solution:

Given f(x) = x/(x-2)(x+3)

It is of the form f(x) = \frac{px+q}{(x-a)(x-b)} 

So it can be written as \frac{x}{(x-2)(x+3)} = \frac{A}{x-2}+\frac{B}{x+3}

Taking LCM on RHS, we get,

\frac{x}{(x-2)(x+3)} = \frac{A(x+3)+B(x-2)}{(x-2)(x+3)}

Comparing numerators on LHS and RHS:

x = A(x+3)+B(x-2)

Put x = 2 to get,

2 = 5A or A = 2/5

similarly put x = -3 to get

-3 = -5B or B = 3/5

Thus f(x) = \frac{2}{5(x-2)}+ \frac{3}{5(x+3)}

\int f(x) = \int[\frac{2}{5(x-2)}+ \frac{3}{5(x+3)}]dx\\ = \frac{2}{5}\int\frac{1}{x-2} + \frac{3}{5}\int\frac{1}{x+3}\\ = \frac{2}{5}\log{(x-2)} + \frac{3}{5}\log(x+3)

Integration of Some Special Functions

In Mathematics, we have some special functions which have pre-defined integration formulas. These functions and their integration are shown below:

• \int \frac{dx}{ (x^2 – a^2)} = \frac{a}{2}log | \frac{x – a}  {x + a} | + c
• \int \frac{dx}{ (a^2 – x^2)} = \frac{a}{2}log | \frac{a + x}  {a - x} | + c
• \int \frac{dx} {x^2 + a^2} =  \frac{tan^{–1}\frac{x}{a}}{a}  + c
• \int \frac{ dx} {√ (x^2 – a^2)} = log| x+√(x^2 – a^2) | + c
• \frac{ dx} {√ (a^2 – x^2)} = sin^{–1} (xa) + c
• \int \frac{dx} {√ (x^2 + a^2)} = log | x + √(x^2 + a^2) | + c

Related Resources,

• Integrals
• Integration Formulas
• Area Between Two Curves
• Area Under the Curve
• Differentiation and Integration

Examples of Methods of Integration

Example 1: Solve \bold{\int \log x~dx} .

Solution:

Given: I = \int \log x~dx = \int \log x.1~dx 

The functions are already written according to ILATE rule. Using integration by parts we get,

I = \log x\int1dx - \int(\frac{d}{dx}(\log x)\int1dx)dx \\ I = x\log x - \int\frac{1}{x}.x~dx \\ I = x\log x - \int1dx \\ I = x\log x - x + c

Example 2: Solve \bold{\int x \sin x~dx}

Solution:

Given: I = \int x \sin x~dx

The functions are already written according to ILATE rule. Using integration by parts we get,

I = x\int \sin xdx - \int(\frac{d}{dx}x\int\sin x)dx \\ I = -x\cos x- \int-\cos x~dx \\ I = -x \cos x + \int\cos xdx \\ I = -x \cos x + \sin x + c

Example 3: Solve \bold{\int (2x^3+1)^7x^2~dx}

Solution:

Given: I = \int (2x^3+1)^7x^2~dx

Substitute (2x³+1) = u 

Differentiate both sides w.r.t x to get 

6x² dx = du

x²dx = du/6

Rewriting the given function as I = \int \frac{u^7}{6}du

I = \frac{1}{6}(\frac{u^8}{8}) + c \\ I = \frac{u^8}{48} + c \\ I = \frac{(2x^3+1)^8}{48}+c

Example 4: Solve \bold{\int \sin^2x ~dx}

Solution:

Given I = \int \sin^2x ~dx

We know that cos(2x) = 1 - 2sin²x

(1-cos2x)/2 = sin²x

Substituting the value of sin²x, we get 

I = \int \frac{(1-\cos(2x))}{2}dx \\ I = \frac{1}{2}[x-\frac{\sin(2x)}{2}] + c \\ I = \frac{x}{2}-\frac{\sin(2x)}{4}+c

Example 5: Solve \bold{\int \frac{dx}{x^2-25}} .

Solution:

Given: I = \int \frac{dx}{x^2-25} = \int \frac{dx}{x^2-5^2} 

Using 

\int \frac{dx}{ (x^2 – a^2)} = \frac{a}{2}log | \frac{x – a}  {x + a} | + c \\ I = \frac{5}{2}log | \frac{x – 5}  {x + 5} | + c

Practice Problems on Methods of Integration

Problem 1: Calculate the following integrals:

• ∫(3x² + 2x - 5) dx
• ∫(e^x + 1) dx
• ∫(sin(x) + cos(x)) dx

Problem 2: Use the substitution method to evaluate the following integrals:

• ∫(2x + 1)³ dx
• ∫(x² + 4x + 3)√(x³ + 6x² + 9x) dx
• ∫2x cos(x²) dx

Problem 3: Apply integration by parts to solve the following integrals:

• ∫x ln(x) dx
• ∫x² sin(x) dx
• ∫e^x cos(x) dx

Problem 4: Use partial fraction decomposition to integrate:

• ∫(3x² + 2)/(x³ + 4x² + 4x) dx
• ∫(x³ - 2x² + 5x - 1)/(x² - 3x + 2) dx
• ∫(4x³ - 2x² + 3)/(x⁴ + 2x² + 1) dx