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Compute the integral xe−x dx. Solution: Considering the integrand, we notice that substitution is unlikely to help, since x and e−x are not closely related. On the other hand, the integrand is a product, and one of the terms, x, is made simpler by differentiation. Therefore, we choose the technique of integration by parts, with x = u and v = e−x . Then u = 1, while v = −e−x , and we get xe−x dx = −xe−x − e−x dx = −xe−x + e−x = (1 − x)e−x . 3. Radicals. If the two most general methods (substitution and in- tegration by parts) are not helpful, then it is quite possible that there is a root sign in the integrand.

Compute tan3 x sec5 x dx. Compute tan5 x by separating one factor of tan2 x in the integrand and expressing it in terms of sec2 x. (7) Compute sec3 x dx using integration by parts, with u (x) = sec2 x and v(x) = sec x. (1) (2) (3) (4) (5) (6) 43. 1. Reversing the Technique of Substitutions. Let us assume that we want to compute the area of a circle by viewing one-fourth of that circle as the domain under a curve. Let r be the radius of the circle, and let us place the center of the circle at the origin.

Let us assume that we want to compute the area of a circle by viewing one-fourth of that circle as the domain under a curve. Let r be the radius of the circle, and let us place the center of the circle at the origin. 1, is just the domain under the graph of the function f (x) = r2 − x2 , where x ranges from 0 to r. In other words, we need to compute the integral r√ r2 − x2 dx. 11) 0 In Chapter 5, we presented the technique of integration by substitution. This technique worked in situations when the best way to compute an integral was to define a simple function of x, such as y(x) = x2 , and then continue the integration in terms of that new variable y.

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