Integration of Functions
[460/99%]Applying formulas 1 and 2 and rules I and II, find the integral .
Applying formulas 1 and 2 and rules I and II, find the integral .
Applying formulas 1 and 2 and rules I and II, find the integral .
Applying formulas 1 and 2 and rules I and II, find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
The following integrals are found directly by applying formulas 1--12, and also by means of the change of variable . Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral , by rationalizing the numerator.
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
In a number of cases it is convenient to transform products and powers of sines and cosines into sums or differences. Find the integral .
Find the integral .
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Find the integral .
Find the integral .
The following integrals are found by simple substitutions. Find the integral .
Find the integral .
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Find the integral .
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Find the integral .
Using one of the trigonometric substitutions , , , find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral , by rationalizing the denominator.
Applying integration by parts, expressed by equality III, find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral , using the identity
Find the integral , using the identity
in which the coefficients and are easily found by equating coefficients of like powers of after clearing the denominator.
Find the integral .
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In the following integrals, to factor the denominator it is convenient to add and subtract the term , where is a suitably chosen coefficient. Find the integral .
Find the integral .
Find the integral .
Find the integral .
If the denominator factors into simple factors of the first degree, then for decomposing the fraction into partial fractions the formula
is convenient, where are the roots of the polynomial . If among the roots there is a pair of complex conjugates, then in the sum on the right-hand side the two corresponding terms can be replaced by a single quantity:
Here is the number complex conjugate to .
Find the integral .
Find the integral .
Find the integral .
The following integrals are reduced to integrals of the previous types after extracting the integer part of the integrand fraction. Find the integral .
Find the integral .
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Using the identity and integrating by parts the integral , one can prove the reduction formula
where .
Find the following integrals using the reduction formula. Find the integral .
Find the integral .
Find the integral .
Find the integral .
Prove the equality
Prove the equality
Find the integrals. Find the integral .
Find the integral .
Find the integral .
Find the integral .
In the following problems the integrand has the form . For these the substitution is useful, where is the greatest common divisor of and . Find the integral .
Find the integral .
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In the following problems the integral is found after the denominator is factored. Find the integral .
Find the integral .
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Find the integral .
Find the algebraic part of the integrals. Find the integral .
Find the integral .
Find the integral .
Find the integral .
A number of the following problems are solved by various special techniques that avoid decomposing the fraction into partial fractions. The linear-fractional substitution conveniently gives the integral .
Find the integral .
Find the integral .
Find the integral .
The substitutions and make it easy to find the following integrals, which contain palindromic polynomials and the factor or . Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral , using the identities
Find the integral , by subtracting and adding the quantity in the numerator.
Find the integral , introducing the auxiliary integral and noting that the integrals and are easily found after the substitutions and .
Find the value of the integral , introducing the integral .
Find the integrals and .
If the integrand contains a root of some degree of a linear-fractional function , then this irrationality can be removed by the substitution , where is a suitably chosen exponent. Using this technique, find the integral .
Find the integral .
Find the integral .
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Find the integral .
The following four integrals, although of a different type, are also found by a substitution of the form . Find the integral .
Find the integral .
Find the integral .
Find the integral .
The following three integrals are found by similar techniques. Find the integral .
Find the integral .
Find the integral .
The following integrals have the form , where , , and are rational numbers. They reduce to integrals of rational functions in three cases: 1) if is an integer, even if negative (substitution , where is the common denominator of the fractions and ); 2) if is an integer (substitution , where is the denominator of ); 3) if is an integer (substitution , where is the denominator of ). If none of the three numbers , , and is an integer, then by Chebyshev's theorem integrals of this form cannot be expressed by a finite combination of elementary functions.
Compute the integral .
Compute the integral .
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The following integrals are solved by a technique similar to that used in the third case of the previous type. Find the integral .
Find the integral .
In many cases, when finding integrals of differential binomials, i.e., integrals of the form , it is useful first to reduce the integral to another one in which the quantities and have more convenient values. This can be done using reduction formulas. The main ones have the following form:
Using the reduction formulas, find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Find the integral .
A large number of the following problems give integrals containing the square root of a quadratic polynomial. All of them can be reduced to integrals of rational functions by one of the three Euler substitutions:
where is a root of the polynomial .
However, in most cases the Euler substitutions lead to longer computations than direct methods. To apply them one should become familiar with three basic types of the integrals under study.
I. Integrals of the form , where is a polynomial. In most cases the method of undetermined coefficients is convenient. Here we write the presumed identity
Differentiating both sides of the equality, we obtain
Clearing the denominator, we obtain polynomials of degree on both sides of the equality. Equating coefficients of like powers of , we obtain equations from which the coefficients are found one after another.
Find the integral .
Find the integral .
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The preceding problems contained integrals of the form , where is a polynomial. In order to be able to integrate the most general integral of the form , where is a rational function of and , and , it is important to be able to find integrals of two further types, which we denote by the numerals II and III.
II. Integrals of the form , where is a polynomial of degree less than . These integrals reduce to integrals of the previous class by the substitution . If the degree of the polynomial is not less than , then the integer part of the fraction should be extracted first.
Using these methods, find the integral .
Find the integral .
Find the integral .
Find the integral .
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Find the integral .
III. The third and most complicated type is represented by integrals
where the roots of the trinomial are imaginary. If , then
Here the first integral is found by the substitution , and the second by the substitution . In the general case the integral is reduced to this simplest case by the substitution . This gives
where is a polynomial of degree , and the coefficients and have the following values:
Setting , , we obtain equations giving real values for and for real , provided at least one of the polynomials has imaginary roots. With this choice of the numbers and , the integral splits into a sum of integrals of two types:
which are integrated respectively by the substitutions:
Find the integral .
Find the integral .
Find the integral .
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Find the integral .
The most general integral of the form , where is a rational function of and , reduces to integrals of the three previous types. For this it suffices, by rationalizing the denominator, to transform the function to the form , where and are rational functions of . Here . Thus the matter reduces to finding the integral of the rational function and of , where is a rational fraction. In the latter integral the rational fraction can be transformed by extracting the integer part and decomposing into partial fractions. This yields integrals of the three types already considered. In simple cases part of these transformations is unnecessary.
Find the integral .
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The substitution is sometimes called the Abel substitution; it is convenient for finding the integral of the function , where is an integer.
Find the integral .
Find the integral .
Find the integral .
Find the integral .
Integrals , where is a rational function and , called elliptic integrals, generally cannot be integrated in closed form unless there are special relations among the coefficients of the function or of the polynomial .
In the following examples with palindromic polynomials, one of the substitutions or makes it comparatively easy to obtain the integral. Find the integral .
Find the integral .
Find the integral .
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Find the integral .
Theorem (Abel). Let be an algebraic function of of degree , i.e., satisfies the equation , where is a polynomial in and of degree in . If the integral of the function , rational in and , can be expressed through algebraic functions, then the equality
must hold, where are rational functions of .
Corollary. If , where is a polynomial of degree , and if the integral is expressed by an algebraic function, then it must equal , where is a polynomial of degree .
Find the following algebraic integrals. Find the integral .
Find the integral .
Find the integral .
Integrals containing the functions , , , where is some algebraic function of , are often integrated by parts, setting , , . This is how the integral is found.
Find the integral .
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Integrals containing the product of a polynomial with one of the functions , , are best evaluated using the formula for repeated integration by parts, which can be given the form
If is a polynomial of degree , then the last term vanishes.
Prove the equalities, in which is a polynomial of degree :
Prove the equality
Prove the equality
Find the integral .
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The following integrals, containing exponential functions, are found by integration by parts, and also by simple substitutions. Find the integral .
Find the integral .
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The following integrals are most conveniently found by using the fact that the integration formulas for functions containing remain valid for complex values of . Separating the real and imaginary parts, we obtain equalities with real integrals. This is how the integral is found.
Find the integral .
Find the integral .
Find the integral .
A number of the following integrals belong to the type , where is a rational function. Here it is useful to keep in mind three rules: 1) if is multiplied by when the sign is changed before one of the quantities or , then it is useful to denote the other of them by ; 2) if does not change when the sign is changed simultaneously before and , then it is useful to set or ; 3) the substitution in all cases turns an integral of the type under consideration into an integral of a rational fraction. Unfortunately, this substitution leads to longer computations than the previous ones, when the latter are applicable.
Find the integral .
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For the integrals there exist reduction formulas that allow reducing the integrals to others in which the indices and are closer to zero. These formulas reduce to four basic ones:
Denoting by the integral and by the integral , we obtain from this two more reduction formulas for the integrals and :
Using the reduction formulas, find the integral .
Find the integral .
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Below is a series of integrals of trigonometric functions found by various techniques. Find the integral , by comparing the numerator with the derivative of the denominator.
Find the integral , by writing the numerator in the form , where .
Find the integral , by reducing the denominator to logarithmic form.
Apply the same technique to the integral .
Find the integral , using the fact that .
Using the substitution , find the integral .
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Find the integral ; .
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Find the integrals. Find the integral .
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Remark. Here the substitution is useful.
Find the integral for and for .
The following integrals contain hyperbolic functions. In these examples it is convenient to express them through exponential functions. Find the integral .
Find the integral .
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Find the integral .
In the following examples it is more convenient not to express the hyperbolic functions through exponentials. Find the integral .
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