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Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 82. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | 25.2 Substitution The Chain Rule in Reverse 791 x2 du 2x dx so x dx 1 du 1 x cos x2 dx j cos x2 x dx 1 cos u du 2 Transform the integral using substitution. _ 1 2 y cos u du This is one of our basic integrals. _ 1 2 sin u C Now return to the original variable. sin x2 C We can verify that this answer is correct by differentiating it.3 1 sin x2 C dx 2 - cos x2 2x x cos x2 2 Why Is the Substitution du u x dx Legitimate In the example above we went from d u x to du u x dx. Why is this valid Suppose that fwdx c That is F x f x . Then I f u du F u C. Is it true that u c Sure F u x F u x u x f u x u x dx Therefore it works to replace u x by u and u x dx by du. Replacing du dx by du is legitimate. This is part of the genius of Leibniz notation du for the derivative. One generally doesn t run into trouble treating du as it appears despite the fact that du is really not a fraction. Now we return to more examples. 3 We can always check an answer to an integration problem by differentiation. 792 CHAPTER 25 Finding Antiderivatives An Introduction to Indefinite Integration Using Substitution to Reverse the Chain Rule The key idea when making a substitution in order to reverse the Chain Rule is to choose u so that the integral is of the form f f u x u x dx. We must choose u so that u is already sitting by in the integrand up to a constant multiple. EXAMPLE 25.4 Evaluate the following integrals. aU 3ÍT2 dx c f dx b f x dx Jxe d f cos2 x sin x dx SOLUTIONS Try these problems on your own and then compare your answers with those below. You can always check your answers by differentiating. a f 3X 2 dx 5 3x 2 dx .We hope we essentially have 1 du. Let u 3x 2. du 3 so dx 1 - du dx. 3 Then Rewrite the original integral in terms of u. 1 3x 2 1 du 3 This is a familiar integral. 5 3 5 3 In 3x 2 C Return to the original variable. b S . xeV dx xX dx. We hope we essentially have eudu. Let u Vx. du 11 dx 2 x 1 2du dx x Rewrite the original integral in terms of u. y e x dx y eu 2 du 2 y eudu This
