7.5, #17. Formula 15 with x = y, p(x) = y2, a = 2 gives

ó
õ 		y2sin2y dy = - 1
2
 y2cos2y+ 1
4
 2ysin2y + 1
8
 2cos2y+C.


7.5, #18. Formula 11 with x = y, a = 2, b = 7 gives

ó
õ 		cos2ycos7y dy = 1
45
 (7cos2ysin7y-2sin2ycos7y)+C.


7.5, #24. Substitute x = 3y, dx = 3dy to get òcos4 3y dy = (1/3)òcos4x dx. Then use formula 18 with n = 4 to get

ó
õ 		cos4x dx = 1
4
 cos3xsinx+ 3
4
 ó
õ 		cos2 x dx
Use formula 18 again with n = 2 to get
ó
õ 		cos2x dx = 1
2
 cosxsinx+ ó
õ 		 dx = 1
2
 cosxsinx+ 1
2
 x+C.
Putting these pieces together gives
ó
õ 		cos4 3y dy = 1
12
 cos3 3ysin3y+ 1
8
 cos3ysin3y+ 3
8
 y+C.


7.5, #28. Make the Substitution w = 2x, dw = 2dx and then use formula 24 to get

ó
õ 		dx
1+4x2
 = 1
2
 ó
õ 		dw
1+w2
 = 1
2
 arctanw+C = 1
2
 arctan2x+C.

Comment: A very common mistake - using formula 24 without the first making the substitution w = 2x. This gives incorrect answer arctan2x +C.


7.5, #37. Formula 26 with x = z, a = 3, and b = 0 gives

ó
õ 		dz
z(z-3)
 = 1
3
 (ln|z-3|-ln|z|)+C.


7.5, #38. Formula 26 with x = y, a = 2, and b = -2 gives

ó
õ 		dy
4-y2
  dy = - ó
õ 		dy
(y-2)(y+2)
 = - 1
4
 (ln|y-2|-ln|y+2|)+C.


7.5, #44. Since y2+4y+5 = (y+2)2+12, we substitute w = y+2 and dw = dy to get

ó
õ 		dy
y2+4y+5
 = ó
õ 		dw
w2+12
 = arctanw+C = arctan(y+2)+C.


7.5, #45. The integrand is an improper rational function, so we use long division to write

x3+3
x2-3x+2
 = x+3+ 7x-3
x2-3x+2
 .
Now x2-3x+2 = (x-1)(x-2), so formula 27 with a = 1, b = 2, c = 7, and d = -3 applies, giving
ó
õ 		x3+3
x2-3x+2
  dx = 1
2
 x2+3x-4ln|x-1| +11ln|x-2|+C.

File translated from TEX by TTH, version 2.00.
On 15 Aug 1999, 15:26.