Integrals of Some Functions

 Table of contents Introduction Some Standard Integrals Integration of Irrational Functions Integrals of the type   where x & y are linear or quadratic expressions Integration Of A Binomial Differential Integration by Substitution Integration Using Trigonometric Identities Solved Examples: Integration by Parts: Integration by Partial Fractions

## Introduction

Consider the integral  dx
If we change the variable from x to θ by the substitution x = a sinθ, then the identity

1 – sin2θ = cos2θ

allows us to get rid of the roots sign because

Notice the difference between the substitution u = a2 – x2 (in which the new variable is a function of the old one) are the substitution x = a sin θ (the old variable is a function of the new one).

In general, we can make a substitution of the form x = g(t) by using the Substitution Rule in reverse. To make our calculations simpler, we assume that g has an inverse function; that is, g is one-to-one.

This kind of substitution is called inverse substitution.
We can make the inverse substitution x = a sin θ provided that it defines a one-to-one function. This can be accomplished by restricting θ to lie in the interval [–π/2, π/2].

## Some Standard Integrals

### Solved Examples

Ex.1 Evaluate  where a > 0

Sol.

We let x = a secθ, where 0 < θ < π/2 or π < θ < 3π/2. Then dx = a sec θ tan θ dθ and

Ex.2 Integrate

Sol.

Ex.3 Integrate (3x + 1) / (2x2 – 2x + 3).

Sol. Here (d/dx) (2x2 – 2x + 3) = 4x – 2.

Ex.4 Integrate

Sol.

Ex.5 Evaluate   dx.

Sol.

## Integration of Irrational Functions

Certain types of integrals of algebraic irrational expressions can be reduced to integrals of rational functions by a appropriate change of the variable. Such transformation of an integral is called its rationalization.

1. If the integrand is a rational function of fractional powers of an independent variable x, i.e. the function Rthen the integral can be rationalized by the substitution x = tm, where m is the least common multiple of the numbers q1, q2, ...., qk.
2. If the integrand is a rational function of x and fractional powers of a linear fractional function of the form  then rationalization of the integral is effected by the substitution  where m has the same sense as above.

### Solved Examples

Ex.1 Evaluate

Sol.

Rationalizing the denominator, we have

Ex.2 Evaluate I =

Sol. The least common multiple of the numbers 3 and 6 is 6, therefore we make the substitution

x = t6, dx = 6t5 dt.

Ex.3 Evaluate I =

Sol. The integrand is a rational function of  therefore we put 2x – 3 = t6, whence

Returning to x, we get

Ex.4 Evaluate

Sol.

Let x = t3 ⇒ dx = 3t2 then

Ex.5 Evaluate I =

Sol. The integrand is a rational function of x and the expression   therefore let us introduce the substitution

## Integrals of the type where x & y are linear or quadratic expressions

Ex.1 Integrate

Sol.

Put 4x + 3 = t2, so that 4dx = 2tdt and (2x + 1)

Ex.2 Evaluate

Sol.

Put (x + 2) = t2, so that dx = 2t dt, Also x = t– 2.

∴

dividing the numerator by the denominator

Ex.3 Integrate

Sol.

Put (x + 1) = t2, so that dx = 2t dt. Also x = t2 – 1.

Ex.4 Integrate

Sol.

Put (1 + x) = 1/t, so that dx = – (1/t2) dx.

Also x = (1/t) – 1.

Ex.5 Evaluate

Sol.

Put x = 1/t, so that dx = – (1/t2) dt.

∴

Now put 1 + t2 = z2 so that t dt = z dz. Then

[∵ t = 1/x]

## Integration Of A Binomial Differential

The integral  where m, n, p are rational numbers, is expressed through elementary functions only in the following three cases :
Case I : p is an integer. Then, if p > 0, the integrand is expanded by the formula of the binomial; but if p < 0, then we put x = tk, where k is the common denominator of the fractions and n.
Case II : is an integer. We put a + bxn = tα, where α is the denominator of the fraction p.
Case III :+ p is an integer we put a + bxn = tαxn, where a is the denominator of the fraction p.
Ex.1 Evaluate I =

Sol.

Here p = 2, i.e. an integer, hence we have case I.

Ex.2 Evaluate I =

Sol.

i.e. an integer.we have case II. Let us make the substitution. Hence ,

Ex.3 Evaluate I =

Sol.

Here p = – 1/2 is a fraction, m+1/2 = -5/2 also a fraction, but m+1/n + p/2 = -5/2 -1/2 = -3 is an integer, i.e. we have case III, we put 1 + x4 = x4/2,

Hence

Substituting these expression into the integral, we obtain

Returning to x, we get I =

## Integration by Substitution

Let g be a function whose range is an interval l, and let f be a function that is continuous on l. If g is differentiable on its domain and F is an antiderivative of f on l, then  f(g(x))g'(x) dx = F(g(x)) + C.

If u = g(x), then du = g'(x) and  f(u) du = F(u) + C .

### Guidelines for making a change of variable

1. Choose a substitution u = g(x). Usually, it is best to choose the inner part of a composite function, such as a quantity raised to a power.

2. Compute du = g'(x) dx.

3. Rewrite the integral in terms of the variable u.

4. Evaluate the resulting integral in terms of u.

5. Replace u by g(x) to obtain an antiderivative in terms of x.

### The General Poser Rule for Integration

If g is a differentiable function of x, then

### Rationalizing Substitutions

Some irrational functions can be changed into rational functions by means of appropriate substitutions.

In particular, when an integrand contains an expression of the form  then the substitution u =  may be effective.

Some Standard Substitutions

## Integration Using Trigonometric Identities

In the integration of a function, if the integrand involves any kind of trigonometric function, then we use trigonometric identities to simplify the function that can be easily integrated.

Few of the trigonometric identities are as follows:

All these identities simplify integrand, that can be easily found out.

## Solved Examples:

Ex.1 Evaluate  (x2 +1)2 (2x) dx .
Sol. Letting g(x) = x2 + 1, we obtain g'(x) = 2x and f(g(x)) = [g(x)]2.

From this, we can recognize that the integrand and follows the f(g(x)) g'(x) pattern. Thus, we can write

Ex.2 Evaluate
Sol.

Ex.3 Evaluate
Sol.

Let u = x+ 2   ⇒   du = 4x3 dx

Ex.4 Evaluate

Sol.

Let u = x3 – 2. Then du = 3x2 dx. so by substitution :

Ex.5 Evaluate

Sol. Let u  =   . Then u= x + 4, so x = u–4 and dx = 2u du.

Therefore

## Integration by Parts:

dx where u & v are differentiable functions.

Note : While using integration by parts, choose u & v such that
(a)  dx v is simple &

(b)  dx is simple to integrate.

This is generally obtained, by keeping the order of u & v as per the order of the letter in ILATE, where

I – Inverse function
L – Logarithmic function
A – Algebraic function
T – Trigonometric function
E – Exponential function

Remember This:

Proof:

Integrating by parts taking sin bx as the second function,

Again integrating by parts taking cos bx as the second function, we get

Transposing the term -a2/b2 I to the left hand side, we get

### Solved Examples

Ex.1 Integrate xlog x

Sol.

Ex.2 Evaluate dx

Sol.

Put sec–1 x = t so that

Then the given integral  =

= t (log t – log e) + c  = sec–1 x (log (sec–1 x) – 1) + c

Ex.3 Evaluate dx.

Sol.

Put x = cos θ so that dx = - sin θ dθ. the given integral

Ex.4 Evaluate

Sol.

We have

[ x3 = x(x2 + 1) - x]

integrating by parts taking x2 as the second function

Ex.5 Evaluate dx.

Sol.

(put, 2x + 2 = 3 tanθ ⇒ 2 dx = 3 sec2θ dθ )

## Integration by Partial Fractions

We know that a Rational Number can be expressed in the form of p/q, where p and q are integers and q≠0. Similarly, a rational function is defined as the ratio of two polynomials which can be expressed in the form of partial fractions: P(x)/Q(x), where Q(x)≠0.

1. Proper partial fraction: When the degree of the numerator is less than the degree of the denominator, then the partial fraction is known as a proper partial fraction.
2. Improper partial fraction: When the degree of the numerator is greater than the degree of denominator then the partial fraction is known as an improper partial fraction. Thus, the fraction can be simplified into simpler partial fractions, that can be easily integrated.

In this section, we show how to integer any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions 2/(x – 1) and 1/(x + 2) to a common denominator we obtain

If we now reverse the procedure, we see how to integrate the function on the right side of this equation
To see how the method of partial fractions works in general, let's consider a rational function
Where P and Q are polynomials. It's possible to express f as sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper. Recall that if P(x) =
where an ≠ 0, then the degree of P is n and we write deg (P) = n.
If f is improper, that is, deg(P)  deg (Q), then we must take the preliminary step of dividing Q into P  (by division) until a remainder R(x) is obtained such that deg (R) < deg(Q). The division statement is

where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required.

### Solved Examples:

Ex.1 Evaluate dx.
Sol. Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write

The next step is to factor the denominator Q(x) as far as possible . It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors

(of the form ax2 + bx + c, where b2 – 4ac < 0).

For instance, if Q(x) = x4 – 16, we could factor it as Q(x) = (x2 – 4) (x2 + 4) = (x – 2) (x + 2) (x2 + 4)
The third step is to express the proper rational function R(x)/Q(x) (from equation 1) as a sum of partial fractions of the form

A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur.

### Case I : The Denominator Q(x) is a product of distinct linear factors.

This means that we can write Q(x) = (a1x + b1) (a2x + b2) ... (akx + bk)
where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1,A2,...,Ak such that.

These constants can be determined as in the following example.

Ex.2 Evaluate  dx.

Sol. Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.

We factor the denominator as 2x3 + 3x2 – 2x = x(2x2 + 3x – 2) = x(2x – 1) (x + 2)

Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form.

To denominator the values of A, B and C, we multiply both sides of this equation by the product of the denominators, x(2x – 1) (x + 2), obtaining.

(4) x2 + 2x – 1 = A(2x – 1) (x + 2) + Bx(x + 2) + Cx(2x – 1)

Expanding the right side of equation 4 and writing it in the standard form for polynomials, we get

(5) x2 + 2x – 1 = (2A + B + 2C) x2 + (3A + 2B – C) x – 2A

The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of xon the right side, 2A + B + 2C, must equal the coefficient of x2 on the left side-namely, 1. Likewise. The coefficients of x are equal and the constant terms are equal. This gives the following system of equation for A, B and C.

2A + B + 2C = 1  , 3A + 2B – C = 2  , – 2A = –1

Solving, we get

In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2dx and dx = du/2.

### Case II : Q(x) is a product of linear factors, some of which are repeated

Suppose the first linear factor (a1x + b1) is repeated r times, that is, (a1x + b1)r occurs in the factorization of Q(x). Then instead of the single term A1/(a1x + b1) in equation 2, we would use

(6)

By way of illustration, we could write

but we prefer to work out in detail a simpler example.

Ex.3 Evaluate  dx

Sol.

The first step is to divide. The result of long division is

The second step is to factor the denominator Q(x) = x3 – x2 – x + 1. Since Q(1) = 0, we know that x – 1 is a factor and we obtain x3 – x2 – x + 1 = (x – 1) (x– 1) = (x – 1) (x – 1)(x + 1) = (x – 1)2 (x + 1) Since the linear factor x – 1 occurs twice, the partial fraction decompositoin is

Multiplying by the least common denominator (x – 1)2 (x + 1), we get

(7) 4x = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)2 = (A + C) x2 + (B – 2x) x + (–A + B + C)

Now we equate coefficients : A + C = 0,B – 2C = 4, –A + B + C = 0

Solving, we obtain A = 1, B = 2, and C = – 1, so

### Case III : Q(x) contains irreducible quadratic factors, none of which is repeated

If Q(x) has the factor ax2 + bx + c, where b2 – 4ac < 0, then in addition to the partial fractions in equation 2 and 6, the expression or R(x)/Q(x) will have a term of the form.

(8)  where A and B are constants to be determined. For instance, the function givenby f(x) = x/[(x – 2)(x2 + 1) (x2 + 4)] has a partial fraction decomposition of the form

The term given in (8) can be integrated by completing the square and using the formula.

Ex.4 Evaluate  dx
Sol.

Since x3 + 4x = x(x2 + 4) can't be factored further, we write

Multiplying by x(x2 + 4), we have 2x2 – x + 4 = A (x2 + 4) + (Bx + C) x = (A + B) x2 + Cx + 4A Equating coefficients, we obtain A + B =2 C = –1 4A = 4

Thus A = 1, B = 1 and C = –1 and

In order to integrate the second term we split it into to parts

We make the substitution u = x2 + 4 in the first of these integrals so that du = 2x dx. We evaluate the second integral by means of Formula 9 with a = 2.

### Case IV : Q(x) Contains A repeated irreducible quadratic factor.

If Q (x) has the factor (ax2 + bx + c)r, where b2 – 4ac < 0, then instead of the single partial fraction

occurs in the partial fraction decomposition of R(x)/Q(x), each of the terms in (10) can be integrated by first completing the square.

Ex.5 Write out the form of the partial fraction decomposition of the function

Sol.

Ex.6 Evaluate

Sol. The form of the partial fraction decomposition is

Multiplying by x(x2 + 1)2, we have –x3 + 2x– x + 1 = A(x2 + 1)2 + (Bx + C) x (x2 + 1) + (Dx + E)x

= A(x+ 2x2 + 1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex = (A + B) x4 + Cx3 + (2A + B + D) x2 + (C + E) x + A

If we equate coefficient, we get the system    A + B = 0 ,C = –1  ,2A + B + D = 2  ,C+E =–1 , A=1

Which the solution A = 1, B = –1, D = 1, and E = 0. Thus

We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral

could be evaluated by the method of case III, it's much easier to observe that if u = x(x+ 3) = x3 + 3x, then du = (3x+ 3) dx and so

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## Mathematics (Maths) for JEE Main & Advanced

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## FAQs on Integrals of Some Functions - Mathematics (Maths) for JEE Main & Advanced

 1. What are some standard integrals that are commonly used in calculus?
Ans. Some standard integrals include integration of basic functions like polynomials, trigonometric functions, exponential functions, and logarithmic functions.
 2. How can we integrate irrational functions in calculus?
Ans. Irrational functions can be integrated by using techniques such as substitution, partial fractions, trigonometric identities, and integration by parts.
 3. How do we integrate expressions of the form ∫(ax + b)(cx + d)dx in calculus?
Ans. Expressions of the form ∫(ax + b)(cx + d)dx can be integrated by expanding the product, simplifying the terms, and then integrating term by term.
 4. What is the process of integration by substitution and when is it used in calculus?
Ans. Integration by substitution involves replacing a variable in the integral with a new variable to simplify the integrand. This technique is used when the integrand can be expressed in terms of the new variable.
 5. Can you explain the concept of integration by parts and provide an example in calculus?
Ans. Integration by parts is a technique used to integrate the product of two functions. It involves applying the formula ∫udv = uv - ∫vdu, where u and v are functions of x. An example of integration by parts is integrating ∫x*sin(x)dx.

## Mathematics (Maths) for JEE Main & Advanced

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