INVERSE TRIGONOMETRIC FUNCTIONS in .NET framework

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[CHAP. 37
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37.24 Find the area under the curve y = 37.25 Find the area under the curve y =
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1 above the x-axis, and between the lines x = 0 and x = 1. 1 + x2 1 1 x2 1 x2 x2 1 above the x-axis, and between the lines x = 0 and x = 1/2. above the x-axis, and between the lines x = 2/ 3 and x = 2, is revolved
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37.26 The region R under the curve y =
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around the y-axis. Find the volume of the resulting solid. 37.27 Use the arc-length formula (32.2) to nd the circumference of a circle of radius r. [Hint: Find the arc length of the part of the circle x 2 + y2 = r 2 in the rst quadrant and multiply by 4.] 37.28 A person is viewing a painting hung high on a wall. The vertical dimension of the painting is 2 feet and the bottom of the painting is 2 feet above the eye level of the viewer. Find the distance x that the viewer should stand from the wall in order to maximize the angle subtended by the painting. [Hint: Express as the difference of two inverse cotangents; see Fig. 37-10.]
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Fig. 37-10
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37.29 For which values of x is each of the following equations true (a) sin 1 (sin x) = x (d) sin (sin 1 x) = x (b) cos 1 (cos x) = x (e) cos (cos 1 x) = x (c) sin 1 ( x) = sin 1 x
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(f ) GC Use a graphing calculator to draw the graph of y = sin 1 (sin x) and review your answer to part (a). 37.30 Find y by implicit differentiation: (a) x 2 x tan 1 y = ln y (b) cos 1 xy = e2y
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37.31 Sketch the graph of y = tan 1 x ln 1 + x 2 . 37.32 Assuming that cot 1 x and csc 1 x are differentiable, use implicit differentiation to derive the formulas for their derivatives. 37.33 Find the average values of f (x) = 37.34 (a) GC Show that = n = 8.
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1 on [ 2, 2]. 4 + x2 . (b) Approximate by applying Simpson s rule to the de nite integral in part (a), with
4 dx
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 McGraw-Hill 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
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