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Section6_7

# Section6_7 - Section 6.7 Integrals involving transcendental...

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(3/20/08) Section 6.7 Integrals involving transcendental functions In this section we derive integration formulas from formulas for derivatives of logarithms, exponential functions, hyperbolic functions, and trigonometric functions. Topics: Integrals of y = x - 1 Integrals of exponential functions Integrals of the hyperbolic sine and cosine functions Integrals involving trigonometric functions Integrals of y = 1 / a 2 - x 2 and y = 1 / ( a 2 + x 2 ) Integrals of y = x - 1 The integration formula integraldisplay x n dx = 1 n + 1 x n +1 + C from Section 6.5 does not give integrals of x - 1 = 1 /x because we cannot set n = - 1 in the fraction 1 / ( n + 1). Instead, we use an integration formula obtained from the formula for the derivative of the natural logarithm, d dx (ln x ) = 1 x for x > 0 . (1) Theorem 1 For x in any interval not containing zero, integraldisplay 1 x dx = ln | x | + C. (2) Proof: The differentiation formula (1) translates into the integration formula integraldisplay 1 x dx = ln x + C for x > 0 which is (2) for positive x . This formula states that y = ln x is an antiderivative of y = 1 /x for positive x . What are the antiderivatives of y = 1 /x for negative x , where ln x is not defined? To answer this, we look at the graph of y = ln x and its tangent line at x = a for a positive constant a , as in Figure 1. The slope of the tangent line to y = ln x at x = a is 1 /a by (1) . The graph of an antiderivative of y = 1 /x for negative x has to be such that its tangent line at x = - a has slope 1 / ( - a ) = - 1 /a . We can obtain the graph of such an antiderivative for x < 0 by taking the mirror image about the y -axis of the graph of y = ln x , as in Figure 2, so that the tangent line at x = - a is the mirror image of the tangent line at x = a . This makes the slope of the tangent line at x = - a be the negative - 1 /a of the slope of the tangent line at x = a since taking the mirror image of a line multiplies the run between two points on the line by - 1, without changing the rise. The slope is multiplied by - 1. 243

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p. 244 (3/20/08) Section 6.7, Integrals involving transcendental functions x 1 - 1 y 1 2 a y = ln x x 1 - 1 y 1 2 a - a y = ln | x | Tangent line of slope 1 a Tangent lines of slopes 1 - a and 1 a FIGURE 1 FIGURE 2 The function that equals ln x for x > 0 and whose graph for x < 0 is the mirror image of the graph of y = ln x is y = ln | x | . Consequently, d dx (ln | x | ) = 1 x for all x negationslash = 0 (3) and this gives (2) . We could also derive (3) for negative x by using (1) and the Chain Rule. For x < 0, | x | = - x and d dx (ln | x | ) = d dx (ln( - x )) = 1 - x d dx ( - x ) = 1 x . QED Example 1 Give the exact and approximate decimal values of integraldisplay 5 2 1 x dx and integraldisplay - 2 - 5 1 x dx . Solution Formula (2) with the Fundamental Theorem gives integraldisplay 5 2 1 x dx = bracketleftBig integraldisplay 1 x dx bracketrightBig 5 2 = bracketleftBig ln | x | bracketrightBig 5 2 = ln | 5 | - ln | 2 | = ln(5) - ln(2) .
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