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TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
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sin 2x 1 sin 4x 1 x x+ 8 2 2 4
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cos 2x dx
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u2 du
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[letting u = sin 2x]
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sin 2x x sin 4x sin 2x 1 sin3 2x 1 x + 8 2 2 8 2 2 3 1 x sin 4x sin3 2x 8 2 8 6 +C
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(c) From Problem 34.6(a), we know how to integrate the rst power of tan x, tan x dx = ln |sec x| + C Higher powers are handled by means of a reduction formula. We have, for n = 2, 3, , tann x dx = = = = Similarly, from sec x dx = ln |sec x + tan x| + C and the reduction formula of Problem 38.13(a), we can integrate all powers of sec x. (d) Antiderivatives of the forms sin Ax cos Bx dx, sin Ax sin Bx dx, cos Ax cos Bx dx can be computed by using the identities tann 2 x(tan2 x) dx = tann 2 x sec2 x dx un 2 du tann 1 x n 1 tann 2 x(sec2 x 1) dx tann 2 x dx [let u = tan x] (39.1)
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1 (sin(A + B)x + sin(A B)x) 2 1 sin Ax sin Bx = (cos(A B)x cos(A + B)x) 2 1 cos Ax cos Bx = (cos(A B)x + cos(A + B)x) 2 sin Ax cos Bx = For instance, sin 8x sin 3x dx = 1 sin 5x sin 11x 1 (cos 5x cos 11x) dx = 2 2 5 11 +C
39.2 TRIGONOMETRIC SUBSTITUTIONS To nd the antiderivative of a function involving such expressions as often helpful to substitute a trigonometric function for x.
a2 + x 2 or
a2 x 2 or x 2 a2 , it is
TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
[CHAP. 39
EXAMPLES
(a) Evaluate None of the methods already available is of any use here. Let us make the substitution x = 2 tan , where /2 < < 1 (x/ 2). Figure 39-1 illustrates the relationship between x and , with interpreted as an angle. /2. Equivalently, = tan We have dx = 2 sec2 d and, from Fig. 39-1, x2 + 2 = sec 2 where sec > 0 (since /2 < < /2). Thus, x 2 + 2 dx = ( 2 sec )( 2 sec2 ) d = 2 sec3 d [by Problem 38.13(b)] 2 sec x 2 + 2 dx.
x2 + 2 =
= sec tan + ln |sec + tan | + C = x2 + 2 x 2 + ln 2 x2 + 2 2
x + +C 2 since 2+C |a| a = |b| b
= = =
x x2 + 2 + ln 2 x x2 + 2 + ln 2 x x2 + 2 + ln 2
x2 + 2 + x +C 2 x 2 + 2 + x ln
x 2 + 2 + x + C1
Note how the constant ln 2 was absorbed in the constant term in the last step. The absolute value signs in the logarithm may be dropped, since x 2 + 2 + x > 0 for all x. This follows from the fact that x 2 + 2 > x 2 = |x| x.
This example illustrates the following general rule: If x = a tan with ( /2) < < ( /2).
x 2 + a2 occurs in an integrand, try the substitution
Fig. 39-1
(b) Evaluate
Fig. 39-2
4 x2 dx. x Make the substitution x = 2 sin , where /2 /2; that is, = sin 1 (x/2). The angle interpretation of is shown in Fig. 39-2. Now dx = 2 cos d and 4 x2 = x 2 1 sin2 cos 4 4 sin2 = = = cot 2 sin 2 sin sin
CHAP. 39]
TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
Note that
1 sin2 =
cos2 = |cos | = cos , since cos 0 when /2 /2. Thus, (cot )(2 cos ) d = 2 cos2 d = 2 sin 2 x cos sin (cos ) d csc d sin d
4 x2 dx = x =2
1 sin2 d = 2 sin 4 x2 2
= 2(ln |csc cot | + cos ) + C = 2 ln = 2 ln 4 x2 + x 4 x2 + x +C
2 x
4 x2 + C
This example illustrates the following general rule: If x = a sin with /2 /2.
(c) Evaluate
a2 x 2 occurs in an integrand, try the substitution
x2 4 dx. x3 Let x = 2 sec , where 0 < /2 or < 3 /2; that is, = sec 1 (x/2). Then dx = 2 sec tan d and x2 4 = 4 sec2 4 = 2 sec2 1 = 2 tan2 = 2 |tan | = 2 tan
Note that tan > 0, since is in the rst or third quadrant. Then x2 4 dx = x3 2 tan (2 sec tan ) d 8 sec3 1 1 1 tan2 sin2 = d = d = 2 2 sec2 2 2 2 sec cos = = 1 x 1 ( sin cos ) + C = sec 1 4 4 2 2 x2 4 1 x sec 1 4 2 x2 +C
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