In the most convenient cases, integrals can be done purely in terms of elementary functions such as exponentials, logarithms and trigonometric functions. In fact, if you give an integrand that involves only such elementary functions, then one of the important capabilities of Integrate is that if the corresponding integral can be expressed in terms of elementary functions, then Integrate will essentially always succeed in finding it.

Beyond working with elementary functions, Integrate includes a large number of algorithms for dealing with special functions. Sometimes it uses a direct generalization of the procedure for elementary functions. But more often its strategy is first to try to write the integrand in a form that can be integrated in terms of certain sophisticated special functions, and then having done this to try to find reductions of these sophisticated functions to more familiar functions. A large book of integral tables will list perhaps a few thousand indefinite integrals.

The Wolfram Language can do essentially all of these integrals. And because it contains general algorithms rather than just specific cases, the Wolfram Language can actually do a vastly wider range of integrals. Particularly if you introduce new mathematical functions of your own, you may want to teach the Wolfram Language new kinds of integrals. You can do this by making appropriate definitions for Integrate. In the case of differentiation, the chain rule allows one to reduce all derivatives to a standard form, represented in the Wolfram Language using Derivative.

But for integration, no such similar standard form exists, and as a result you often have to make definitions for several different versions of the same integral. Changes of variables and other transformations can rarely be done automatically by Integrate. Enable JavaScript to interact with content and submit forms on Wolfram websites. Learn how. This integral can be evaluated using the same kind of functions that appeared in the input:. But for this integral the special function LogIntegral is needed:. It is not difficult to find integrals that require all sorts of functions:.

This integral involves an incomplete gamma function. Note that the power is carefully set up to allow any complex value of x :. Integrals of rational functions are straightforward to evaluate, and always come out in terms of rational functions, logarithms and inverse trigonometric functions:. The integral here is still of the same form, but now involves an implicit sum over the roots of a polynomial:.

This is a strong indication that that the processes of integration and differentiation are interconnected. The following tables list the formulas for antidifferentiation.

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These formulas allow us to determine the function that results from an indefinite integral. Since the formulas are for the most general indefinite integral, we add a constant C to each one. With these formulas and the Fundamental Theorem of Calculus, we can evaluate simple definite integrals.

## Lists of integrals

Note: After finding an indefinite integral, you can always check to see if your answer is correct. Since integration and differentiation are inverse processes, you can simply differentiate the function that results from integration, and see if it is equal to the integrand. The total change theorem is an adaptation of the second part of the Fundamental Theorem of Calculus. The Total Change Theorem states: the integral of a rate of change is equal to the total change.

If we know that the function f x is the derivative of some function F x , then the definite integral of f x from a to b is equal to the change in the function F x from a to b. With our current knowledge of integration, we can't find the general equation of this indefinite integral. There are no antidifferentiation formulas for this type of integral. We must also account for the chain rule when we are performing integration. To do this, we use the substitution rule.

If we substitute u into the left side of the equation for g x and du for g' x dx, then we get the integral on the right side of the equation. If we substitutite these values into the integral, we get an integral that can be solved using the antidifferentiation formulas. However, this answer is still in terms of u. The substitution rule also applies to definite integrals. If f x is continuous on [-a, a] and f is an even function , then. If f x is continuous on [-a, a] and f is an odd function , then. These properties of integrals of symmetric functions are very helpful when solving integration problems.

Some of the more challenging problems can be solved quite simply by using this property.

Similar to integrals solved using the substitution method, there are no general equations for this indefinite integral. However there do not appear to be any clear substitutions that could be made to simplify this integral. This brings us to an integration technique known as integration by parts , which will call upon our knowledge of the Product Rule for differentiation.

## Integration Rules

The Product Rule states: If f and g are differentiable functions, then. To make it easier to remember it is commonly written in the following notation. If we substitute these values into the formula we have:. This equation is correct, but the integral is more difficult than the one we started with. So in order to integrate powers of sine we need an extra cosx factor. Similarily, in order to integrate powers of cosine we need an extra sinx factor. If we were to use the method from the previous example and separate one cosine factor we would be left with a factor of cosine of odd degree which isn't easily converted to sine.

We must now consider the half angle formulas. Now that we have learned strategies for solving integrals with factors of sine and cosine we can use similar techniques to solve integrals with factors of tangent and secant.

### Chapter 8 Solutions

Now we will consider two examples to illustrate two common strategies used to solve integrals of the form. Thus we have:. When we separate out a factor of secxtanx we are left with an odd power of tangent which is not easily converted to secant. When we separate out a factor of sec 2 x we are left with an odd power of secant which is not easily converted to tangent.

Note: If the power of secant is even and the power of tangent is odd then either method will suffice, although there may be less work involved to use method a if the power of secant is smaller, and method b if the power of tangent is smaller. Unlike integrals with factors of both tangent and secant, integrals that have factors of only tangent, or only secant do not have a general strategy for solving. Use of trig identities, substitution and integration by parts are all commonly used to solve such integrals.

For example,.

### Tables of Indefinite Integrals

Another problem that may be encountered when solving trigonometric integrals are integrals of the form. Sometimes trigonometric substitutions are very effective even when at first it may not be so clear why such a substitution be made. So we must consider a change in variables where the old variable is a function of the new one. This is where trigonometric identities are put to use.

## Integral (Antiderivative) Calculator with Steps - eMathHelp

Then using the trig identity we can simplify the integral by eliminating the root sign. By changing x to a function with a different variable we are essentially using the The Substitution Rule in reverse. In the example above we would require to assure has an inverse function. Integration of rational functions by partial fractions is a fairly simple integrating technique used to simplify one rational function into two or more rational functions which are more easily integrated.

Think back to the steps taken when adding or subtracting fractions that do not have the same denominator. First you find the lowest common multiple of the two denominators and then cross multiply with the numerators accordingly. Now by reversing this process we can simplify a function such as into two fractions which are more easily integrated. This process is possible when the function is proper; that is the degree of the numerator is less than the degree of the denominator. If the function is improper; that is the degree of the numerator is greater than or equal to the degree of the denominator, then we must first use long division to divide the denominator into the numerator until we obtain a remainder, such that it's degree is less than the denominator.

Then if possible the above process is used to simplify the proper function.

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In general there are 4 cases to consider to express a rational function as the sum of two or more partial fractions. Case 1 The denominator is a product of distinct linear factors no factor is repeated or a constant mulptiple of another.