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Chapter 7.  Techniques of Integration

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Chapter 7.  Techniques of Integration

7.1 Integration by Parts   http://youtu.be/WX-6C9tCneE

문제풀이 by 이인행  http://youtu.be/jKCAGJ4HqvQ

7.2 Trigonometric Integrals   http://youtu.be/sIR0zNGQbus

문제풀이 by 김태현  http://youtu.be/ytETYf1wLbs

7.3 Trigonometric Substitution   http://youtu.be/avTqiEUi8u8

문제풀이 by 이훈정  http://youtu.be/utTQHIabTyI

7.4 Integration of Rational Functions by the Method of Partial Fractions

문제풀이 by 장재철  http://youtu.be/SkNW_bax0YI

7.5 Guidelines for Integration   http://youtu.be/Fgn8U4We60o

문제풀이 by 김대환 http://youtu.be/-N9Fe_Arp2c

7.6 Integration Using Tables    http://youtu.be/tn9jLkgTMp8

문제풀이 by 조건우  http://youtu.be/EnEQ9ZS3B_k

7.7 Approximate Integration     http://youtu.be/hg2pw1n1cZI

7.8 Improper Integrals      http://youtu.be/rquxbYrC0Yc

문제풀이 by 이송섭  http://youtu.be/C3kb4c9nLXM

문제풀이 by 이인행 http://youtu.be/dfSkjvmSXYo

7.1 Integration by Parts

There are many integrals that cannot be easily evaluated using basic formulas or integration by substitution. For example, cannot be evaluated with what we know at this point. The integral can be solved by introducing an integration technique called integration by parts.

We have already observed that every differentiation rule gives rise to a corresponding integration rule. For example, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. Integration by parts is one of the integration techniques that comes from the Product Rule for differentiation. The Product Rule for differentiation is given by

Integrating with respect to gives us

or

.

Note that the constant of integration is not ignored! It is still there, we simply did not write it! Solving for the first integral on the left-hand side yields the formula for integration by parts;

1.

Integration by parts is thus a technique for evaluating integrals whose integrand is the product of two functions. If we choose and , this formula then takes a simpler form:

2.

To apply integration by parts, we need to make wise choices for and . The integral on the right-hand side should be the one that is simpler than the original integral and is easier to evaluate. A rule of thumb is that is chosen to be a function that becomes simpler when it is differentiated (or at least not more complicated) as long as can be readily integrated to give . (It is customary to write as .)

EXAMPLE 1

Evaluate .

Solution.  1.Take and .

Then and .

Using formula 1. , we have .

2.Let and . Then and and so

.

Remark  If we choose and . Then and and we have . This is more complicated than the original one. That is why we need to make an appropriate choice for and . In particular, if then . Thus we have,

EXAMPLE 2

Find .

Solution.  Let and , then and .

Integrating by parts, we get

.

EXAMPLE 3

Evaluate .

Solution.  Let , then , .

Thus . We have

.

Hence, .

var('x, A')

assume(A>0)

f(x)= sqrt(x^2 + A)

integral(f(x), x)

Answer : 1/2*A*arcsinh(x/sqrt(A)) + 1/2*sqrt(x^2 + A)*x             ■

EXAMPLE 4

Evaluate .

Solution. Let and . Then and Integrating by parts gives

var('x')

f(x)= (x^3)/((1- x^2)^(-1/2))

print integral(f(x), x)

Answer : -1/5*(-x^2 + 1)^(3/2)*x^2 - 2/15*(-x^2 + 1)^(3/2)

EXAMPLE 5

Evaluate .

Solution. Let and , then and .

Hence, .

EXAMPLE 6

Evaluate .

Solution.  Let and , then and .

EXAMPLE 7

Evaluate , . (See Figure 1.)

Solution.

Figure 1

Let and , then , .

Integrating by parts, we obtain .

Using integration by parts to evaluate yields

,

.

Substituting this into the equation above, we get

.

var('a, b, x')

f(x)= e^(a*x) * cos(b*x)

print integral(f, x)

Answer :  (a*cos(b*x) + b*sin(b*x))*e^(a*x)/(a^2 + b^2)

Using FTC 2 and integration by parts, definite integrals can be evaluated as follows:

3.

In particular, the following is a special e of the above.

EXAMPLE 7

Evaluate .

Solution.  Let , , then , .

So, .

Since , we finally get. ■

Recall that

4.

Proof  Multiply and divide by and use the Substitution Rule

.

Set , then we find .

By substituting and , we have

after substituting .          ■

EXAMPLE 8

Evaluate .

Solution.

f=sec(x)^3

integral(f, x)

Answer :  -1/2*sin(x)/(sin(x)^2 - 1) - 1/4*log(sin(x) - 1) + 1/4*log (sin(x) + 1)

Another method: Let , , then , . Using integrating by parts gives

.

Solving for from this equation, we obtain

.

EXAMPLE 9

Evaluate the integral using integration by parts.

Solution.

integral(x*log(x-1),x)

var('x')

u(x)=x; V(x)=log(x-1)

v(x)=diff(V(x), x)

U(x)=integral(u(x), x)

I1=U(x)*V(x)-integral(U(x)*v(x), x)  # integration by parts

I2=integral(u(x)*V(x), x)

print I1

print I2

Answer :  1/2*x^2*log(x - 1) - 1/4*x^2 - 1/2*x - 1/2*log(x – 1)

1/2*x^2*log(x - 1) - 1/4*x^2 - 1/2*x - 1/2*log(x – 1)

bool(I1==I2)

7.1 EXERCISES (Integration by Parts)

1.

Solution.  .

.

2.

Solution.

integral((x^2+2*x+1)*e^(7*x), x)

Answer :  2/49*(7*x - 1)*e^(7*x) + 1/343*(49*x^2 - 14*x + 2)*e^(7*x) + 1/7*e^(7*x)

3.

Solution.

.

integral(x^2*cos(x),x,0,pi/2)

4.

Solution.

integral(x*sin(x), x)

5.

Solution.

.

6.

Solution.

var('x')

integral(x^3 * sin(x^2), x)

Answer : -1/2*x^2*cos(x^2) + 1/2*sin(x^2) + C

7.

Solution.

.

8.

Solution.

.

integral(ln(1+sqrt(x)), x)

Answer : x*log(sqrt(x) + 1) - 1/2*x + sqrt(x) - log(sqrt(x) + 1)

9.

Solution.

.

integral(ln(x)^2, x)

Answer :  (log(x)^2 - 2*log(x) + 2)*x

10.

Solution.

.

11.

Solution.  .

12.

Solution.

.

13.

Solution.

,

.

integral(x*sin(x), x)

14.

Solution.

15.

Solution.

.

(by Example 1 Section 7.3)

16.

Solution.  , , ,

… ①

Since

.

Therefore, from ①,

.

17.

Solution.

.

integral(x*(arccos(x)), x)

Answer :  1/2*x^2*arccos(x) - 1/4*sqrt(-x^2 + 1)*x + 1/4*arcsin(x)

18.

Solution.  .

integral (e^(-x)*(sin(x))^2, x)

Answer :  -1/10*(2*sin(2*x) - cos(2*x) + 5)*e^(-x)

19.

Solution.

integral (x*cos(x/2), x, 0, pi)

20.

Solution. .

var('x, a');

integral(sqrt(a^2-x^2), x)

21.

Solution. .

integral (x*e^(x), x, 0, 1)

Ansewr : 1

22.

Solution.  .

integral((x^3)*(ln(x)^2), x, 1, 2)

23.

Solution.  .

integral (arctan(x), x, 0, 1)

24.

Solution.

.

25. (a) Establish the reduction formula.

.

(b) Use part (a) to evaluate .

(c) Use parts (a) and (b) to evaluate .

Solution.  (a).

.

Therefore, .

(b) .

(c)

.

26. (a) Establish the reduction formula.

(b) Find the integral explicitly for .

Solution.  .

(a) .

(b) .

27. (a) Establish the reduction formula

.

(b) Find the integral for .

Solution. (a)

.

.

.

(b)

.

7.2 Trigonometric Integrals

A trigonometric integral is an integral whose integrand contains powers of one or more trigonometric functions. This integral can be evaluated by making a clever substitution of trigonometric identities to turn the integrand into a new integrand that we can integrate.

There are guidelines to evaluate integrals of the form ,

where and are non-negative integers.

E I : is an odd positive integer.

If the power of cosine is odd , , factor out one cosine term and use to express the remaining expression in terms of the sine:

Then use substitution with .

EXAMPLE 1

Find .

Solution.  Let , then we have and so

.    ■

E II : is an odd positive integer.

If the power of sine is odd , , factor out one sine term and use to express the remaining factors in terms of cosine:

.

Then use substitution with .

EXAMPLE 2

Evaluate .

Solution.  Let , then we have and so

.    ■

E III : and are even positive integers.

For the integrals whose the integrand contains even powers of both sine and cosine functions, we can apply the following half-angle formulas

and ,

possibly several times, to reduce the powers in the integrand.

EXAMPLE 3

Find .

Solution.  Rewriting the integrand gives

.       ■

EXAMPLE 4

Evaluate .

Solution.  Rewrite and use the half-angle formula for cosine:

.

There are also guidelines to evaluate the integrals of the form .

E I : is an even positive integer.

If the power of secant is even , factor out a single and use the trigonometric identity to express the remaining factors in terms of

tan:

.

Then choose .

EXAMPLE 5

Find .

Solution.  Separate the term, and express the remaining factor in terms of tangent using the trigonometric identity . Let , then we have . We have

.     ■

E II : is an odd positive integer.

If the power of tangent is odd , in the integral , collect the term of and use the trigonometric identity to express the remaining factors in terms of :

. Then choose .

EXAMPLE 6

Evaluate .

Solution.  Separate the term, and convert the remaining power of tangent to an expression involving only the secant using the trigonometric identity . Let , so . We have

.    ■

Recall that

1.

EXAMPLE 7

Evaluate .

Solution.

.          ■

E Ⅲ : is an even integer and is an odd positive integer.

If the power of tangent is even and the power of secant is odd , in the integral ,

collect the term of and use the trigonometric identity to express the remaining factors in terms of :

EXAMPLE 8

Evaluate .

Solution.

.        ■

Note  If an even power of tangent appears with an odd power of secant, express the integrand completely in terms of sec and use integration by parts.

Note Using the trigonometric identity , integrals of the form  can be evaluated using similar methods. The integrals

(a)   (b)    or   (c)

can be evaluated using the following trigonometric identities

(a)

(b)

(c) .

EXAMPLE 9

Find .

Solution.

7.2 EXERCISES (Trigonometric Integrals)

1.

Solution.  .

2.

Solution.

var('t')

integral(sin(t)^5,t)

Answer :  -1/5*cos(t)^5 + 2/3*cos(t)^3 – cos(t)

.

Thus, .

.

Thus, .

.

Therefore,

.

3.

Solution.

integral(sin(x)^5*cos(x)^4,x)

Answer :   -1/9*cos(x)^9 + 2/7*cos(x)^7 - 1/5*cos(x)^5

4.

Solution.

integral(sin(x)^6*cos(x)^3, x)x)

Let then .

5.

Solution.

integral(sin(x)^5/cos(x)^6,x)

Answer :  1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5

Let , then . Therefore,

.

6.

Solution.  .

7.

Solution.

integral(sin(x)^2*cos(x)^4,x)

Answer :  1/48*[sin(2*x)]^3 + 1/16*x - 1/64*sin(4*x)

8.

Solution.

integral(sin(3*x)^7*cos(3*x)^2,x)

Answer :  1/27*cos(3*x)^9 - 1/7*cos(3*x)^7 + 1/5*cos (3*x)^5 – 1/9*cos(3*x)^3

Let then .

9.

Solution.

integral(sin(4*x)^2,x)

Since .

.

10.

Solution.

var('t')

integral(sin(2*t)^4*cos(2*t)^4, t)

Answer :  3/128*t - 1/256*sin(8*t) + 1/2048*sin(16*t)

11.

Solution.

integral(sin(2*x)*cos(3*x), x)

12.

Solution.

var('y')

integral(sin(4*y)*cos(5*y), y)

13.

Solution.

integral(sin(x)^3*cos(x)^5, x, 0, pi/2)

14.

Solution.

integral(cos(x)^3*sin(x)^4, x, -pi/2, pi/2)

15.

Solution.

integral(x*cos(x)^2*sin(x), x)

Answer :  -1/12*x*cos(3*x) - 1/4*x*cos(x) + 1/36*sin (3*x) + 1/4*sin(x)

16.

Solution.

integral(cot(x)^4, x)

Answer :  1/3*(3*tan(x)^2 - 1)/tan(x)^3 + arctan(tan(x))

17.

Solution.

integral(tan(x)^3*sec(x)^(-1/2), x)

18.

Solution.

var('y')

integral(tan(3*y)^3*sec(3*y)^3, y)

19.

Solution.

var('t')

integral(tan(t)^(-3)*sec(t)^2, t)

20.

Solution.

integral(cot(2*x)^4, x)

Answer :  x + 1/6*(3*tan(2*x)^2 – 1)/tan(2*x)^3

21.

Solution.

integral(tan(x)^3*sec(x)^4, x, 0, pi/4)

22.

Solution.

integral(cot(x)*csc(x)^3, x)

23.

Solution.

integral(x*sec(x)*tan(x),x)

Answer : 1/2*(4*x*sin(2*x)*sin(x)+4*x*cos(2*x)*cos(x)+ (sin(2*x)^2 +cos(2*x)^2 + 2*cos(2*x) + 1)*log(sin (x)^2 + cos(x)^2 - 2*sin(x) + 1) -(sin(2*x)^2+ cos(2*x)^2 + 2*cos(2*x) + 1)*log(sin(x)^2 + cos(x)^2 + 2*sin(x) + 1) + 4*x*cos(x))/(sin(2*x)^2 + cos(2*x)^2 + 2*cos(2*x) + 1)

7.3 Trigonometric Substitutions

An integral that contains terms of the form , or can sometimes be evaluated by considering a new variable representing an angle, which may eliminate the square root function. We substitute a new variable for the original variable and transform the integral into a trigonometric one (hence the name trigonometric substitution). Once the trigonometric integral has been found, we reverse the substitution in order to find the original integral in terms of the original variable. The three typical expressions, the corresponding substitution and trigonometric identities to use for each e are given in the following table.

Table of Trigonometric Substitutions

 Expression Substitution Identity , , , or

EXAMPLE 1

Find .

Solution.

Substitute , where . (See Figure 1.)

Figure 1

Then and

.

Note that because . Thus the Substitution Rule gives

.

Since  , , and .

We obtain

.           ■

EXAMPLE 2

Find .

Solution.  Substitute (). Then . Since , and for ,

we have

.

Since  , , and , we obtain

.

var('x, a')

integral(sqrt(a^2-x^2), x)

Answer :  1/2*a^2*arcsin(x/sqrt(a^2)) + 1/2*sqrt(a^2 – x^2)*x     ■

EXAMPLE 3

Find .

Solution. Completing the square under the square root we have

.

This suggests that we make the substitution . Then and , so.

Let where . Then and , so

.     ■

EXAMPLE 4

Evaluate .

Solution.

Substitute (). (See Figure 2.)

Figure 2

Then and

.

Since for

Thus, we have.

Drawing a right triangle such that :

we observe from this triangle that     so

,

where .          ■

EXAMPLE 5

Evaluate .

Solution.  We put .   Then . As a definite integral, we can convert the limits of into values of . So,

.     ■

EXAMPLE 6

Find .

Solution.  Substitute , where or .

Then and .

Since for or , we have

.

A right triangle with allows us to conclude that , , and , so we obtain.

EXAMPLE 7

Evaluate the integral using a trigonometric substitution. ().

Solution.  Let , then

and .

Thus

Hence .

var('x')

f=1/(x*sqrt(x^2-1))

assume(x>0)

integral(f, x)

Note  The following will apply in example 8 and 9. Let . Then,

because .

,

and   .

EXAMPLE 8

Find .

Solution.  Let , then and .

.

EXAMPLE 9

Find .

Solution. Let , then and .

.

var('x')

integral(1/(sin(x) + cos(x)), x)

Answer : -1/2*sqrt(2)*log(-(2*sqrt(2) - 2*sin(x)/(cos(x) + 1)+ 2) / ((2*sqrt(2)) + 2*sin(x)/(cos(x) + 1) - 2))

7.3 EXERCISES (Trigonometric Substitutions)

1.

Solution.  ,

.

integral((sqrt(1-x^2))/(x^2), x)

Answer : -sqrt(-x^2 + 1)/x – arcsin(x)

2.

Solution.

integral(1/(sqrt(4+x^2)*(4-x^2)), x)

Answer :  1/16*sqrt(2)*arcsinh(2*x/abs(2*x + 4) - 4/abs

(2*x +4))+1/16*sqrt(2)*arcsinh(2*x/abs(2*x - 4) + 4/abs(2*x – 4))

3.

Solution.

.

integral(1/((sqrt(9+x^2))*(x^2)), x)

4.

Solution.

integral(1/(-x^2 + 1)^(3/2), x)

5.

Solution.

integral(1/sqrt(3*x^2 –2*x-1), x)

6.

Solution.

integral(1/3*x^2*sqrt(9-x^2), x)

Answer :  -1/12*(-x^2 + 9)^(3/2)*x + 3/8*sqrt(-x^2 + 9)*x + 27/8*arcsin(1/3*x)

7.

Solution.

integral(1/(x^2+4*x+13)^(3/2), x)

Answer :  1/9*x/sqrt(x^2 + 4*x + 13) + 2/9/sqrt(x^2 + 4*x + 13)

8.

Solution.

integral(x^2/(4-x^2)^(3/2), x)

9.

Solution.

integral(1/(x*sqrt(9+x^2)), x)

10.

Solution.

integral(1/(x*sqrt(9-x^2)), x)

11.

Solution.  .

.

var('x, a');

integral(1/sqrt(x^2-a^2), x)

Answer :  -1/2*a^2*log(2*x + 2*sqrt(-a^2 + x^2)) + 1/2*sqrt(-a^2 + x^2)*x

12.

Solution.

integral((sqrt(2*x^2-2)/x^4), x)

13.

Solution.

integral(x /((-x^2+1)^2), x)

14.

Solution.

integral(1/(sqrt(-x^2+1)), x)

15.

Solution.

integral(1/sqrt(x^2+3*x+1), x)

Answer : log(2*x + 2*sqrt(x^2 + 3*x + 1) + 3)

16.

Solution.

.

integral((3*x+1)/sqrt(-x^2-x+5), x)

Answer :  -3*sqrt(-x^2-x+5) +1/2*arcsin(-1/21*(2*x + 1)* sqrt(21))

17.

Solution.

integral(sqrt(x^2-1)/x, x)

Answer :  sqrt(x^2 - 1) + arcsin(1/abs(x))

18.

Solution.

.

integral((e^x)*sqrt(9-e^(2*x)), x)

Answer :  1/2*sqrt(-e^(2*x) + 9)*e^x + 9/2*arcsin(1/3*e^x)

19. (a) Prove that .

(b) Use to prove that

.

Solution.  (a) , .

.

var('x, a');

integral(sqrt(x^2+a^2), x)

Answer :  1/2*a^2*arcsinh(x/sqrt(a^2)) + 1/2*sqrt(a^2 + x^2)*x

(b) Use to prove that   .

.

7.4 Integration of Rational Functions

using Partial Fractions and Computer Algebra System

It is not obvious how one should evaluate the integral of a complicated rational function (a ratio of polynomials), for instance, .

However, the integral can be evaluated by rewriting the rational function as a sum of simpler fractions, called partial fractions, which are easily integrable.

The decomposition of a rational function into a sum of its partial fractions is called a partial fraction decomposition of the rational function.

To integrate a rational function defined by  ,

where and are polynomials, by its partial fraction decomposition, it is required that the numerator of the rational function has degree less than the denominator.

For example, the antiderivative is in the correct form for one to proceed with the partial fraction decomposition.

If it is not in the desired form, we divide by using long division until a remainder is obtained such that to acquire the following form of

1.

where  and are also polynomials. We now have

.

To evaluate the second integral, , on the right-hand side of the final expression, one must find the partial fraction decomposition of the integrand, which depends on the factored form of , then integrate.

It is known that any polynomial can be factored as a product of linear factors (of the form ) and irreducible quadratic factors (of the form , where ). So the next step is to factor the denominator, , completely. To rewrite the rational function  using partial fractions, we consider four possible es:

E I : The denominator is a product of distinct linear factors.

In this e we can write

where no factor is repeated (and no factor is a constant multiple of another).

In this e the partial fraction theorem states that there exist constants such that

2.           .

EXAMPLE 1

Evaluate .

Solution. Since the degree of the numerator is larger than the degree of the denominator, perform long division to obtain

.

EXAMPLE 2

Find .

Solution.  The degree of the numerator is smaller than the degree of the denominator, thus the long division is not needed. The denominator factors as follows: .

Since the denominator has three distinct linear factors, the integrand decomposes into three partial fractions of the form

3.          .

To determine the constants , , and , multiply both sides of this equation 3 by the product of the denominators, , resulting in

4.          .

Rewriting the right-hand side of  4 in the standard form for polynomials, we obtain

5.          .

Comparing the coefficients of powers of on both sides of 5, we observe that the coefficient of on the right side, , must equal the coefficient of on the left sidenamely, 1. Likewise, the coefficients of are equal and the constant terms are equal. This gives the following system of equations for , , and :

.

Solving it, we get

.

var('x')

exp= (x^2 +3*x-4)/(3*x^3 -4*x^2 - 5*x +2)

print integral (exp, x) exp.partial_fraction()

Answer :  2/5*log(x - 2) - 1/2*log(x + 1) + 13/30*log(3*x – 1)2/5/(x - 2) - 1/2/(x + 1) + 13/10/(3*x – 1)   ■

Note   Alternatively the coefficients , , in 4 can be determined by choosing values of that simplify the equation 4. If we substitute in 4, then the equation becomes , . Likewise, gives and gives , so and .

EXAMPLE 3

Find where .

Solution.  The method of partial fractions gives

and therefore . Put in this equation and get
, so . If we put , we get , so .

Thus.

var('a, b, x')

exp= 1/(x^2 + (b-a)*x – a*b)

print integral (exp, x)

exp.partial_fraction()

Answer :  -log(b + x)/(a + b) + log(-a + x)/(a + b)-1/((b + x)*(a - x))             ■

Example 3 provides us with an integration formula.

6.          .

E II : is a product of linear factors, some of which are repeated.

Suppose the first linear factor is repeated times; that is, occurs in the factorization of . Then instead of the single term in Equation 2, we use

7.

EXAMPLE 4

Evaluate .

Solution.  Since the degree of the numerator is larger than the degree of the denominator, we perform the long division to obtain

.

Next we factor, the denominator , which gives

.

Since the linear factor occurs twice, the partial fraction decomposition is

.

Multiplying by the least common denominator, , we get

8.

.

Now we equate coefficients to obtain:

(or put in 8: , put : , put : ).

Solving, we obtain , , , so

.

var('a, b, x')

exp= (2*x^4 + x^3 - 6*x^2 - 6*x -2)/(x^3 - 3*x – 2)

print exp.partial_fraction()      # partial fraction print

print integral (exp, x)

Answer :  2*x + 2/9/(x - 2) - 2/9/(x + 1) + 1/3/(x + 1)^2 + 1-1/3/(x + 1) + x^2 + x + 2/9*log(x - 2) - 2/9*log(x + 1)

E III : contains irreducible quadratic factors,

none of which is repeated.

If has the factor , where , then, in addition to the partial fractions in Equations 2 and 7, the expression for will have a term of the form

9.

where and are constants to be determined. The term given in 9 can be integrated by completing the square and using the formula

10.          .                                    ■

EXAMPLE 5

Find .

Solution.  Since cannot be factored further, we may write .

Multiplying by , we have

.

Equating coefficients, we obtain

, , , .

Thus , , , and and so

.

var('a, b, x')

exp= 1/(x^4 – 1)

print integral (exp, x)

exp.partial_fraction()      # partial fraction

Answer :  1/4*log(x - 1) - 1/4*log(x + 1) - 1/2*arctan(x)1/4/(x - 1) - 1/4/(x + 1) - 1/2/(x^2 + 1)

EXAMPLE 6

Find .

Solution. The partial fractions decomposition of the integrand, we will be of the form .

Note that is an irreducible quadratic factor. Multiplying the above equation by , we have

.

Equating coefficients, we obtain         .

Thus , and so

.

var('a, b, x')

exp= 1/(2*x^3 -3*x^2 + 2*x –1)

print integral (exp, x)

exp.partial_fraction()     # partial fraction

Answer :  1/2{ln}(x-1) - 1/4{ln}(2 x^2 - x + 1) - {3 sqrt7}/14   {tan^-1} (sqrt7/7  (4x - 1))-1/2*(2*x + 1)/(2*x^2 - x + 1) + 1/2/(x – 1)

Note  The general procedure for integrating a partial fraction of the form

where  is to complete the square of the denominator and then make a substitution that brings the integral into the form

.

Then the first integral is a logarithm and the second is expressed in terms of inverse tangent function.

E IV : contains a repeated irreducible quadratic factor.

If has the factor , where , then instead of the single partial fraction 9 the sum

11.

occurs in the partial fraction decomposition of . Each of the terms in 11 n be integrated by first completing the square.

EXAMPLE 7

What is the form of the partial fraction decomposition of the function?

Solution.

.

Working carefully by hand, or by employing Sage, we may find the coefficients to be , , , ,, , , , , . .

var('x')

exp=(x^4+x^3+x^2+x+2)/((x+1)*(x-2)*(x^2 – x+4)^4)

exp.partial_fraction()

Answer : -1/648*(5*x + 6)/(x^2 - x + 4) + 2/243/(x - 2) - 1/1944/(x + 1) - 1/108*(5*x + 6)/(x^2 - x + 4)^2 - 1/18*(5*x - 12)/(x^2 - x + 4)^3 + 1/3*(4*x - 3)/(x^2 - x + 4)^4

EXAMPLE 8

Find .

Solution.  The partial fraction decomposition of the integrand has the form

.

Multiplying this equation by , we have

.

If we equate coefficients, we get the following system of equations

,

which has the solution , , , , and .

Thus

.

Put .

Let . Then . Thus .

Put , .

Let . Then . Thus

.

Observe that.

Thus .

Let , ,

. Thus, .

Therefore,   and

.

Hence

.

var('x')

exp= (x^2 + 3)/((x+1)*(x^2 + 2*x + 3)^2 )

print integral (exp, x)

exp.partial_fraction()

Answer :  -1/4*sqrt(2)*arctan(1/2*(x + 1)*sqrt(2)) - 1/2*x/(x^2 + 2*x + 3) + log(x + 1) - 1/2*log(x^2 + 2*x + 3) -(x + 1)/(x^2 + 2*x + 3) + 1/(x + 1) - (x + 3)/(x^2 + 2*x + 3)^2  ■

Substitution Techniques

Some nonrational functions can be transformed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form , then the substitution of the form may be effective.

EXAMPLE 9

Find . , .

Solution. Let . Then and .

Hence we have

.

var('x')

exp= (1/x)* ( ((x+1)/(x-1))^(-1/2) )

print integral (exp, x)

Answer :  -log(sqrt((x + 1)/(x - 1)) - 1) + log(sqrt((x + 1)/(x - 1)) + 1) + 2*arctan(sqrt((x + 1)/(x – 1)))

EXAMPLE 10

Evaluate the integral.

Solution.  Let and use Sage to get

.

var('x')

f= x/((x+1)*(x+2)*(x+3))

print integral (f, x)

f.partial_fraction()

Answer :  -1/2*log(x + 1) + 2*log(x + 2) - 3/2*log(x + 3),-1/2/(x + 1) + 2/(x + 2) -

3/2/(x + 3)     ■

7.4 EXERCISES (Integration of Rational Functions by the Method of Partial Fractions and Computer Algebra System)

1.

Solution.

.

integral((x^3+1)/(x*(x-1)^3), x)

Answer :  -x/(x^2 - 2*x + 1) + 2*log(x - 1) - log(x)

2.

Solution.

integral(1/(x^3-x), x)

.

3.

Solution.

integral((x-1)/(x^2-4*x-5), x)

Answer : 2/3*log(x - 5) + 1/3*log(x + 1)

4.

http://matrix.skku.ac.kr/cal-lab/cal-7-4-4.html

Solution.

integral((5*x^3+2*x^2-12*x-8)/(x^4-8*x^2+16),

Answer :  -4/(x^2 - 4) + 3*log(x - 2) + 2*log(x + 2)

5.

Solution.

integral((x^2)/((x^2+1)*(x^2+4)), x)

6.

Solution.

integral((x^3+2*x^2)/(x^2-1), x)

Answer : 1/2*x^2 + 2*x + 3/2*log(x - 1) - 1/2*log(x + 1)

7.

Solution.

integral((1+2*x^2)/(x*(x^2+1)), x)

Answer :  1/2*log(x^2 + 1) + log(x)

8.

Solution.

.

integral(1/(x^3+1), x)

9.

Solution.

integral((3*x-4)/(x^2-6*x+9), x)

Answer :  -5/(x - 3) + 3*log(x - 3)

10.

Solution.  integral(1/(x^6-1), x)

(1/3*(2*x+1)*sqrt(3))+1/6*log(x-1)-1/6*log(x+1)+ 1/12*log (x^2-x+1)-1/12*log(x^2+x+1)

11.

Solution.

integral(1/(x^4+1), x)

1/4*sqrt(2)*arctan(1/2*(2*x-sqrt(2))*sqrt(2))+1/4*sqrt(2)*arctan(1/2*(2*x+sqrt(2))*sqrt(2))

12.

Solution.

.

integral((x^2)/((x^2+4)*(x^2+9)), x)

13.

Solution.

integral(x^2/(x^2+1)^2, x)

Answer :  -1/2*x/(x^2 + 1) + 1/2*arctan(x)

14.

Solution.

integral(1/((x-1)*(x^2+1)^2), x)

15.

Solution.

integral((2*x^2-x+1)/((x-1)^3*(x-2)^2), x)

Answer :  -(14*x^2 - 34*x + 19)/(x^3 - 4*x^2 + 5*x - 2) - 14*log(x - 2) + 14*log(x – 1)

16.

Solution.

integral((5*x+1)/(2*x^2+5*x+3), x)

Answer :  -4*log(x + 1) + 13/2*log(2*x + 3)

17.

Solution.

integral((sqrt(x^2+9))/(x), x)

Answer :  sqrt(x^2 + 9) - 3*arcsinh(3/abs(x))

18.

Solution.

integral(1/((x^3)*(-x+1)^3), x)

Answer :  -1/2*(12*x^3 - 18*x^2 + 4*x + 1)/(x^4 - 2*x^3 + x^2) - 6*log(x - 1)+ 6*log(x)

19.

Solution.

var('x, a, n');

integral(1/x*(x^n+a)^3, x)

.

.

.

20.

Solution.

integral(x^4/(x^3+1), x, 0, 1)

Answer :  -1/9*pi*sqrt(3) + 1/3*log(2) + 1/2

21.

Solution.

integral((2*x^2-x+4)/(x^3+4*x), x)

Answer :  1/2*log(x^2 + 4) + log(x) - 1/2*arctan(1/2*x)

22.

Solution.

integral(1/(1+x^2)^2, x)

Answer :  1/2*x/(x^2 + 1) + 1/2*arctan(x)

23.

Solution.

integral(1/((x-1)*sqrt(x+2)), x)

Answer :  1/3*sqrt(3)*log((sqrt(x + 2) - sqrt(3))/(sqrt(x + 2) + sqrt(3)))

24.

Solution.

integral(1/((sqrt(x))*(1-x^(1/3))), x)

25.

Solution.

.

26.

[Hint: Substitute ]

Solution.

integral(1/(x*(sqrt(x^2+2*x+2))), x)

7.5 Formulas for Integration  http://youtu.be/Fgn8U4We60o

Most functions can be differentiated by applying the formulas for differentiation. However, integrating a given function may not be as obvious as finding the derivative of a function. We may find integrals without much difficulty when an appropriate integration technique is recognized and correctly employed. So far, the method of substitution, integration by parts, trigonometric substitution and the method of partial fraction decomposition have been applied and their effectiveness in dealing with a variety of integration problems has been shown. In this section a collection of miscellaneous integrals (there is no particular order), is given. No hard and fast rules can be given as to which technique applies in a given situation, but one must be aware of the following basic integration formulas.

Table of Integration Formulas [Constant of integration is omitted.]

1.        2.

3. ,         4.

5. ,          6.

7.             8.

9.           10.

11.     12.

13.

14.   15.

16.   17.

18.         19.

20.

21.

22. ,

23. ,

24. ,

25. ,

26.

27.

There is no universal set of rules that indicate which method to employ for every integral. A guide to which techniques you might try for a particular integrand is given below.

Strategy for integration

1.Use algebraic manipulation and/or trigonometric identities to simplify the integrand as much as possible.

2.Determine a suitable substitution in the integrand whose differential also occurs, baring a constant factor.

3.Classify the integrand according to its form.

(a) If is a product of powers of and , and , and , then try trigonometric integrations.

(b) If is a rational function, the method of partial fractions  may be used.

(c) If is a product of a power of (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then consider integration by parts.

(d) When certain radicals appear as part of the integrand, the following substitutions may by helpful.

(ⅰ) If occurs, use a trigonometric substitution.

(ⅱ) If occurs, use the rationalizing substitution . Sometimes, this also works for .

4. If the first three es do not lead to an answer:

(a) When no substitution is obvious, try an appropriate substitution.

(b) Integration by parts should performed as many times as it is necessary.

(c) Relate the problem to previous problems. Use a method on a given integral that is similar to a method you have already used by expressing the given integral in terms of a previous one.

(d) Use several methods, a combination of two or more methods may be required to evaluate an integral, involving several successive substitutions of different types, or it may combine integration by parts with one or more substitutions etc.

EXAMPLE 1

Find .

Solution.  Rewrite the integral: .

The integral is now of the form with even. So,

.   ■

EXAMPLE 2

Find .

Solution.  Let . Then , , and

.          ■

Some of the following examples are not solved completely but some hints are given:

EXAMPLE 3

Find .

Solution.  Since the degree of the numerator is greater than the degree of the denominator, we perform long division and apply method of partial fractions.

EXAMPLE 4

Find .

Solution.  Using integrate by parts.

.           ■

EXAMPLE 5

Find .

Solution. Multiplying the numerator and denominator by , we have

.          ■

Note  The rationalizing substitution may be used even if it makes the problem more cumbersome.

, . So, we get

.

Polynomials, rational functions, power functions , exponential functions , logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions and all functions that can be obtained from these by the five operations of addition, subtraction, multiplication, division and composition are called elementary functions and are the objects differentiated or integrated in calculus course. If is an elementary function, then is also an elementary function but need not to be an elementary function. For instance, the indefinite integral can not be expressed in terms of elementary functions. Similarly the following integrals cannot be expressed in terms of elementary functions.

Remark  The majority of elementary functions that one may encounter in various branches of science and engineering do not possess elementary antiderivatives.

EXAMPLE 6

Evaluate the integral   .

Solution.  Let .

Use Sage to find and the integral.

var('x')

f= 1/((x^2 +1)*(x^2 + x))

print integral (f, x)

f.partial_fraction()

Answer : -1/2*log(x + 1) - 1/4*log(x^2 + 1) + log(x) - 1/2*arctan(x)-1/2*(x + 1)/(x^2 + 1)-1/2/(x + 1) + 1/x

7.5 EXERCISES (Formulas for Integration)

1.

Solution.

var('x, a');

integral(sqrt(a-x/a+x),x)

Answer :  -2/3*(a + x - x/a)^(3/2)/(1/a – 1)

2.

Solution.

Let .Then and .

Therefore we get:

.

Therefore, .

integral((1/x)*sqrt((x+1)/(x-1)),x)

Answer : -log(sqrt((x + 1)/(x - 1)) - 1) + log(sqrt((x + 1)/(x - 1)) + 1) -2*arctan(sqrt((x + 1)/(x- 1)))

3.

Solution. Let . Then , so .

.

integral(1/(sqrt(x)*(1-x^(1/3))),x)

Answer :  -6*x^(1/6)-3*log(x^(1/6) - 1) + 3*log(x^(1/6) + 1)

4.

Solution.

.

integral(1/(1+(x+1)^(1/3)), x)

Answer :  3/2*(x + 1)^(2/3) - 3*(x + 1)^(1/3) + 3*log((x + 1)^(1/3) + 1)

5.

Solution. Let . Then , so .

.

integral(1/(sqrt(x)*(1+x^(1/3))), x)

6.

Solution.

integral((x^(1/4))/(1+x), x)

Answer :  1/2*sqrt(2)*log(-sqrt(2)*x^(1/4) + sqrt(x) + 1)-1/2*sqrt(2)*log(sqrt(2)*x^(1/4) + sqrt(x) + 1) -sqrt(2)*arctan(-1/2*(sqrt(2) - 2*x^(1/4))*sqrt(2)) -sqrt(2)*arctan(1/2*(sqrt(2) +2*x^(1/4))*sqrt(2)) + 4*x^(1/4)

7.

Solution.

integral(sqrt(3+ln(x))/(x*(ln(x))^2), x)

Answer : 1/6*sqrt(3)*log((sqrt(log(x) + 3) - sqrt(3))/(sqrt (log(x) + 3) +sqrt(3))) - sqrt(log(x)+3)/log(x)

8.

Solution.

integral(sqrt(3*x-2-x^2), x)

Answer : 1/2*sqrt(-x^2 + 3*x - 2)*x - 3/4*sqrt(-x^2 + 3*x - 2) + 1/8*arcsin(2*x -3)

9.

Solution.

.

integral(x^3/sqrt(1-x^2), x)

Answer :  -1/3*sqrt(-x^2 + 1)*x^2 - 2/3*sqrt(-x^2 + 1)

10.

Solution.

.

integral(tan(x)^5/cos(x)^5, x)

11.

Solution.

.

integral(cos(x)^2/sin(x)^6, x)

12.

Solution.

integral(x*(x+1)^(1/5), x)

Answer :   5/11*(x + 1)^(11/5) - 5/6*(x + 1)^(6/5)

13.

Solution.

integral(x^2/sqrt(x^2+1), x)

Answer :   1/2*sqrt(x^2 + 1)*x - 1/2*arcsinh(x)

14.

Solution.

integral(x/sqrt(3-x^4), x)

15.

Solution.

.

16.

Solution.

integral((3*x^2-x+6)/((x+1)^2)*(x^2+1),x)

Answer : -20/(x+1) + x^3 - 7/2*x^2 + 20*x - 34*log(x + 1)

17.

Solution.

.

integral(e^(sqrt(x+5)),x)

Answer :  2*(sqrt(x + 5) - 1)*e^(sqrt(x + 5))

18.

Solution.

.

integral(e^(x^(1/4)),x)

Answer : 4*(x^(3/4)-3*sqrt(x) + 6*x^(1/4) - 6)*e^ (x^(1/4))

7.6 Integration Using Formulas

We may need to use an integral table to evaluate an integral. In this e we need to use substitution or algebraic manipulation to transform the given integral into one of the integrals listed in the table. (See Section 7.5.)

EXAMPLE 1

Find the volume of the solid obtained by rotating about the -axis the region bounded by the curves , , and . (See Figure 1.)

Figure 1

Solution.

The volume by the method of cylindrical shells is .

We know from the table .

Thus, the volume is

.  ■

EXAMPLE 2

Evaluate .

Solution.  We know from the table

.

Thus,

.                   ■

EXAMPLE 3

Evaluate .

Solution.  We know from the table that .

Using the reduction formula repeatedly, we have

.                  ■

EXAMPLE 4

Evaluate .

Solution. First complete the square:

If we make the substitution (so ), the integrand will involve the form .

.

We know from the table that.

Applying the formula with gives

.   ■

EXAMPLE 5

Evaluate the integral

.

Let then .

.

var('x')

f=arctan(exp(x))/exp(x)

integral(f, x)

Answer :   -e^(-x)*arctan(e^x) + x - 1/2*log(e^(2*x) + 1)

7.6 EXERCISES (Integration Using Formulas)

1.

Solution.  Let , then .

Since , .

Therefore,

.

integral(x^2/(sqrt(7-3*x^2)), x)

Answer :  -1/6*sqrt(-3*x^2 + 7)*x + 7/18*sqrt(3)*arcsin (1/7*sqrt (21)*x)

2.

Solution.

integral(1/((2*x-3)*(sqrt(-x^2+3*x-2))), x)

Answer: -log(sqrt(-x^2 + 3*x - 2)/abs(2*x - 3) + 1/2/abs(2*x - 3))

3.

Solution.

integral(x*(sqrt(x^2-4*x+7)), x)

Answer :  1/3*(x^2 - 4*x + 7)^(3/2) + sqrt(x^2 - 4*x + 7)*x - 2*sqrt(x^2 - 4*x + 7) + 3*arcsinh(1/3*(x –2)*sqrt(3))

4.

Solution.

integral(x^5*(sin(x)), x)

Answer :   5*(x^4 - 12*x^2 + 24)*sin(x) - (x^5 - 20*x^3 + 120*x)*cos(x)

5.

Solution.

integral(sin(x)^5*cos(x)^4, x)

Answer :  -1/9*cos(x)^9 + 2/7*cos(x)^7 - 1/5*cos(x)^5

6.

Solution.

integral((sec(x))^7, x)

Answer :  -1/48*(15*sin(x)^5 - 40*sin(x)^3 + 33*sin(x))/ (sin(x)^6 - 3*sin(x)^4 +3*sin(x)^2 – 1) - 5/32* log(sin(x) - 1) + 5/32*log(sin(x) + 1)

7.

Solution.

integral(x^3*arctan(3*x), x)

Answer :  1/4*x^4*arctan(3*x) - 1/36*x^3 + 1/108*x – 1/324*arctan(3*x)

8.

Solution.

integral(1/(e^x*sqrt(1+e^(2*x))), x)

9.

Solution.

integral(e^x*(2-3*e^(2*x))^(3/2), x)

Answer :  1/4*(-3*e^(2*x) + 2)^(3/2)*e^x + 3/4*sqrt (-3* e^(2*x) + 2)*e^x+1/2*sqrt(3)*arcsin(1/2* sqrt(6)*e^x)

10.

Solution.

.

integral(1/(x*((ln(x))^2+(ln(x)))), x)

Answer :   -log(log(x) + 1) + log(log(x))

11.

Solution.  Let , then .

.

integral(1/((sin(x))^3*(cos(x))^3), x)

Answer :  -1/2*(2*sin(x)^2 - 1)/(sin(x)^4 - sin(x)^2) - log(sin(x)^2 - 1) +2*log(sin(x))

12.

Solution.

integral(e^(2*x)*cos(3*x), x)

7.7 Approximate Integration

There are many definite integrals which are difficult or even impossible to evaluate since we are unable to find an antiderivative of the integrand or the function is given as a table of values of collected data. For instance, it is impossible to evaluate the integrals like and exactly. In these es we have no choice but to find approximate values of these definite integrals.

From the definition of the definite integral, we can obtain approximate integration formulas. Recall the definite integral is defined as a limit of Riemann sums, and any Riemann sum could be used as an approximation to the integral.  If we divide the interval into subintervals of equal length , we then have

where is any point in the subinterval . A choice for results in an approximate integration formula. Among all possible choices for , the e where is chosen to be the midpoint of the subinterval is known as the ‘midpoint approximation’.

1. Midpoint Rule

where and midpoint of .

(a) Left endpoint approximation   (b) Right endpoint approximation   (c) Midpoint approximation

We could see intuitively that using trapezoids instead of rectangles over each subinterval will give a better approximation to the area of the corresponding subregion. Based on this observation, we can construct another approximate integration formula

.

This approximation is called the Trapezoidal Rule.

2. The Trapezoidal Rule

where and . (See Figure 2.)

Figure 2 Trapezoidal approximation

EXAMPLE 1

Approximate the integral using (a) the Trapezoidal Rule and (b) the Midpoint Rule with .

Solution.

(a) With , and , we have , and so the Trapezoidal Rule gives

.

This approximation is illustrated in Figure 3

Figure 3

(b) The midpoints of the five subintervals are 1.1, 1.3, 1.5, 1.7, and 1.9, so the Midpoint Rule gives

.

This approximation is illustrated in Figure 4.

Figure 4

By the FTC, compute explicitly the exact value of the integral:

value of the approximationerror

error = exact value – approximate value

Now the errors in the Trapezoidal and Midpoint Rule approximations for are

and .

In general, we have and .

The left end point and right end point rules may be found in

http://en.wikipedia.org/wiki/Riemann_sum. The following tables show the results of calculations(similar to those in Example 1) for and 20 for the left, right endpoint approximations, Trapezoidal and Midpoint Rules.

 5 10 20 0.580783 0.538955 0.519114 0.430783 0.463955 0.481614 0.505783 0.501455 0.500364 0.497127 0.499273 0.499817

 5 10 20 0.080783 0.038955 0.019114 0.069216 0.036044 0.018385 0.005783 0.001455 0.000364 0.002872 0.000726 0.000182
http://sage.luther.edu:8443/home/pub/98/

It should be mentioned that  in all of the above approximation methods, more accurate approximations are obtained when the value of is increased, but this results in an accumulated round-off error. The Trapezoidal and Midpoint Rules are known to be much more accurate than the endpoint approximations which can be observed from the tables above. The error estimates for the approximate integration values provided by the Trapezoidal and Midpoint Rules are established from text books on numerical analysis, which are given as follows:

3. Error Bounds

Suppose for . If and are the errors for the Trapezoidal and Midpoint Rules, respectively then

and   .

Here, can be any number larger than all the values of , but smaller values of give better error bounds.

In the error estimates given above, observe that the Midpoint Rule is more accurate than the Trapezoidal Rule. Let us apply this error estimate, for the Trapezoidal Rule approximation in Example 1. If then and     . Since , we have , so  .

Therefore, taking , , , and in the error estimate 3, we find that .

If we compare this error estimate of with the actual error of around , then we can see that the actual error is substantially less than the upper bound for the error provided by 3.

EXAMPLE 2

Determine how large should be so that the Trapezoidal and Midpoint Rule approximations for are accurate to within 0.0001.

Solution. We know (from the above calculations) that for , so take , , and in 3. Since the size of the error should be less than 0.0001, therefore choose so that .

Solving the inequality for , we get or .

Thus, will ensure the desired accuracy.    ■

Note  A larger value for would suffice, but 71 is the smallest value for which the error bound formula will guarantee us accuracy to within 0.0001.

For the same accuracy with the Midpoint Rule we choose so that

, which gives .

EXAMPLE 3

(a) Approximate the integral using the Midpoint Rule with .

(b) Find an upper bound for the error involved in this approximation.

Solution.

(a) With , and , the Midpoint Rule gives

.

This approximation is illustrated in Figure 5.

Figure 5

(b) Since , we have and .

Also, since , we have and so

Taking , , , and in the error estimate 3, an upper bound for the error is  .

Simpson’s  Rule-Approximation using Parabolas

By approximating a curve using parabolas instead of straight line segments that produce trapezoids, another well-known rule for approximate integration is obtained. As earlier, we make a partition of the interval into subintervals of equal length , but this time we require that is an even number.  We approximate the curve by a parabola on each consecutive pair of intervals as shown in Figure 6. If , then is the point on the curve lying above . A typical parabola passes through three consecutive points , , and .

Figure 6                                                   Figure 7

We consider the e where , , and . See Figure 7. The parabola through , , and has an equation of the form . So the area under the parabola from to is

.

Since the parabola passes through , , and , we have

and therefore .

Thus, the area under the parabola is  .

Similarly, the area under the parabola through , , and from to
is  .

Computing the areas under all the parabolas in this manner and adding the results, we obtain

Note the pattern of coefficients in Simpson’ rule: 1, 4, 2, 4, 2, 4, 2, , 4, 2, 4, 1.

4. (Composite) Simpson’s Rule

where is even and .

The approximations in Simpson’s Rule are weighted averages of those in the Trapezoidal and Midpoint Rules: .

* Run the following Simpson’s Rule code in http://sage.skku.edu/

def integrate(y_vals, h):

i=1

total=y_vals[0]+y_vals[-1]

for y in y_vals[1:-1]:

if i%2 == 0:

total+=2*y

else:

total+=4*y

i+=1

y_values=[13, 45.3, 12, 1, 476, 0]

interval=1.2

area=integrate(y_values, interval)

print "The area is", area   # The area is 469.680

EXAMPLE 4

Approximate using Simpson’s Rule with .

Solution. Here , , and Simpson’s Rule, we obtain

.

Observation In Example 4, observe that Simpson’s Rule gives a much better approximation to the exact value of the integral than does the Trapezoidal Rule or the Midpoint Rule .

EXAMPLE 5

Given the values of the function at the following points,

 1 1.2 1.4 1.6 1.8 2.71828 2.30097 2.04272 1.86824 1.7429

approximate using Simpson’s Rule.

Solution. With and the estimate of the integral by Simpson’s Rule is given by

.

5. Error Bound for Simpson’s Rule

Assume that for . Let be the error involved in using Simpson’s Rule, then .

But it is often very difficult to compute the fourth derivatives and obtain a good upper bound for by hand. It is recommended to use  when it is needed.

EXAMPLE 6

Determine how large should be so that the Simpson’s Rule approximation for is accurate to within 0.0001?

Solution. Here then .  Since , we have and so .

So take in 5. Thus, for an error less than 0.0001 we should choose such that .

This gives or .

Therefore, ( must be even) gives the desired accuracy.

Observation Compare this with Example 2, where we obtained for the Trapezoidal Rule and for the Midpoint Rule.

EXAMPLE 7

(a) Approximate the integral using Simpson’s Rule with .

(b) Determine the error involved using this approximation.

Solution. (a) If , then and Simpson’s Rule gives

.

(b) The fourth derivative of is .

Since , we have .

Therefore, taking , , , and in 4, we observe that the error is at most .

Thus, correct to three decimal places, we have .  ■

EXAMPLE 8

Approximate the given integral with the specified by the Trapezoidal Rule.

, .

Figure 8

Solution.

With , and , we have , and so the Trapezoidal Rule gives

.

0.3*(exp(-1)+2*exp(-0.4^2)+2*exp(-0.2^2)+2*exp(-0.8^2)+2*exp(-1.4^2)+exp(-2^2)).n()

var('x')

f=exp(-x^2)

integral(f, x, -1, 2).n()

We see some differences in these results.         ■

We may exercise numerical integrals with various rules in

Here procedures for left end point rule, right end point rule, mid point rule, trapezoidal rule and Simpson’s rule are given.

Remark  One can increase degree of the polynomial of approximation to get even better approximation rules. Approximation using 3rd degree polynomials is called Simpson’s Rule for Numerical Integration1). The gives Boole’s rule of integration and for gives called Weddle's rule of integration.

7.7 EXERCISES (Approximate Integration and )

1-6. Find the integral using the midpoint, trapezoidal and Simpson’s rule for the given .

1. ,

Solution.

Use sin(x^3) in http://matrix.skku.ac.kr/cal-lab/Area-Sum.html  with midpoint and trapezoidal rules.

(a) 0.232771 with midpoint rule

(b) 0.2359771 with trapezoidal rule

(c) 0.233760

2. , (2.58862863250718)

Solution.

Use e^(3*x)*sin(2*x) in http://matrix.skku.ac.kr/cal-lab/Area-Sum.html  with midpoint and trapezoidal rules.

(a)2.57670 with midpoint rule

(b)2.612462 with trapezoidal rule <Error = 0.0238343387259241>

(c) 2.588559

3. ,

Solution. (a) 0.919952  (b) 0.927027  (c) 0.925237

4. ,

Solution. (a) 0.272198 (b) 0.272198 (c) 0.272198

5. ,

Solution. (a) 0.457277 (b) 0.458528 (c) 0.458114

6. ,

Solution. (a) 1.182973 (b) 1.160116 (c) 1.169130

7. (a) Determine the approximations and for .

(b) Find the errors involved in the approximations of part (a).

(c) Determine how large must be so that the approximations and to the integral in part (a) are accurate to within 0.00001?

Solution. (a) ,

(b)

For , we must choose so that solving this, so that . For , we must choose so that solving this gives, so that .

8.(a) Determine the approximations and for and for and the corresponding errors and .

(b) Compare the actual errors in part (a) with the error estimates given by 3 and 4.

(c) Determine how large must we choose so that the approximations , , and for the integral in part (a) are accurate to within 0.00001?

Solution. (a) , ,

(b)Since 3 and 4 gives  ,

.

(c) For , find so that , . For , find so that . For , find so that .

9. Given the function at the following values, approximate using Simpson’s Rule.

 1.8 1.9 2 2.1 2.2 2.3 2.4 0.028561 0.020813 0.015384 0.011525 0.008742 0.006709 0.004079

Solution. Plot and use Simpson’s Rule

.

7.8 Improper Integrals  http://youtu.be/rquxbYrC0Yc

For the definite integral, , (so far) we have assumed that the length of the integration interval is finite and that the range of the integrand remains finite on this finite interval. Sometimes, however, we may be curious about what happens if either one or both of these assumptions is not satisfied. In fact, these situations occur frequently in applied science fields and probability theory. This question requires a new interpretation to the symbol . If either or both the limits of integration becomes infinite, or if  the function becomes infinite somewhere in the interval , then the symbol is called an improper integral.

Type I: Improper Integrals with an Infinite Interval

Consider an infinite region that lies under the curve above the -axis, and on the right of the line . We would like to find the area of this infinite region .  Since is infinite region, one might be tempted to say the area is infinite, but its area is in fact finite, which will be justified by what follows. To find the area of an infinite region ,  our approach is based on the notion of limit. First, we find the area of the part of that lies to the left of the line (shaded in Figure 1.):

Figure 1

, which is a definite integral on a finite interval. We next find the limit of as .

Since the limit exists, it is natural to assign this value to the area of the infinite region and therefore we write Area of .

The discussion given above can be extended to any function . Therefore, we

make the following definition.

DEFINITION1  Definition of an Improper Integral of Type 1

(a) The symbol is defined by the equation   .

(b) The symbol is defined by the equation

.

The improper integrals and are called convergent if the

corresponding limit exists; otherwise they are called divergent.

(c) If both and are convergent, then we can define

.

EXAMPLE 1

Is the integral convergent?

Solution.  We know that .

p1=plot(1-cos(x), x, -10,10, color='blue')

print limit(1-cos(x), x=00)

show(p1, ymax=3, ymin=-0.5, aspect_ratio=1)

Figure 2

The improper integral is divergent since the limit does not exist.

EXAMPLE 2

Find .

Solution.

■

EXAMPLE 3

Find .

Solution.  We know that .

Evaluating the integrals on the right-hand side separately:

and

.

The given integral is convergent since both of these integrals are convergent.

Therefore,.            ■

EXAMPLE 4

Determine the values of for which the integral is convergent.

Solution. When ,

.

When ,

If , then , so as and .

Therefore if , and so the integral converges.

But if , then so as and .

Therefore if and the integral diverges. Thus

.       ■

As a consequence of Example 4, we know converges, but diverges.

Type 2: Improper Integrals for which the Integrand becomes Infinite

Consider the integral . Observe that this is not a definite integral since the integrand is not continuous at and becomes infinite as approaches . Yet we can find the area under the graph of the  integrand on the interval . How we do this is presented in the following definition.

Figure 3

2. Definition of an Improper Integral of Type 2

(a) If becomes infinite  at , then we define (See Figure 4.)

.

(b) If  becomes infinite  at , then  (See Figure 5.)

.

The improper integral is called convergent if the corresponding limit exists and divergent if the limit does not exist.

(c)If becomes infinite for some value of , where , and both and are convergent, then we define (See Figure 6.)

.

Figure 4                  Figure 5                 Figure 6

EXAMPLE 5

Evaluate .

Solution. Since has a vertical asymptote at , the given integral is improper. We use part (b) of Definition 1 since the infinite discontinuity occurs at . We have

.

Thus, the given improper integral converges and the value is .

EXAMPLE 6

Is convergent or divergent?

Solution. Since , then the given integral is improper.

But .

Thus, the given improper integral is convergent.

EXAMPLE 7

Find . (See Figure 7.)

Figure 7

Solution. The graph of the integrand has a vertical asymptote at , which lies in the middle of the interval .

Then

.             ■

EXAMPLE 8

Find . (See Figure 8.)

Figure 8

Solution.  Since , the function has a vertical asymptote at 0. Thus, .

Integrating by parts gives

.

Using L’Hospital’s Rule, we obtain . Therefore,

.             ■

A Comparison Test for Improper Integrals

Often it is not possible to find the exact value of an improper integral directly from the definition. Even in this e, the convergence or divergence of the improper integral may be determined by comparing with another improper integral whose convergence or divergence can be easily determined.

THEOREM 2 Comparison Theorem

Assume that and are continuous functions with for . Then

(a) if is convergent, then is convergent.

(b) if is divergent, then is divergent.

Figure 9

The Comparison Theorem makes sense if we interpret it in terms of area. If the area under the top curve is finite, then so is the area under the bottom curve . Similarly if the area under is infinite, then so is the area under (see Figure 9).  The integral is divergent by the Comparison Theorem because and is divergent by Example 4.  However, the converse of  the Comparison Theorem  is not necessarily true: If is convergent, may or may not be convergent, and if is divergent, may or may not be divergent.

EXAMPLE 9

Prove that is convergent.

Solution. Because the integrand is unbounded near we decompose the integral into two parts: .

In the first integral we have , and the integral converges by the Comparison Theorem. In the second integral we have , and once again the Comparison Theorem yields the result since is convergent. Therefore, is convergent.

EXAMPLE 10

Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent. .

Solution.

var('x')

f=e^(x*3)*sin(2*x)

plot(f,(x, -5, 0))

Figure 10

integrate(f, x, -infinity, 0)

7.8 EXERCISES (Improper Integrals)

1.

Solution.  .

integral(ln(x)/x, x, e, infinity)

ValueError : Integral is divergent

2.

Solution.

integral(x*e^(-x), x, 0, infinity)

3.

Solution.

.

integral(1/(x^2+1), x, -infinity, infinity)

4.

Solution.

integral(sec(x), x, 0, pi/2)

ValueError : Integral is divergent.

5.

Solution.

integral(1/(x*(ln(x))^2), x, 2, infinity)

6.

Solution.

integral(1/(x*(ln(x))^2), x, 2, infinity)

7.

Solution.

var('x, n');

integral(e^(-x)*x^(n-1), x)

8.

Solution.

integral(1/(2*x-1)^3, x, -infinity, 0)

9.

Solution.

integral((e^(-(x^2)/2))*x, x, -infinity, infinity)

10.

Solution.

integral(x*sin(x), x, 0, 1)

11.

Solution.

integral(1/sqrt(x), x, 0, 1)

12.

Solution.

integral(1/(sqrt(1-x^2)), x, -1, 1)

13.

Solution.

integral(ln(x), x, 0, 1)

14.

Solution.

integral(1/(x*(sqrt(x^2-1))), x, 1, infinity)

15.

Solution.

integral(1/(x*(x^2+1)), x, 1, infinity)

16.

Solution.

integral(1/(x*sqrt(x^2-1)), x, 1, infinity)

17.

Solution.

integral((sqrt(x^2+1)-x)^2, x, 0, infinity)

18.

Solution.

integral((1+x)/(1+x^2), x, 0, infinity)

Answer :  ValueError : Integral is divergent

19.

Solution. .

(: positive integer)

var('x, n');

integral(x^(2*n-1)*e^(-x^2), x)

20.

Solution.

convergent.

21.

Solution.

integral(1/(sin(x)*cos(x)), x, 0, pi/4)

Answer : ValueError: Integral is divergent

22.

Solution.

integral(1/(x^(1/4)), x, 0, 16)

23.

Solution.  =

24.

Solution. .

convergent.

25.

Solution.

integral(sec(x)-tan(x), x, 0, pi/2)

26.

Solution.

integral(sin(ln(x)), x, 0, 1)

27.

Solution.

integral((1+x)^3/sqrt(1-x^2), x, -1, 1)

28.

Solution.

integral(x^2/(1+x^2), x, 0, infinity)

Answer :   ValueError: Integral is divergent

29.

Solution. .

integral(1/sqrt(x^48), x, 1, infinity)

30.

Solution.

integral((x^5)/sqrt(1-x^2), x, 0, 1)

31.

Solution.

Therefore,.

integrate(1/x, x, 1, infinity)

Answer :  ValueError: Integral is divergent.

32.

Solution.  .

integrate(1/(sqrt(x)), x, 0, 1)

33.

Solution.

integral(e^(-x)/sqrt(x),x,0,infinity)

34. Show that .

Proof.

Therefore, .

35. Prove that   converges.  (Hint: let

Proof.

.

Calculus