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INVERSE TRIGONOMETRIC FUNCTIONS in .NET framework
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0 1 + x2
Integration by Parts
In this chapter, we shall learn one of the most useful techniques for nding antiderivatives. Let f and g be differentiable functions. The product rule tells us that d (f (x)g(x)) = f (x)g (x) + g(x)f (x) dx or, in terms of antiderivatives, f (x)g(x) = (f (x)g (x) + g(x)f (x)) dx = f (x)g (x) dx + g(x)f (x) dx The substitutions u = f (x) and v = g(x) transform this into1 uv = from which we obtain u dv = uv v du integration by parts u dv by an easy integration v du. u du + v du The idea behind integration by parts is to replace a dif cult integration EXAMPLES
(a) Find xex dx. This will have the form u dv if we choose u=x and dv = ex dx
Since dv = v (x) dx and dv = ex dx, we must have v (x) = ex . Hence, v= and we take the simplest case, C = 0, making v = ex . 1 For
ex dx = ex + C
example, f (x)g (x) dx =
u dv, where, in the result of the integration on the right, we replace u by f (x) and v by g(x). In fact, by the chain f (x)g (x) dx rule, d d dv d d u dv = u dv = u [g(x)] = f (x)g (x) = dx dv dx dx dx f (x)g (x) dx. Similarly, v du = g(x)f (x) dx. Hence, u dv =
Copyright 2008, 1997, 1985 by The McGrawHill Companies, Inc. Click here for terms of use.
INTEGRATION BY PARTS
[CHAP. 38
Since du = dx, the integration by parts procedure assumes the following form: u dv = uv xex dx = xex v du ex dx = xex ex + C = ex (x 1) + C
Integration by parts can be made easier to apply by setting up a box such as the following one for example (a) above: u=x du = dx dv = ex dx v = ex In the rst row, we put u and dv. In the second row, we write du and v. The result uv v du is obtained from the box by rst multiplying the upper left corner u by the lower right corner v and then subtracting the integral v du of the two entries in the second row. Notice that everything depends on a wise choice of u and v. If we had instead picked u = ex and dv = x dx, then v = x dx = x 2 /2 and we would have obtained xex dx = ex x2 2 x2 x e dx 2 xex dx which is true enough, but of little use in evaluating xex dx. We would have replaced the dif cult integration by the even more dif cult integration (x 2 /2)ex dx. (b) Find x ln x dx. Let us try u = ln x and dv = x dx: u = ln x du = Thus, u dv = uv x ln x dx = (ln x) = = = (c) Find ln x dx. Let us try u = ln x and dv = dx: u = ln x 1 du = dx x dv = dx v=x v du x2 2 x2 1 dx 2 x x dx 1 dx x dv = x dx v= x2 2

