The EASIEST Way to Solve cos(x)sin(sin(x)) Integrals!

The integral symbol in the previous definition should look familiar. We have seen a similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter, we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral. Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered to be the co-discoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval [a, b]. The numbers a and b are x-values and are called the limits of integration; specifically, a is the lower limit and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as n → ∞. Second, the boundaries of the region are called the limits of integration. We call the function f (x) the integrand, and the dx indicates that f (x) is a function with respect to x, called the variable of integration.
Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:
Functions that are not continuous on [a, b] may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable. It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.
The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback
of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section, we
examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us
find antiderivatives when the integrand is the result of a chain-rule derivative.
At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task-that
is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to
see?
Problem-Solving Strategy: Integration by Substitution
1. Look carefully at the integrand and select an expression g(x) within the integrand to set equal to u. Let’s select
g(x). such that g′ (x) is also part of the integrand.
2. Substitute u = g(x) and du = g′ (x)dx. into the integral.
3. We should now be able to evaluate the integral with respect to u. If the integral can’t be evaluated we need to
go back and select a different expression to use as u.
4. Evaluate the integral in terms of u.
5. Write the result in terms of x and the expression g(x).

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