ΙΝΤΕΓΡΑΤΙΟΝ ΒΨ ΠΑΡΤΣ
233
28.9
28.10
28.11
28.12
28.13
28.14
28.15
28.16
28.17
28.18
Let M = cos fee, dv = e""dx, du=-bsinbx, v = (\la)e". Then J e" cos bxdx = (\la)e cos bx +
(b/a) I e°* sin bx dx. Apply integration by parts to the latter: a = sin bx, dv = e°* dx, du = b cos bx,
v = (\ld)e°*. So / e" sin bxdx = (l/a)ea" sin bx - (b/«) JV* cos bxdx. Hence, by substitution,
J e°* cos bx dx = (l/a)e'"'cosbx + (b/a)[(l/a)e'"smbx-(b/a)$ e°*cosbxdx] = (l/a)e" cos fee +
(bla2)eax sin fee - (62/a2) J e* cos fee dx. Thus, (1 + b2/a2) J eaf cos fee dx = (e"/a2)(a cos bx + b sin fee) + C,
f e" cos bx dx = [e"/(a
2 + fc" Wo cos fcx + b sin fee) + C,.
Let w = sinx, du = sinxdx, du=cosxdx, u = —cosx. Then J sin
2 x dx = -sin xcosx + J cos
2 x -sin x cos jc + J (1 - sin
2 ;c) dx = —sin x cos * 4- AT - J sin
2 x dx.
So
2 J sin
2 JT dx = x — sin jr cos x + C,
f sin
2 x dx= 5 (x - sin jc cos x) + C,.
Then J" x cos
2 x dx =
Let
2* = y
and use Problem 28.5:
/ * sin 2x dx
(-2x cos 2x + sin 2*) + C.
Use a substitution
M = x
2 ,
du = 2x dx.
Then
/ x sin x
2
J e"' cos fee <&.
/ sin
2 x dx.
f cos
3 x dx.
J cos
3 je dx = J cos jc (1 - sin
2 *) dc = J cos .* dx — / sin
2 x cos x dx = sin
sin
3 x + C.
| cos4 x dx.
/ cos
4 A:
$xe
3 * dx.
J A: sec
2 x dx.
J je cos
2 x dx.
J (In x)
2 dr.
Let M = AT, rfu = e
31 dx, du — dx,
Then
Let u = x, dv = sec je dx, da = dx, v = tan x. Then J x sec
2 x dx = x tan x - J tan x dx = x tan x —
Inlsecxl + C.
(1 + 2 cos 2x + cos
2 2x)
y sin y dy
Let
x = e"
and use Problem 28.1:
J (Inx)
2 dx = -/ fV dt = e~'(t
2 + 2t + 2) + C = x[(lnx)
2 -
2 In x + 2] + C.
J x sin 2x dx.
Jx sin (x
2 ) dx.
cos x + C.
sin u du =
cos u + C =
(-y cos y + sin y) + C =
Let M = x,
dv = cos
2 A: etc, du = dx,
(2 sin 2x +
cos 2*) + C.