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

510 Chapter 7 Techniques of Integration
Then we use Formula 34 with a 2 − 5 (so a − s5 ):
y
x 2
s5 2 4x 2 dx − 1 8 y
u 2
s5 2 u 2 du − 1 8
S2 u 2 s5 2 u 2 1 5 2 sin21
u
s5
D 1 C
− 2 x 8 s5 2 4x 2 1 5 16 sin2 1S 2x
s5
D 1 C n
Example 3 Use the Table of Integrals to evaluate y x 3 sin x dx.
SOLUtion If we look in the section called Trigonometric Forms, we see that none of
the entries explicitly includes a u 3 factor. However, we can use the reduction formula in
entry 84 with n − 3:
y x 3 sin x dx − 2x 3 cos x 1 3 y x 2 cos x dx
85. y u n cos u du
− u n sin u 2 n y u n21 sin u du
We now need to evaluate y x 2 cos x dx. We can use the reduction formula in entry 85
with n − 2, followed by entry 82:
y x 2 cos x dx − x 2 sin x 2 2 y x sin x dx
Combining these calculations, we get
− x 2 sin x 2 2ssin x 2 x cos xd 1 K
y x 3 sin x dx − 2x 3 cos x 1 3x 2 sin x 1 6x cos x 2 6 sin x 1 C
where C − 3K.
n
Example 4 Use the Table of Integrals to find y xsx 2 1 2x 1 4 dx.
SOLUtion Since the table gives forms involving sa 2 1 x 2 , sa 2 2 x 2 , and sx 2 2 a 2 ,
but not sax 2 1 bx 1 c , we first complete the square:
x 2 1 2x 1 4 − sx 1 1d 2 1 3
If we make the substitution u − x 1 1 (so x − u 2 1), the integrand will involve the
pattern sa 2 1 u 2 :
y xsx 2 1 2x 1 4 dx − y su 2 1d su 2 1 3 du
− y usu 2 1 3 du 2 y su 2 1 3 du
The first integral is evaluated using the substitution t − u 2 1 3:
y usu 2 1 3 du − 1 2 y st dt − 1 2 ? 2 3 t 3y2 − 1 3 su 2 1 3d 3y2
21. y sa 2 1 u 2 du − u 2 sa 2 1 u 2
1 a 2
2 lnsu 1 sa 2 1 u 2 d 1 C
For the second integral we use Formula 21 with a − s3 :
y su 2 1 3 du − u 2 su 2 1 3 1 3 2 lnsu 1 su 2 1 3 d
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