174 Chapter 6Methods of integration
and (6.14) becomes
= 1 x 1 sin 1 x 1 + 1 cos 1 x 1 + 1 C
The art of integration by parts is the ability to make the correct choice of uand v.
Thus, the choice ofu 1 = 1 cos 1 xanddv 2 dx 1 = 1 xin our example gives
and the problem has become moredifficult. This demonstrates the rule that if one of
the factors is a polynomial then, with only one important exception, the polynomial
must be chosen as the function uin equation (6.14).
EXAMPLE 6.11Integrate by parts.
Letu 1 = 1 x
2and Then andv 1 = 1 sin 1 x, and
The degree of the polynomial under the integral sign has been decreased by one;x
2has
been replaced by x. Integrating the new integral by parts, withu 1 = 1 xand , gives
Therefore
= 1 x
21 sin 1 x 1 + 12 x 1 cos 1 x 1 − 1 2sin 1 x 1 + 1 C
The results may be verified by differentiation.
0 Exercises 40–45
Zxxdxxx xx xC
22cos =−−+sin 2 cos sin
ZZxxdxx x xdxx x xsin =−cos + cos =−cos sin+
d
dx
x
v
=sin
ZZx x dx x x x x dx
22cos =−sin 2 sin
du
dx
= 2 x
d
dx
x
v
=cos.
Zxxdx
2cos
=+
1
2
1
2
22xx xxdxcos Z sin
ZZx x dx xcos =−×−cosx x sinx dx
()
1
2
1
2
22=−xx xdxsin Zsin
ZZx
d
dx
xdx x x x
d
dx
(sin ) =−sin sin ( )xdx