Integrals of Some Functions | Mathematics (Maths) Class 12 - JEE PDF Download

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 = t^m, where m is the least common multiple of the numbers q₁, q₂, ...., q_k.
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 = t² – 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 = t^k, where k is the common denominator of the fractions and n.
Case II : is an integer. We put a + bxⁿ = t^α, where α is the denominator of the fraction p.
Case III :+ p is an integer we put a + bxⁿ = t^αxⁿ, 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 = x^4/2,

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 = x⁴ + 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 u² = x + 4, so x = u² –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 xⁿ log 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:

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) ... (a_kx + 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,...,A_k 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 x² on 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

(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) (x² – 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

(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.

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 + 2x² – x + 1 = A(x² + 1)² + (Bx + C) x (x² + 1) + (Dx + E)x

= A(x⁴ + 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(x² + 3) = x³ + 3x, then du = (3x² + 3) dx and so