Reverse Chain Rule Explained with Simple Animations

In this lesson, we'll explore integration by substitution, also known as the reverse chain rule.To understand integration by substitution, let's first recall the chain rule for derivatives.For example, if y equals sine of x squared, then dy dx equals cosine of x squared times two x.Integration by substitution works in reverse of the chain rule. It helps us solve complex integrals by making a strategic substitution.When we see an integral that looks like the integral of f prime of g of x, times g prime of x, dx...We can substitute u equals g of x, which gives us du equals g prime of x dx.This transformation changes our complex integral into the simpler form: integral of f prime of u, du.Which equals f of u plus a constant of integration. Finally, we substitute back to get f of g of x plus C.Let's visualize how the substitution process works with a simple example.Consider the integral of two x times cosine of x squared, dx.We identify that we can let u equal x squared, which means du equals two x dx.This transforms our integral to the cosine of u, du, which equals sine of u plus C, or sine of x squared plus C.Let's visualize how the variable u 'absorbs' the complexity of g of x.We start with a complex integral involving f prime of g of x times g prime of x.Through substitution, u absorbs both g of x and its derivative.This results in a much simpler integral of f prime of u, du, which we can solve directly.Let's apply the substitution method to solve the integral of cosine two x dx.First, we need to identify the inner and outer functions. The inner function is g of x equals two x, while the outer function's derivative is cosine of two x.Now, we'll make a substitution. Let u equal two x. Then, the differential du equals two dx, which we can rearrange to get dx equals du over two.We can now transform our original integral by substituting u for two x and dx with du over two.The integral of cosine u du is sine u plus a constant. So we have one half sine u plus C.Finally, we substitute back u equals two x to get our answer in terms of x: one half sine of two x plus C.To summarize the substitution method: we identified the inner and outer functions, made our substitution u equals two x, found the relationship between dx and du, transformed and solved the integral, and finally substituted back to get our answer.This method works for many types of integrals where we can identify an inner and outer function.In this section, we'll explore common patterns and problem-solving strategies for integration by substitution.Recognizing patterns is crucial for applying the reverse chain rule effectively.We'll examine three common patterns that you'll frequently encounter.Pattern one occurs when we have the derivative of a function composed with another function, multiplied by the derivative of the inner function.For example, when integrating cosine of x squared times two x dxWe can identify that g of x is x squared, its derivative g prime of x is two x, and f prime of u is cosine of u.Using the first pattern, we can directly write the answer as sine of x squared plus C.The second pattern involves a function composition multiplied by the derivative of the inner function.For example, when integrating x times sine of x squared dxWe make the substitution u equals x squared, which gives us du equals two x dx, or dx equals du over two x.Substituting into the integral, the x terms cancel, and we get one-half times the integral of sine of u du.Which gives us negative one-half cosine of u plus C, or negative one-half cosine of x squared plus C.The third pattern involves nested functions where we need to identify if a multiple of the derivative appears.For example, when integrating tangent squared of three x times six dxWe identify that g of x equals three x, its derivative is three, and h of x equals six, which is two times g prime of x.We substitute u equals three x, which gives us du equals three dx, or dx equals du over three.Substituting into our integral, we get two times the integral of tangent squared of u du.Now let's discuss a practical strategy for applying substitution effectively.First, look for a function and its derivative appearing together in the integrand. Then identify the inner function g of x and check if its derivative appears. Try setting u equal to g of x to see if the substitution simplifies the integral. And always check for constant multiples of the derivative.Let's compare solving a problem with and without substitution.On the left, we'll solve without substitution. On the right, we'll use our pattern recognition approach.Let's integrate e to the two x times sine of e to the two x.Without substitution, we might try integration by parts, but we quickly run into complex expressions and circular reasoning.With substitution, we recognize pattern two. We set u equal to e to the two x, find du, and the integral transforms into a standard form that's easy to solve.The key takeaway is to recognize patterns that let you transform complex integrals into familiar ones through clever substitution.