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The Indefinite Integral and Basic Formulas of Integration. Table of Integrals.

Definition of the Antideri vative and Indefini te Integral

The function F ( x) is called an antiderivative of f ( x), if

The family of all antiderivatives of a function f ( x) is called the indefinite integral of the function f ( x) and is denoted by

Thus, if F is a particular antiderivative, we may write

where C is an arbitrary constant.

Propert ies of the Indefin ite Integral

In formulas given below f and g are functions of the variable x, F is an antiderivative of f , and a, k, C are constants.

Table of Integrals

It's supposed below that a, p ( p ≠ 1), C are real constants, b is the base of the exponential function (b ≠ 1, b > 0).



Example 1

Calculate .

Solution.

Example 2

Calculate the integral .



Solution.

Transforming the integrand and using the formula for integral of the power function, we have

Example 3

Calculate .

Solution.

We use the table integral . Then

Example 4

Find the integral .

Solution.

Using the table integral , we obtain

Example 5


Solution.

Since , the integral is

Example 6



Find the integral without using a substitution.

Solution.

Using the double angle formula sin 2 x = 2 sin x cos x and the identity sin2 x + cos2

x = 1, we can write:

Change of Variable

Let F(x) be an indefinite integral or antiderivative of f(x). Then

where x = g (u) is a substitution. Accordingly, the inverse function u = g −1(x) describes the dependence of the new variable ovariable.

It's important to remember that the differential dx also needs to be substituted. It must be replaced with the differential ofthe nFor definite integrals, it is also necessary to change the limits of integration. See about this on the page "The Definite IntegralFundamental Theorem of Calculus".

Example 1


Solution.

Let . Then . Hence, the integral is

Example 2

Find the integral .

Solution.

We make the substitution . Then or .The integral is easy to calculate with the new variable:

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Example 3


Solution.

Rewrite the integral in the following way:

Noting 2e = a (This is not a change of variable, since x still remains the independent variable), we get the table integral:

Example 4


Solution.We can write the integral as

Changing the variable

we get the answer

Example 5

Find the integral .

Solution.We make the following substitution:

Hence,

Integration by Parts

Let u(x) and v(x) be differentiable functions. By the product rule,

Integrating both sides of the expression, we obtain



or, rearranging terms,

Integration by using this formula is called integration by parts.

Example 1

Compute .

Solution.

We use integration by parts: . Let . Òhen

Hence, the integral is

Example 2

Integrate .

Solution.

We are to integrate by parts: u = ln x, dv = dx. The only choices we have for u and dv are . Then

Example 3


Solution.

Use integration by parts: . Then , and the integral



To calculate the new integral, we substitute . In this case, , so that the integral in th

Then the result is

Example 4

Find the integral .

Solution.

We use integration by parts: . Let . Then integral can be written in the form:

Apply integration by parts one more time. Now let . Hence, integral we started with is

Solving this equation for this integral, we obtain

Example 5

Find a reduction formula for .

Solution.

We apply integration by parts: . Let . Then

Hence,



Solve the equation for . As a result, we obtain

Integration of Rational Functions

A rational function , where P(x) and Q(x) are both polynomials, can be integrated in four steps:

1. Reduce the fraction if it is improper (i.e. degree of P(x) is greater than degree of Q(x));

2. Factor Q(x) into linear and/or quadratic (irreducible) factors;

3. Decompose the fraction into a sum of partial fractions;

4. Calculate integrals of each partial fraction.

Consider the specified steps in more details.Step 1. Reducing an improper fractionIf the fraction is improper (i.e. degree of P(x) is greater than degree of Q(x)), divide the numerator P(x) by the denominatorQ

where is a proper fraction.Step 2. Factoring Q(x) into linear and/or quadratic factorsWrite the denominator Q(x) as



where quadratic functions are irreducible, i.e. do not have real roots.Step 3. Decomposing the rational fraction into a sum of partial fractions.Write the function as follows:

The total number of undetermined coefficients Ai , Bi , Ki , Li , Mi , Ni , ... must be equal to the degree of the denominatorQ(

Then equate the coefficients of equal powers of x by multiplying both sides of the latter expression by Q(x) and write the systequations in Ai , Bi , Ki , Li , Mi , Ni , .... The resulting system must always have a unique solution.Step 4. Integrating partial fractions.Use the following 6 formulas to evaluate integrals of partial fractions with linear and quadratic denominators:

1.

2. For fractions with quadratic denominators, first complete the square:

where Then use the formulas:

3.

4.



5.

The integral can be calculated in k steps using the reduction formula:

6. Example 1

Evaluate the integral .

Solution.Decompose the integrand into partial functions:

Equate coefficients:

Hence,

Then

The integral is equal to

Example 2

Evaluate .



Solution.First we divide the numerator by the denominator, obtaining

Then

Example 3


Solution.We can write:

Example 4


Solution.Decompose the integrand into partial functions:


Hence,

Then




Example 5

Evaluate .

Solution.Decompose the integrand into the sum of two fractions:


Hence

The integrand can be written as

The initial integral becomes

Example 6

Find the integral .

Solution.

We can factor the denominator in the integrand:

Decompose the integrand into partial functions:




Hence,

Then

Now we can calculate the initial integral:

Example 7


Solution.



Rewrite the denominator in the integrand as follows:

The factors in the denominator are irreducible quadratic factors since they have no real roots. Then


This yields

Hence,

Integrating term by term, we obtain the answer:



We can simplify this answer. Let

Then

Hence, . The complete answer is



Example 8


Solution.We can factor the denominator in the integrand:

Decompose the integrand into partial functions:


Hence,

Thus, the integrand becomes

So, the complete answer is

Example 9


Solution.



Decompose the integrand into partial functions, taking into account that the denominator has a third degree root:


We get the following system of equations:

Hence,

The initial integral is equal to

Example 10

Find the integral .

Solution.Since is reducible, we complete the square in the denominator:

Now, we can compute the integral using the reduction formula

Then



Integration of Irrational Functions

To integrate an irrational function containing a term we make the substitution .

To integrate an irrational function involving more than one rational power of x, make a substitution of the form , whto be the least common multiple of the denominators of all the fractional powers that appear in the function.

An expression rational in x and is integrated by the substitution .

See Trigonometric and Hyperbolic Substitutions about integration of irrational functions involving a

Example 1

Find the integral .

Solution.We make the substitution:

Then

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Example 2


Solution.We make the following substitution:

Then the integral (we denote it by I ) becomes

Divide the fraction:

As a result, we have

Example 3



Since the least common multiple (LCM) of the denominators of the fractional powers is equal to n = LCM(1,3) = 3, we make

Thus

Make the new substitution:



The final answer is

Example 4



Make the substitution:

The integral becomes

Since the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator:


Example 5

Find the integral .


As can be seen, the least common multiple (LCM) of the denominators of the fractional powers is equal ton = LCM(3,4) = 12substitution:



This yields

The degree of the numerator is greater than the degree of the denominator, therefore, we divide the fraction:

After simple transformations we obtain the final answer:

Example 6

Evaluate .

Solution.Make the substitution:

This yields

Example 7


Solution.We make the substitution



As a result, we have

Integration of Rational Expressions of Trigonometric Functions

Any rational expression of trigonometric functions can be always reduced to integrating a rational function by makingthe

universal trigonometric substitution x = 2arctan t (or ).

The following trigonometric formulas are used to transform rational expressions of sin x, cos x, tan x, cot x, sec x and csc xin

functions of t :

To calculate an integral of the form , where R is a rational function, use the substitution .

Similarly, to calculate an integral of the form , where R is a rational function, use the substitution

If an integrand is a function of only tan x, the substitution t = tan x transforms this integral into integral of a rational function.

To calculate an integral of the form , where both functions sin x and cos x have even powers, usethe sub= tan x and the formulas

Example 1




Solution.We use the universal trigonometric substitution:

Since , we have

Example 2


Solution.Make the universal trigonometric substitution:

Then the integral becomes

Example 3

Find the integral .

Solution.As in the previous examples, we will use the universal trigonometric substitution:

Since we can write:



Example 4

Evaluate .

Solution.We can write the integral in form:

Use the universal trigonometric substitution:

This leads to the following result:

Example 5


Solution.Since , we can write

Hence,



so the integral can be transformed in the following way:

Make the substitution . Then use the identity

We get the final answer:

Example 6


Solution.We solve this integral by making the trigonometric substitution

Taking into account that we find the integral:



Example 7

Find the integral .


As a result, the integral becomes

Using the method of partial fractions, we can write the integrand as

Calculate the coefficients A, B and C:

Hence,



So the integral is

Integration of Some Classes of Trigonometric Functions

In this section we consider 8 classes of integrals with trigonometric functions. Special transformations and subtitutions used fclasses allow us to obtain exact solutions for these integrals.

1. Integrals of the formTo find integrals of this type, use the following trigonometric identities:

2. Integrals of the formIt's assumed here and below that m and n are positive integers. To find an integral of this form, use the following substitutions

a. If the power n of the cosine is odd (the power m of the sine can be arbitrary), then the substitution is used

b. If the power m of the sine is odd, then the substitution is used.



c. If both powers m and n are even, then first use the double angle formulas

to reduce the power of the sine or cosine in the integrand. Then, if necessary, apply the rules a) or b).

3. Integrals of the form

The power of the integrand can be reduced by using the trigonometric identity and the reduction formul


We can reduce the power of the integrand using the trigonometric identity and the reduction formula

5. Integrals of the formThis type of integrals can be simplified with help of the reduction formula:

6. Integrals of the formSimilarly to the previous examples, this type of integrals can be simplified by the formula


a. If the power of the secant n is even, then using the identity the secant function is expressed asthtangent function. The factor is separated and used for transformation of the differential. As a result,the entir(including differential) is expressed as the function of tan x.

b. If both the powers n and m are odd, then the factor sec x tan x, which is necessary to transform the differential, is sep

the entire integral is expressed through sec x.

c. If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is expressed as the secan

identity . Then the integrals of the secant are calculated.




a. If the power of the cosecant n is even, then using the identity the cosecant function is expressedcotangent function. The factor is separated and used for transformation of the differential. As a result,the int

differential are expressed through cot x.

b. If both the powers n and m are odd, then the factor csc x cot x, which is necessary to transform the differential, is septhe integral is expressed through csc x.

c. If the power of the cosecant n is odd, and the power of the cotangent m is even, then the cotangent is expressed asthe

the identity . Then the integrals of the cosecant are calculated.

Example 1


Solution.Let u = cos x, du = − sin xdx. Then

Example 2


Solution.Making the substitution u = sin x, du = cos xdx and using the identity , we obtain

Example 3

Find the integral .

Solution.



Using identities and , we can write:

Calculate the integrals in the latter expression.

To find the integral , we make the substitution u = sin 2x, du = 2cos 2xdx. Then

Hence, the initial integral is

Example 4


Solution.We can write:

Transform the integrand using the identities

We get

Example 5


Solution.Making the substitution u = cos x, du = − sin xdx and expressing the sine through cosine with help of the formulaobtain



Example 6


Solution.Transform the integrand by the formula

Hence,


Example 7


Solution.

We use the identity to transform the integral. This yields

Example 8


Solution.

Using the identity , we have



Example 9


Solution.We use the reduction formula

Hence,

The integral is a table integral which is equal to . (It can be easily found usingt

universal trigonometric substitution .) As a result, the integral becomes

Example 10


Solution.We use the reduction formula

Hence,

Example 11

Compute .



Solution.

Example 12

Compute .

Solution.

Use the identity . Then

Since (see Example 9) and is a table integral equal

to , we obtain the following complete answer:

Integration of Hyperbolic Functions

The 6 basic hyperbolic functions are defined by

There are the following differentiation and integration formulas for hyperbolic functions:



We provide here a list of useful hyperbolic identities:

When an integrand contains a hyperbolic function, the integral can be reduced to integrating a rational function by usingthe

substitution .Example 1


Solution.

We make the substitution: u = 2 + 3sinh x, du = 3cosh xdx. Then . Hence, the integral is

Example 2

Evaluate .

Solution.

Since , and, hence, , we can write the integral as

Making the substitution u = cosh x, du = sinh xdx, we obtain

Example 3


Solution.



We use integration by parts: . Let . Then, integral is

Example 4


Solution.

Since , we obtain

Example 5

Find the integral .

Solution.

By definition of the hyperbolic cosine, . Hence, the integral is equal

Example 6

Find the integral .

Solution.

By definition, and . Hence,

We make the substitution: u = e x, du = e xdx and calculate the initial integral:



Example 7


Solution.

Applying the formulas and , we get

Example 8


Solution.We use integration by parts. Let

Then

Apply integration by parts again to the latter integral. Let

Then we have

Solving this equation for , we obtain the complete answer:








substitution .

Example 1


Solution.




Example 2

Evaluate .

Solution.



Example 3


Solution.


Example 4


Solution.

Since , we obtain

Example 5

Find the integral .

Solution.




Example 6

Find the integral .

Solution.



Example 7


Solution.


Example 8





Then


Then we have









substitution .

Example 1


Solution.


Example 2

Evaluate .

Solution.



Example 3


Solution.


Example 4




Solution.

Since , we obtain

Example 5

Find the integral .

Solution.


Example 6

Find the integral .

Solution.



Example 7


Solution.




Example 8



Then


Then we have


Trigonometric and Hyperbolic Substitutions

In this section we consider integrals of the form , where R is a rational function of x and the

radical .

To calculate such an integral, we need first to complete the square in the quadratic expression:

Making the substitution , we can obtain one of the three possible integrals, depending on the values o



coefficients a, b and c:

1.

2.

3. Now, we use the following trigonometric or hyperbolic substitutions to simplify the integrals:

1. Integrals of the formTrigonometric substitution:


Hyperbolic substitution:


Hyperbolic substitution:

Remarks:

Instead of the trigonometric substitutions in cases 1, 2, 3 you can use the substitutionsx = r cos t, x = r cot t, x = r csc

Using the formulas given above, we consider only positive values of the root. For example, in strict writing



We suppose here that .

Example 1



Then

Here, we used the trigonometric formula to simplify the integral.

Example 2


Solution.

We make the hyperbolic substitution: x = a sinh t, dx = a cosh tdt. Using the hyperbolic identity , we c

Example 3


Solution.



To find the integral, we make the substitution: . Using the identity

Express sin t in terms of x:


Example 4


Solution.

We make the substitution: . Hence,


Returning back to the variable x :

we obtain the complete answer:

Example 5


Solution.



We make the trigonometric substitution: x = a sec t, dx = a tan t sec tdt. Calculate the integral using the identity

For t, we have the following expression:

Hence, returning back to the variable x, we obtain:

Example 6



Make the trigonometric substitution:

Now we can calculate the integral:

Example 7


Solution.

Here we make the hyperbolic substitution (for diversity): . Since , th becomes



Use the double-angle formula to reduce the power of the integrand. Then

Example 8


Solution.First we complete the square in the integrand.

Then, making the substitution and using the identity , we can wri

The integral was already found in Problem 9 on the page Integration of some classes of trigonometric functions,answer is

Example 9

Find the integral .

Solution.Make the substitution:

Then

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Express and in terms of x:

Hence,

Example 10


Solution.

Using the trigonometric substitution , we get

Now, we make the substitution . The integral becomes

Use the method of partial fractions to transform the integrand:

Equate the coefficients:



Then

Hence, we can write the integrand as

So, the initial integral is equal to

Returning back to the variable x, we have

Hence, the final answer is

The Definite Integral and Fundamental Theorem of Calculus

Let f (x) be a continuous function on the closed interval [a, b]. The definite integral of f (x) from a to b is defined to bethe lim

where

Properties of the Definite IntegralWe assume below that f (x) and g (x) are continuous functions on the closed interval [a, b].



1.

2. where k is a constant;

3.

4.

5. If for all , then .

6.

7.

8. If in the interval [a, b], then



The Fundamental Theorem of CalculusLet f (x) be a function, which is continuous on the closed interval [a, b]. If F (x) is any antiderivative of f (x) on [a, b], then

The Area under a Curve

The area under the graph of the function f (x) between the vertical lines x = a, x = b (Figure 1) is given by the formula

Fig.1 Fig.2 Let F (x) and G (x) be indefinite integrals of functions f (x) and g (x), respectively. If f (x) ≥ g (x) on the closed interval[a, b],

between the curves y = f (x), y = g (x) and the lines x = a, x = b (Figure 2) is given by

The Method of Substitution for Definite Integrals

The definite integral of the variable x can be changed into an integral with respect to t by making the substitutio

The new limits of integration for the variable t are given by the formulas

where g -1 is the inverse function to g, i.e. t = g -1(x).

Integration by Parts for Definite IntegralsIn this case the formula for integration by parts looks as follows:



where means the difference between the product of functions uv at x = b and x = a.

Example 1


Solution.Using the fundamental theorem of calculus, we have

Example 2


Solution.

Example 3


Solution.First we make the substitution:

Determine the new limits of integration. When x = 0, then t = −1. When x = 1, then we have t = 2. So, the integral withthe new be easily calculated:

Example 4


Solution.We can write



Apply integration by parts: . In this case, let


Example 5

Find the area bounded by the curves and .

Slution.First we find the points of intersection (see Figure 4):

As can be seen, the curves intercept at the points (0,0) and (1,1). Hence, the area is given by



Fig.3 Fig.4

Example 6

Find the area bounded by the curves and .

Solution.First we find the points of intersection of the curves (Figure 4):

The upper boundary of the region is the parabola , and the lower boundary is the straight line . The are

Example 7

Find the area of the triangle with vertices at (0,0), (2,6) and (7,1).

Solution.First we find an equation of the side OA (refer to Figure 5):



Similarly, we find an equation of the side OB:

Next, find an equation of the side AB:

As seen from the Figure 5, the area of the this triangle can be calculated as the sum of two integrals:

Fig.5 Fig.6

Example 8



Find the area inside the ellipse .

Solution.By symmetry (see Figure 6), the area of the ellipse is twice the area above the x-axis. The latter is given by

To calculate the last integral, we use the trigonometric substitution x = asin t, dx = acos tdt. Refine the limits of integration. W

a, then sin t = −1, and . When x = a, then sin t = 1, . Thus we get

Hence, the total area of the ellipse is πab.

The definite integral is called an improper integral if one of two situations occurs:

The limit a or b (or both the limits) are infinite;

The function f (x) has one or more points of discontinuity in the interval [a,b].

Infinite Limits of IntegrationLet f (x) be a continuous function on the interval [a, ∞). We define the improper integral as

Consider the case when f (x) is a continuous function on the interval (−∞, b]. Then we define the improper integral as

If these limits exist and are finite then we say that the improper integrals are convergent.Otherwise the integrals are divergent.

Let f (x) be a continuous function for all real numbers. We define:



If, for some real number c, both of the integrals in the right side are convergent, then we say that the integral is otherwise it is divergent.Comparison Theorems

Let f (x) and g (x) be continuous functions on the interval [a, ∞). Suppose that for all x in the interval[a,

1. If is convergent then is also convergent;

2. If is divergent then is also divergent;

3. If is convergent then is also convergent. In this case, we say thatt

integral is absolutely convergent.

Discontinuous IntegrandLet f (x) be a function which is continuous on the interval [a,b), but is discontinuous at x = b. We define the improper integra

Similarly we can consider the case when the function f (x) is continuous on the interval (a,b], but is discontinuous at x = a.Th

If these limits exist and are finite then we say that the integrals are convergent; otherwise the integrals are divergent.

Let f (x) be a continuous function for all real numbers x in the interval [a,b], except for some point . We define:

and say that the integral is convergent if both of the integrals in the right side are also convergent. Otherwisetheimproper integral is divergent.

Example 1



Determine for what values of k the integral converges.

Solution.By the definition of an improper integral, we have

As seen from the expression, there are 2 cases:

If 0 < k < 1, then as and the integral diverges;

If k > 1, then as and the integral converges.

Example 2


Solution.

Hence, the given integral converges.

Example 3

Determine whether the integral converges or diverges?

Solution.

Note, that for all values x ≥ 1. Since the improper integral is convergent according to the results inExam

integral is also convergent by Comparison Theorem 1.

Example 4




Solution.There is a discontinuity at x = 0, so that we must consider two improper integrals:

Using the definition of an improper integral, we obtain

For the first integral,

Since it's divergent, the initial integral also diverges.

Example 5

Determine whether the improper integral converges or diverges?

Solution.We can write this integral as

By the definition of an improper integral, we have

Since both the limits exist and are finite, the given integral converges.

Example 6


Solution.We can write the obvious inequality for the absolute values:



It's easy to show that the integral converges (see also Example 1). Indeed

Then we conclude that the integral also converges by Comparison Theorem 1. Hence, the given integral (absolutely) by Comparison Theorem 3.

Example 7


Solution.There is a discontinuity in the integrand at x = 2, so that we must consider two improper integrals:

Using the definition of an improper integral, we obtain

For the first integral,

Since it is divergent, the given integral is also divergent.

Example 8

Determine for what values of k the integral converges?



Solution.The integrand has discontinuity at the point x = 0, so that we can write the integral as

As seen from this expression, there are 2 cases:

If 0 < k < 1, then and the integral converges;

If k > 1, then and the integral diverges.

Example 9

Find the area under the curve y = ln x between x = 0 and x = 1.

Solution.The given region is sketched in Figure 1. Since it is infinite, we calculate the improper integral to find the area:

Use integration by parts. Let u = ln x, dv = dx. Then . Thus

We can apply L'Hopital's rule to find the limit:

Hence, the improper integral is

As seen from the figure, the required area is .



Fig.1 Fig.2

Example 10

Find the circumference of the unit circle.

Solution.We calculate the length of the arc of the circle in the first quadrant between x = 0 and x = 1 and then multiply the result by 4.

The equation of the circle centered at the origin is

Then the arc of the circle in the first quadrant (Figure 2) is described by the function

Find the derivative of the function:

Since the length of an arc is given by , we obtain

Now we calculate the improper integral :



Thus, the circumference of the unit circle is .

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