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

Section 7.1 Integration by Parts 475
Figure 1 illustrates Example 4 by showing
the graphs of f sxd − e x sin x and
Fsxd − 1 2 e x ssin x 2 cos xd. As a visual
check on our work, notice that f sxd − 0
when F has a maximum or minimum.
12
This can be regarded as an equation to be solved for the unknown integral. Adding
y e x sin x dx to both sides, we obtain
2 y e x sin x dx − 2e x cos x 1 e x sin x
Dividing by 2 and adding the constant of integration, we get
f
F
y e x sin x dx − 1 2 e x ssin x 2 cos xd 1 C
n
_3
_4
6
If we combine the formula for integration by parts with Part 2 of the Fundamental
Theorem of Calculus, we can evaluate definite integrals by parts. Evaluating both sides
of Formula 1 between a and b, assuming f 9 and t9 are continuous, and using the Fundamental
Theorem, we obtain
FIGURE 1
6
y b
f sxdt9sxd dx − f sxdtsxdg b
a 2 y b
tsxd f 9sxd dx
a a
Since tan 21 x > 0 for x > 0, the integral
in Example 5 can be interpreted as the
area of the region shown in Figure 2.
y
0
y=tan–!x
1
x
Example 5 Calculate y 1
tan 21 x dx.
SOLUTION Let
0
u − tan 21 x
dv − dx
Then du − dx
1 1 x 2 v − x
So Formula 6 gives
y 1
0 tan21 x dx − x tan 21 xg 1 2
0 y 1
0
x
1 1 x 2 dx
− 1 ? tan 21 1 2 0 ? tan 21 0 2 y 1
− 4 2 y 1
0
x
1 1 x 2 dx
0
x
1 1 x 2 dx
To evaluate this integral we use the substitution t − 1 1 x 2 (since u has another meaning
in this example). Then dt − 2x dx, so x dx − 1 2 dt. When x − 0, t − 1; when x − 1,
t − 2; so
y 1
0
x
1 1 x dx − 1 2 2 y 2 dt
− 1 2
1 t
ln | t 2
|g 1
FIGURE 2
Equation 7 is called a reduction formula
because the exponent n has been
reduced to n 2 1 and n 2 2.
Therefore y 1
0 tan21 x dx − 4 2 y 1
0
Example 6 Prove the reduction formula
7
where n > 2 is an integer.
− 1 2 sln 2 2 ln 1d − 1 2 ln 2
x
1 1 x dx − 2 4 2 ln 2
2
y sin n x dx − 2 1 n cos x sinn21 x 1 n 2 1
n
y sin n22 x dx
n
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