19.40
Find J sin
4 x cos x dx.
Let w = sinjc, du = cosxdx. Then Jsin
4 jccosjt dx = J u
4 du = ^u
5 + C = 5 sin
5 x + C.
19.41
Suppose that a particle moves along the *-axis and its velocity at time fis given by v = t
2 -t — 2 for l
v = (t-2)(t + 1). Hence, u=0 when t = 2 or /=-!. Since a = D,v = 2t-l is equal to 3
when t = 2, the position s of the particle is a relative minimum when t = 2. So, the particle moves to the left
from t=\ to t = 2, and to the right from t = 2 to / = 4. Now, 5 = J v dt = $t* - ^t2 - 2t + C. By
direct computation, s(l) = -•£ + C, s(2) = - f + C and s(4) = f + C. Hence, the distance traveled
from t=l to t = 2 is \s(l) - s(2)\ = I and the distance traveled from t = 2 to f = 4 is |s(2)s(4)| = T. Thus, the total distance traveled is f.
19.42
Evaluate
19.43
Find J esc
6 x cot x dx.
Let K = CSCX. Then du =-cscx cotx dx. Hence, J esc6 x cotxdx = -J u5 du = - gw6 + C =
-1 esc
6 A: + C.
19.44
A particle moving along a straight line is accelerating at the rate of 3 ft/s
2 . Find the initial velocity if the distance
traveled during the first 2 seconds is 10 feet.
v = $3dt = 3t+C. C is the initial velocity i> 0 when t = Q. Thus, v = 3t + v a . The position 5 =
Judf= It
2 + v0t+ C,. Hence, s(2) = | -4 + 2v0 + Cl =6 + 2v0 + C,. On the other hand, s(0)=C,. So,
10 = 5(2) -5(0) = 6 + 2u0, i>0 = 2ft/s.
19.45
Evaluate J sin 3;t cos 3x dx.
Let w = sin 3*, d« = 3 cos 3x dx. Then J sin 3x cos SA: rfjc = 5 J w rfu = £ • 4 w
2 + C = g sin
2 3x + C.
19.46 Find J (x4 -4x3)\x3 -3x2) dx.
Note that, if we let g(x) = x
4 - 4x\ then g'(x) = 4jc
3 - Ux
2 =4(x
3 - 3jc
2 ). Thus, our integral has
the form J (g(x))3 • ig'(x) dx = | J (g«)VW dr = J • i • (gW)" + C = &(x4 - 4jr3)4 + C, using the result of
Problem 19.1.
19.49
Find
19.50
Find J (f + l)(f- 1) dr.
19.51
Find J(Vjf + I)
2 dx.
19.47
Evaluate J sec
2 4x tan 4x dx.
Let u = tan 4x. Then du = sec2 4x-4dx. So, J sec2 4x tan 4* dx = \ J M du = | • | u2 + C = g tan2 4.v + C.
19.48
Evaluate J (cos x sin jt)Vl + sin
2 x dx.
Let w = 1 + sin2 x. Then du = 2 sin * cos x dx. So, / (cos jc sin ;c)Vl + sin2 x dx = \ J Vu du = \ •
lu*'2 + C=ii(l + sin2x)3'2 + C.
146
CHAPTER 19
$(,+ \)(t-l)dt = $(t
2 -l)dt=
i ,t>~t + C.
/(VI + I)
2 dx = / (x + 2Vx + 1) dx = J (x + 2x
112 + 1) dx = ±x
2 + 2(i)*
3 '
2 + x + C= \x
2 + I*
3 '
2 + .v + C.
Find J sin
4 x cos x dx.
Let w = sinjc, du = cosxdx. Then Jsin
4 jccosjt dx = J u
4 du = ^u
5 + C = 5 sin
5 x + C.
19.41
Suppose that a particle moves along the *-axis and its velocity at time fis given by v = t
2 -t — 2 for l
v = (t-2)(t + 1). Hence, u=0 when t = 2 or /=-!. Since a = D,v = 2t-l is equal to 3
when t = 2, the position s of the particle is a relative minimum when t = 2. So, the particle moves to the left
from t=\ to t = 2, and to the right from t = 2 to / = 4. Now, 5 = J v dt = $t* - ^t2 - 2t + C. By
direct computation, s(l) = -•£ + C, s(2) = - f + C and s(4) = f + C. Hence, the distance traveled
from t=l to t = 2 is \s(l) - s(2)\ = I and the distance traveled from t = 2 to f = 4 is |s(2)s(4)| = T. Thus, the total distance traveled is f.
19.42
Evaluate
19.43
Find J esc
6 x cot x dx.
Let K = CSCX. Then du =-cscx cotx dx. Hence, J esc6 x cotxdx = -J u5 du = - gw6 + C =
-1 esc
6 A: + C.
19.44
A particle moving along a straight line is accelerating at the rate of 3 ft/s
2 . Find the initial velocity if the distance
traveled during the first 2 seconds is 10 feet.
v = $3dt = 3t+C. C is the initial velocity i> 0 when t = Q. Thus, v = 3t + v a . The position 5 =
Judf= It
2 + v0t+ C,. Hence, s(2) = | -4 + 2v0 + Cl =6 + 2v0 + C,. On the other hand, s(0)=C,. So,
10 = 5(2) -5(0) = 6 + 2u0, i>0 = 2ft/s.
19.45
Evaluate J sin 3;t cos 3x dx.
Let w = sin 3*, d« = 3 cos 3x dx. Then J sin 3x cos SA: rfjc = 5 J w rfu = £ • 4 w
2 + C = g sin
2 3x + C.
19.46 Find J (x4 -4x3)\x3 -3x2) dx.
Note that, if we let g(x) = x
4 - 4x\ then g'(x) = 4jc
3 - Ux
2 =4(x
3 - 3jc
2 ). Thus, our integral has
the form J (g(x))3 • ig'(x) dx = | J (g«)VW dr = J • i • (gW)" + C = &(x4 - 4jr3)4 + C, using the result of
Problem 19.1.
19.49
Find
19.50
Find J (f + l)(f- 1) dr.
19.51
Find J(Vjf + I)
2 dx.
19.47
Evaluate J sec
2 4x tan 4x dx.
Let u = tan 4x. Then du = sec2 4x-4dx. So, J sec2 4x tan 4* dx = \ J M du = | • | u2 + C = g tan2 4.v + C.
19.48
Evaluate J (cos x sin jt)Vl + sin
2 x dx.
Let w = 1 + sin2 x. Then du = 2 sin * cos x dx. So, / (cos jc sin ;c)Vl + sin2 x dx = \ J Vu du = \ •
lu*'2 + C=ii(l + sin2x)3'2 + C.
146
CHAPTER 19
$(,+ \)(t-l)dt = $(t
2 -l)dt=
i ,t>~t + C.
/(VI + I)
2 dx = / (x + 2Vx + 1) dx = J (x + 2x
112 + 1) dx = ±x
2 + 2(i)*
3 '
2 + x + C= \x
2 + I*
3 '
2 + .v + C.