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1=p    2 1 x sin 1 x2 1 1 2 2 1 2 2  p dx sin x  : sin 1 144 4 4 2 1 x4 0
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x dx 1 2x 1 1 1 2x 1 1 dx p p dx p dx p 2 2 x2 x 1 x2 x 1 2 x2 x 1 x2 x 1 1 1 dx q x2 x 1 1=2 d x2 x 1 2 2 x 1 2 3 q p x2 x 1 1 ln jx 1 x 1 2 3j c 2 2 2 4
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p p dx . Let x 1 3 tan u, dx 3 sec2 u du. x 1 2 3 3=2 p u tan 1 0 0; when x 2, u tan 1 1= 3 =6. Then the integral becomes Write the integral as =6 =6 p 2 =6 p 2  3 sec u du 3 sec u du 1 =6 1 1 cos u du sin u  2 u 3=2 2 u 3=2 3 0 3 6 3 sec 0 3 3 tan 0 0
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When x e, y 1; when x e2 , y 2. 2 dy y  3 3 2 1 8 1 y
2 2 
Let ln x y, dx =x dy.
Then the integral becomes
5.17. Find xn ln x dx if (a) n 6 1, (b) n 1.
CHAP. 5]
INTEGRALS
(a) Use integration by parts, letting u ln x, dv xn dx, so that du dx =x, v xn 1 = n 1 . n 1 xn 1 x dx ln x xn ln x dx u dv uv v du n 1 n 1 x xn 1 xn 1 ln x c n 1 n 1 2 b 1 x 1 ln x dx ln x d ln x ln x 2 c: 2
p 2x 1
Then
5.18. Find 3
Then dx y dy and the integral becomes 3y y dy.
p 2x 1 y, 2x 1 y2 .
Integrate by parts, letting u y, dv 3y dy; then du dy, v 3y = ln 3 , and we have y y 3y 3 y 3y 3y 3y y dy u dv uv v du dy c ln 3 ln 3 ln 3 ln 3 2
1 5.19. Find
x ln x 3 dx.
dx x2 , v . Hence on integrating by parts, x 3 2  2  1 x dx x2 1 9 x 3 ln x 3 ln x 3 dx 2 x 3 2 x 3 2 ( ) 1 x2 ln x 3 3x 9 ln x 3 c 2 2 1 5 9 4 ln 4 ln 3 4 2
Let u ln x 3 , dv x dx. Then du x ln x 3 dx x2 2 x2 2
Then
x ln x 3 dx
5.20. Determine
6 x dx. x 3 2x 5
Use the method of partial fractions.
6 x A B . x 3 2x 5 x 3 2x 5
Method 1: To determine the constants A and B, multiply both sides by x 3 2x 5 to obtain 6 x A 2x 5 B x 3 or 6 x 5A 3B 2A B x 1
Since this is an identity, 5A 3B 6, 2A B 1 and A 3=11, B 17=11. Then 6 x 3=11 17=11 3 17 dx dx dx ln jx 3j ln j2x 5j c x 3 2x 5 x 3 2x 5 11 22 Method 2: Substitute suitable values for x in the identity (1). For example, letting x 3 and x 5=2 in (1), we nd at once A 3=11, B 17=11.
5.21. Evaluate
dx by using the substitution tan x=2 u. 5 3 cos x
1 +
From Fig. 5-7 we see that u sin x=2 p ; 1 u2 1 cos x=2 p 1 u2
Fig. 5-7
INTEGRALS
[CHAP. 5
Then
cos x cos2 x=2 sin2 x=2 1 du sec2 x=2 dx or 2
1 u2 : 1 u2 2 du : 1 u2
Also
dx 2 cos2 x=2 du
Thus the integral becomes
  du 1 1 1 tan 1 u=2 c tan 1 tan x=2 c: 2 2 u2 4 2
 5.22. Evaluate
x sin x dx. 1 cos2 x
Let x  y. Then     x sin x  y sin y sin y y sin y I dx dy  dy dy 2x 2y 2y 2 0 1 cos 0 1 cos 0 1 cos 0 1 cos y  d cos y  I  tan 1 cos y j I 2 =2 I 0 2 0 1 cos y i.e.; I 2 =2 I or I 2 =4:
=2 5.23. Prove that
p  sin x p p dx . 4 sin x cos x
Letting x =2 y, we have p p =2 =2 =2 p cos y cos x sin x p p dx I p p dx p p dy sin x cos x cos x sin x cos y sin y 0 0 0 Then I I p p =2 cos x sin x p p dx p p dx cos x sin x sin x cos x 0 0 =2 p p =2  sin x cos x p p dx dx 2 sin x cos x 0 0 =2
from which 2I =2 and I =4. The same method can be used to prove that for all real values of m, =2 sinm x  dx m m 4 0 sin x cos x (see Problem 5.89). Note: This problem and Problem 5.22 show that some de nite integrals can be evaluated without rst nding the corresponding inde nite integrals.
NUMERICAL METHODS FOR EVALUATING DEFINITE INTEGRALS 1 dx approximately, using (a) the trapezoidal rule, (b) Simpson s rule, where the 5.24. Evaluate 2 01 x interval 0; 1 is divided into n 4 equal parts.
Let f x 1= 1 x2 . Using the notation on Page 98, we nd x b a =n 1 0 =4 0:25. Then keeping 4 decimal places, we have: y0 f 0 1:0000, y1 f 0:25 0:9412, y2 f 0:50 0:8000, y3 f 0:75 0:6400, y4 f 1 0:50000. (a) The trapezoidal rule gives
CHAP. 5]
INTEGRALS x 0:25 fy 2y1 2y2 2y3 y4 g f1:0000 2 0:9412 2 0:8000 2 0:6400 0:500g 2 0 2 0:7828:
(b) Simpson s rule gives x 0:25 fy 4y1 2y2 4y3 y4 g f1:0000 4 0:9412 2 0:8000 4 0:6400 0:5000g 3 0 3 0:7854: The true value is =4 % 0:7854:
APPLICATIONS (AREA, ARC LENGTH, VOLUME, MOMENT OF INTERTIA) 5.25. Find the (a) area and (b) moment of inertia about the y-axis of the region in the xy plane bounded by y 4 x2 and the x-axis.
(a) Subdivide the region into rectangles as in the gure on Page 90. A typical rectangle is shown in the adjoining Fig. 5-8. Then Required area lim
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