Integral Calculus

Topic explanation 5Integral Calculus

Topic 5. Integration by substitution5.1 Method of integration by substitution

The method of integration by substitution, also known as change of variable, is just like using the chain rule, but inversely; the power rule is applied (the exponent of the compound base is raised one and divided by the new exponent) only when the derivative of such compound base is considered a factor.

On the following table there are some examples of functions. Here we can identify the main function, which is the “u,” and the change of variable.

Function Main function uChange ofvariable orsubstitution

Power(polynomial)

Squareroot (power)Exponentialbase “e”Sine

The method of integration by substitution is used when we have a function elevated to an exponent by derivation of the base. This is

completed in the original function, and then integrated using any of the previously learned methods.

Example 1. Integrate the following compound functions.

The compound base is selected; normally it is the one with the highest exponent in the variable.

The base composed by “u” is equalized and derived.

First you check if the derivative is the factor of the function. In this case it is, so we can say that the function is completed.

When our function is completed, wethen integrate. On this situation the power rule will be applied.Example 2.

The compound base is selected; regularly, it is the one with the highest exponent in the variable.

The “u” base is equalized and derived.

First you check if the derivative is the factor of the function. In this case it is not, so we can saythat the function is incomplete. Complete it by adding the missing number.

We must add the number 3, and 1/3 outside the function so it cannot be modified.

5(1/3) is multiplied.

Once the function is completed, wecan integrate. In this case, we will use the power rule.

At the end, if it is possible, thenumerical coefficient must be reduced to its minimal expression.

In this case, “5” can be reduced since it is found in the numeratorand denominator.

Example 3.

The compound base is selected; regularly it is the one with the highest exponent in the variable.The “u” base is equalized and derived.

We must check if the derivative is the factor of the function. Inthis case it is not; our functionis incomplete and must be completed by adding the missing number.

We must add the number 2 and 1/2 outside the function so it cannot be modified.

Once the function is completed, we can integrate. In this case, we will use the power rule.

At the end, if it is possible, the numerical coefficient must bereduced to its minimal expression.

The result can be expressed as a binomial raised to an exponent as

radical.Example 4.

The compound base is selected; regularly, it is the one with thehighest exponent in the variable.

The “u” base is equalized and derived.

We must add the number 4, and 1/4 outside the function so it cannot be modified.

Once the function is completed, we can integrate. In this case, we will use the power rule.

Example 5.

The compound base is selected; regularly it is the one with the highest exponent in the variable.The “u” base is equalized and derived.

We must check if the derivative isthe factor of the function. In this case it is not; our function is incomplete and must be completed by adding the missing number.

We must add the number 6 and 1/6 outside the function so it cannot be modified.

Once the function is completed, wecan integrate. In this case, we will use the natural logarithmic formula.


Topic 6. Integration by parts6.1 Method of integration by parts

Integration by parts is the techniqueused when the integrand is differentfrom the ones already studied.

In the method of substitution, thederivative is performed and then wehave to check that the result isincluded as the factor within thefunction. If any constant is missing,a number can be included, but whathappens when there is an extravariable? In this case, the method of integration by parts must be used.

How could you integrate the function ?Do you think it would be by substitution? If you derive x it does not appear on its derivative ex; if you derive ex, it does not appear x on the derivative. It is in this type of integrals when the method of integration by parts is required.

It is easy to identify it because the variable has the same degree on the exponent of the base and on the factor.

It is called by parts because it is divided into two; one part is derived and the other one integrated. When we perform this movement, thetwo easiest parts to perform remain; the derivative and the integral aredirect. After performing these two operations, we get all the parts of the formula and then it is integrated. The formula is:

As we previously mentioned, it is necessary to derive one part and integrate the other one. How to choose which one to be derived from the one to be integrated? Usually, practice will help us decide, but for starters we can choose an easy or direct integral in which we will only have to use the formula, we derive the other part, and by applying the formula by parts it should be easier than the original one, but if it doesn’t it means that we made the wrong choice and we will have to startagain.

There will be occasions in which the formula must be applied more than once until no integral is left, the result will always remain with a minimum of two terms plus the integration constant.

Answer the following integrals using the method of integration by parts.

Example 1. As you can see on this example, when the derivative of the exponent of “e” is done, as we do on the integrals base “e”, the result is “1” and it is useful since we have other function “x”.As we have said, we have to derive“x” and integrate “e”.

Derivative: Integral:

The three pieces of information weneed to substitute on the formula are “u”, “v”, and “du”, in such way that we already have them. Now, we substitute them as indicated, and we have:

As you can see, the integral is simpler than the original.

The first part remains the same, it is already solved, only the second part of the function is

integrated, but this one is easier.The results are two terms, plus the integration constant.

Example 2. In this example we have an integral of a natural logarithm, such we cannot perform because there is no formula for this case.

We will solve it using the method of integration by parts.As we have mentioned before, thereis no formula of integration for “natural logarithm,” therefore this part is derived and “dx” is integrated the following way:


The three pieces of information weneed to substitute on the formula are “u”, “v”, and “du”, in such way that we already have them. Nowwe substitute them as indicated, it is multiplied and divided whatever is possible and we have: As you can see, the integral is simpler than the original.

The first part remains the same, it is already solved, only the second part of the function is integrated, but this one is easier.

Example 3. In this example we have an integral of a natural logarithm, such we cannot perform because there is no formula for this case.

We will solve it using the method of integration by parts.As we have Derivative: Integral:

mentioned before, there isno formula of integration for “natural logarithm;” therefore this part is derived and “dx” is integrated the following way:

The three piecesof information we need to substitute on the formula are “u”, “v”, and “du”, in such way that we already have them. Now, we substitute them as indicated, itis multiplied and divided whatever is possible and we have:

As you can see, the integral is simpler than theoriginal.

The first part remains the same, it is already solved, only the second part of the function is integrated, but this one is easier.

Example 4. As you can see on this example, when the derivative of the exponent of “e” is done, as we do on the integrals base “e”, the result is “1” and it is not useful since we have other function “x”.

Just as before, we are going to derive “x2” and integrate “e5x”.

We must remember the integralshould be easier than the original; for this reason, weselect (what) the following way:


The three pieces of information we need to substitute on the formula are“u”, “v”, and “du”, in such way that we already have them. Now, we substitute themas indicated, and we have:

As previously stated we must derive “x” and integrate “e5x”.

It is important to remember that when we are integrating,the derivative of the function must be complete.


The three pieces of information we need to substitute on the formula are“u”, “v”, and “du”, in such way that we already have them. Now we substitute them as indicated, and we have:

The results are the three terms plus the constant of integration.


Topic 7. Integration of powers of trigonometric functions7.1 Method of trigonometric functions,integration

To integrate powers of trigonometric functions we will use as a strategy the substitution method with the use of trigonometric identities and factorization.

To solve the integrals, it is necessary to use the basic trigonometric identities, Pythagorean, half-angle and product.Pythagorean:

Half angle:

Product:

The powers of trigonometric functions can be classified in the followingway:

For trigonometric powers of Sine and Cosine with a pair or odd exponent we have:

Example 1. Calculate the following integral with odd

exponent: As the exponent is odd, we can re-write it the followingway:

Using the identity

, isolatingwe

have:

substituting in the function

we have:

Multiplying , by the

binomial , the following is obtained:

Separating the integral in two terms: Solving the second term by substitution we obtain:

And the first one directly with formula we have:

Example 2. Calculate the following integral with pair

exponent: On this case we will use the medium angle identities: Substituting, in the function we have:

Raised to square:

Separating the terms:

Integrating the first two terms directly with formulas and the third is used with other identity of half angle:

For the trigonometric powers of sine and cosine products with combinations of pair and odd exponents.

Example 3. Calculate the integral with odd “m” or

“n” On these cases it is necessary to factorize or separate the trigonometricfunctions:

Substituting in the function: Multiplying the binomial we have:

Both terms are integrated by substitution for which we derive . Fixing the integral we have:

Now it is integrated with the power formula:

Example 4. Calculate the integral with pairs “n” and

“m”: We will use the half angle identities on each of them:

Binomials are multiplied:

We will separate both terms:

We will apply again the identity of medium angle and the terms are separated:

Solving the integral we have:

Adding similar terms:

For the trigonometric powers of sine and cosine products with different quotients for each of them.

, ,

Example 5. To solve it we will use the product

identities:As you can observe, m=3 andn=2, applyingthe identity we have:Adding and separating in two the terms wehave left:

We complete the first term:

Integrating we obtain:


Topic 8. Integration by trigonometric substitution8.1 integration method

Let’s summarize the integrals:

Even though functions are similar, they cannot be integrated using the previous methods, since the derivative of the function is not out of theradical. Furthermore, the best way to recognize this method is because there is always a radical and a square addition or subtraction within.

In such cases, the integration by trigonometric substitution is used. The procedure is based on the use of the table. Click on the table to print it out.

Let’s take a look at an example of each case:

Case 1.

http://bbsistema.tecmilenio.edu.mx/bbcswebdav/institution/UTM/semestre/preparatoria/pm/pm09126/anexos/explica8.htm

Integrate the

function: Review the radical to find out thecase. This iscase 1 (square number minus square variable).

=

Make even a2 with the number of theradical.

Substitute the value of “a” in:

Substitute in the original functions and solve the integral.

This is the result; it should keepthe function of “x”.

In the section “Substitutions to be used” in the previous table, locate the function, find the value of “ ” and substitute the value in the result.Substitute the value of “ ”.

Case 2.

Integrate the

function Review the radical to find out thecase. This iscase 2 (square number plus square variable).

=

Make even a2 with the number of theradical.

Substitute the value of “a”

in:



In the section “Substitutions to be used” in the previous table, locate the function y

, find the value of “ ” and “”, substitute the value in the result.Substitute in the results the values found.

The result is gotten with logarithms properties.

Case 3.

Integrate the function Review the radical to find out thecase. This iscase 3 (square variable minus square number).Make even a2 with the number of theradical.

Substitute the value of “a” in:



In the section “Substitutions to be used” in the previous table,

locate the functionand , find the value of “

” and “ ”, substitute the value in the result.Substitute in the results the values found.

The result is gotten with logarithms properties.

Resuelve las siguientes integrales directas y por sustitución:

1.∫ zdze−5z2

2.∫x√x2−8dx

3.∫ dx√5x+2

4.∫x2cos5x3 (sen5x3 )6dx

5.esen2xcos2xdx

6.∫sec24xtan64xdx

7.∫xex2dx

8.∫ e2x−5

2dx

9.∫e3x(4+e3x)3dx

∫cos−5xsenxdx

Resuelve las siguientes integrales aplicando la integración por

partes

1.∫x3lnx2dx

2.∫0

1

√xe2xdx

∫sec (Lnx)dx

3.∫xLn3xdx

4.∫x√x+1dx

∫Lnx∙2xdx

5.∫√xsen2xcosxdx

6.∫ ln (2x3 )√4x

dx

∫x2e5xdx

Resuelve las siguientes integrales con exponentes trigonométricos:

1.∫cos25xdx

∫sen3 x3dx

2.∫sen5xcos3xdx

3.∫sen23xcos23xdx

4.∫sen4xcos2xdx

5.∫cos8xcos2xdx

Resuelve las siguientes integrales mediante sustitución

trigonométrica

1.∫ x2dx√4−16x2

2.∫ xdx√5x2+25

3.∫ √4x2−165x

dx

4.∫√x2+81dx

INSTRUCCIONES: En la tabla siguiente, indicaren la segunda columna la letra correspondienteal método que se debe utilizar para resolverla integral y en las demás columnas llenar losvalores que se necesitan para dicho método.

A) Sustituciónalgebraica

B) Por partesC) Sustitucióntrigonométrica

D) PotenciaTrigonométrica

Función Método

Ecuación paracambiode

variable

trigonométrica

.

Indicar tipode

identidad

trigonométrica autilizar

a u du v dv

∫❑

❑

3cos(3x)dx

∫sen25xdx

∫❑

❑ x2dx❑√4+x2

∫❑

❑

xexdx

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Integral Calculus