Integration by Substitution

Integration by Substitution Substitution is a very powerful tool we can use for integration. It is the analog of the chain rule for differentation, and will be equally useful to us. If we have functions F (u) and u = g(x) the chain rule tells us that the derivative of F (g(x)) with respect to x is dF du dF (g(x)) = · dx du dx It is not immediately clear how we can apply this to integration. However, we integrate both sides of the equation with respect to x, we find that Z Z dF (g(x)) dF du F (g(x)) + c = dx = · dx dx du dx We can simplify the form of this equation if we define dF = f (u) du which yields Z F (g(x)) + c = f (u) du dx dx Finally, we build on the concept of the differential. If we have a function u = g(x), then the differential du is given by du du = dx dx The above relationship says that an infintessimal change in u (du) corresponds to an infintessimal change in x (dx) times the rate of change of u with respect to x ( du ). Substituting dx this into the above equation we find Z F (g(x)) + c = f (u)du This completes the derivativation of the method of substitution, which we summarize as follows. When we are faced with an integral of the form Z f (g(x)) · g 0 (x)dx we can define a function u = g(x), and substitute du = g 0 (x)dx. Then we have an integral of the form Z f (u)du If F (u) is an antiderivative of f (u), we find that Z f (u)du = F (u) + c = F (g(x)) + c The key to integration by substitution is proper choice of u, in order to transform the integrand from an unfamiliar form to a familiar form. Let us consider some examples to gain further insight into this process. Z 2 Example 1 Evaluate 2xex dx 2 Solution Although we don’t know how to integrate 2xex , we do know how to integrate eu . Thus, our goal is to use substitution to change the integrand to the form of eu . We can do this by defining u = x2 Differentiating u we find du = 2x dx which yields du = 2xdx Now we replace any factors of x2 with u in the above integral, and replace 2xdx with du. This yields Z Z x2 2xe dx = eu du We evaluate this as Z 2 eu du = eu + c = ex + c Note that the last step is to replace u with x2 , to write the solution in terms of the original variable. Differentiating we find d x2 2 e + c = ex · 2x dx which verifies that we have successfully evaluated the above integral. It is important to note that substitution will not always work. If we were faced with an integral of the form Z 2 2x2 ex dx substitution would not have simplfied the problem (there is no useful choice for u). In fact, it is impossible to evaluate the above integral in terms of elementary functions. Above is an example of a function that is a perfect candidate for numerical integration, using a computer. Z 4x dx Example 2 Evaluate 4x2 + 2 Solution Faced with a fraction of this form, it is our goal to use substitution to reduce it to the form u1 which we can integrate to ln(|u|). If we include the second power of x in our definition of u, when we differentitate u with respect to x, we will be left with the first power of x, which will replace the x in the denominator nicely. Thus, we choose u = 4x2 + 2 This yields du = 8x dx so du = 8xdx or du = 4xdx 2 Substituting into the original equation we find Z Z 1 1 1 1 1 4x dx = du = ln(|u|) + c = ln(|4x2 + 2|) + c = ln(4x2 + 2) + c 2 4x + 2 2 u 2 2 2 Z Example 3 Evaluate cos(t) sin(t)dt Solution In integrating this function, both u = sin(t) and u = cos(t) are reasonable choices for substitution. We’ll consider both of them, to see how the process varies depending on the choice of u. Let us begin with u = cos(t) we find du = − sin(t) dt so du = − sin(t)dt or − du = sin(t)dt substituting into the integral Z Z cos(t)2 u2 +c cos(t) sin(t)dt = − udu = − + c = − 2 2 Alternatively, if we designate u = sin(t) we find du = cos(t) dt so du = cos(t)dt substituting into the integral Z Z u2 sin(t)2 +c= +c cos(t) sin(t)dt = udu = 2 2 It is curious that we received two different results by using different representatives for substitution. In order for substitution to be a valid method, surely valid substitutions should lead to the same result. In fact, they do in this case, if we recognize that cos(t)2 + sin(t)2 = 1 or cos(t)2 = 1 − sin(t)2 Substituting into our first result − cos(t)2 1 − sin(t)2 sin(t)2 1 sin(t)2 +c=− +c= − +c= +d 2 2 2 2 2 Where d is an arbitrary constant. Absorbing the − 21 into the arbitrary constant, we see that these two results are in fact consistent. The following example involves the application of substitution to solve a pure-time differential equation. Example 4 Suppose fish size begins at L(0) = 0 (measured from fertilization) and follows the differential equation dL = 6.48e−0.09t cm dt Find the length of the fish as a function of time. Solution We can solve this pure-time differential equation using integration, but we will also have to apply the method of substitution. We would like to choose u such that our integrand is of the form eu , which we know how to integrate. Define u = −0.09t and note du = −0.09dt or du = dt −0.09 Using substitution Z Z 6.48 −0.09t 6.48e dt = eu du = −72eu + c = −72e−0.09t + c −0.09 Now we can use the initial condition to solve for an exact value for c. L(0) = −72e−0.09·0 + c = −72 + c = 0 which implies c = 72. Thus, our solution to the initial value problem is L(t) = 72(1 − e−0.09t ) cm This is called the von Bertalanffy growth equation. As the fish matures it nevers stops growing, but the rate at which it grows continues to decrease. The length of the fish approaches but never reaches 72 cm. We say that such a function has a horizontal asymptote at 72 cm, because the function approaches 72 as t → ∞.