240
CHAPTER 29
29.18
29.19
29.20
29.21
29.22
29.23
Calculate
when n is a positive integer (the exceptional case in Problem 29.17).
By Problem 29.2,
In Problems 29.19-29.29, evaluate the given antiderivative.
is in the integrand, we let
Since
Hence,
Then(Fig. 29-1),
Fig. 29-1
Fig. 29-2
present,
Since
Then
is present, we let
Then (Fig. 29-3),
Fig. 29-3
A trigonometric substitution is not required here.
Since
Then
is present, let
and (Fig. 29-4)
/,7 sin
2 nx dx
J 0 " sin
2 nxdx =
x = sec 8, dx = sec 6 tan 6 dO.
sec 0 tan 0 d0 = J tan
2 0 d0 = J (sec
2 0 - 1) d0 = tan 0 - 0 + C =
we let x = 2 sin 6, dx = 2 cos 6 dO.
(Fig. 29-2), V4-x
2 =
2 cos 0.
So
2(0 -sin 0 cos 0) + C = 2(
•f C = 2 sin '
x = tan 6, dx = sec
2 0 dO.
Vl + *
2 = sec 0. So
(1 + tan
2 0) dO = J (esc 0 = sec 0 tan 0) de = In |csc 0 - cot 0| +
sec 0 + C = In
x = 3 sec 0, dor = 3 sec 0 tan 0 d0,
= 3 tan 0.