James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 7.1 Integration by Parts 473
It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as
expected.
It is helpful to use the pattern:
u − dv −
du − v −
SOLUTION USING FORMULA 2 Let
u − x dv − sin x dx
Then du − dx v − 2cos x
and so
u d√ u √ √ du
y x sin x dx − y x sin x dx − x s2cos xd 2 y s2cos xd dx
− 2x cos x 1 y cos x dx
− 2x cos x 1 sin x 1 C n
Note Our aim in using integration by parts is to obtain a simpler integral than
the one we started with. Thus in Example 1 we started with y x sin x dx and expressed
it in terms of the simpler integral y cos x dx. If we had instead chosen u − sin x and
dv − x dx, then du − cos x dx and v − x 2 y2, so integration by parts gives
y x sin x dx − ssin xd x 2
2 2 1 2 y x 2 cos x dx
Although this is true, y x 2 cos x dx is a more difficult integral than the one we started
with. In general, when deciding on a choice for u and dv, we usually try to choose
u − f sxd to be a function that becomes simpler when differentiated (or at least not more
complicated) as long as dv − t9sxd dx can be readily integrated to give v.
Example 2 Evaluate y ln x dx.
SOLUTION Here we don’t have much choice for u and dv. Let
u − ln x
dv − dx
Then du − 1 x dx v − x
Integrating by parts, we get
y ln x dx − x ln x 2 y x dx
x
It’s customary to write y 1 dx as y dx.
Check the answer by differentiating it.
− x ln x 2 y dx
− x ln x 2 x 1 C
Integration by parts is effective in this example because the derivative of the function
f sxd − ln x is simpler than f.
n
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