6.(2)Jo In(1-2a cos +a2)d.c (a|< 1) fif: i f(a, a)=In(1-2a cos +a2), I(a)=o In(1-2a cos x+a2)da 因为1-2ac0sx+a2≥1+a2-2a=(1-|a)2>0,所以f(x,a)在 R×(-1,1)上连续,且fa(x,a)=1=2+在R×(-1,1)上连续利用 定理2可得: T'(a)=Jo 1-20 cos x+a24. aJ7(1+1=2+14)d Cs工 arctan(a tan 5)2=0 0. 于是I(a)=C(常数),而I(0)=0,故I(a)=C=0.口 7.利用xdy=求/sin(m)==d 解:不妨设0<a<b,因为sin(mn)在0,1;a,b上连续(注意在 {0}×{a,上定义sin(mn)xy为0).所以 f o sin(In -)mrdr sin(In )dx dyo sin(In i)r 作变换x=e-t可得 yd c-(+) sin tdt(注:这是一个含参广义积分 H++uF[-(y+1)sint-cos t]e-(+)1=t 1+(1+y 于是 Jo sin(In i)=ned x dy= arctan(1+yI arctan(1+b)-arctan(1 +a) rotan b-a +b)(1+a6.(2)
π 0 ln(1 − 2α cos x + α2)dx (|α| < 1). !"< f(x, α) = ln(1−2α cos x+α2), I(α) =
π 0 ln(1−2α cos x+α2)dx. &# 1 − 2α cos x + α2  1 + α2 − 2|α| = (1 − |α|)2 > 0, $' f(x, α) Æ R × (−1, 1)   fα(x, α) = −2 cos x+2α 1−2α cos x+α2 Æ R × (−1, 1)  . 5* )* 2 S" I
(α) =
π 0 −2 cos x+2α 1−2α cos x+α2 dx = 1 α
π 0 (1 + α2+1 1−2α cos x+α2 )dx = π α − 1−α2 α(1+α2)
π 0 1 1+ −2α 1+α2 cos x dx = π α − 2 α arctan( 1+α 1−α tan x 2 )| x=π x=0 = π α − 2 α · π 2 = 0.  I(α) = CTÆ7UV I(0) = 0, : I(α) = C = 0. 7. 5*
b a xydy = xb−xa ln x
1 0 sin(ln 1 x ) xb−xa ln x dx. !";W< 0 <a<b, &# sin(ln 1 x )xy Æ [0, 1; a, b]  T78Æ {0} × [a, b] )2 sin(ln 1 x )xy # 0U$'
1 0 sin(ln 1 x ) xb−xa ln x dx =
1 0 sin(ln 1 x )dx
b a xydy =
b a dy
1 0 sin(ln 1 x )xydx 9X x = e−t S
1 0 sin(ln 1 x )xydx =
+∞ 0 e−(y+1)t sin tdt( 7".:YZ[\2 ) = 1 1+(1+y)2 [−(y + 1) sin t − cost]e−(y+1)t | t=+∞ t=0 = 1 1+(1+y)2 . 
1 0 sin(ln 1 x ) xb−xa ln x dx =
b a 1 1+(1+y)2 dy = arctan(1 + y)| b a = arctan(1 + b) − arctan(1 + a) = arctan b−a 1+(1+b)(1+a) . 3