第三章积分计算及应用

求下列各定积分1I11

5.5u,13,11.54xu2,x1(5u5.5 (5 12(5 1I 1udu13(5u2)dx15u1u3

1

8 8

ln2xexdxln2xdexxexln2ln2exdxln2exln21(1ln 10

1x2

/ 2/2sin2t(1sin2t)dt20

1313142

0xsinxdx0xdcosxxcosx00cosxdxsinx0 x29dx x299ln(x x29)4109 1 1 3 dx6sin2tdt 6(1cos2t)dt (t sin2t)6 1 2 1

2 4 314x2dx 2arcsinx 34 4

. 2 13 19

34 310 dx21

dx2 1udu8(1

808 cosxcosxdx2 cosxcosxdx2 2 332 cosxdcosx cos2 3300 1

2

2xdx

2

2xd2x1

udu1

(t

0

2 0,n2k2

2/ 2 /2sinn2

(t)dt(n(n1)!!

sinn(t)dt

2 (n

(ax)dx(xasint)0

tdt

/2sin11xdx10!!156 642 642 20 dx sinudu2

5 304(xsinx)2dx1x2(1cos2x)dx11x31x2dsin304 2 3

1

xsin 2 1 0640xdcos2x64xcos 40cos031sin2x3 4 /4tan4xdx/4tan2x(sec2x 0/4tan2xsec2xdx/4tan20/4tan2xdtanx/4(sec2x33tan

tanx|/

11 1

0arcsinxdxxarcsinx|00xdarcsin1 12

dx11x

x2x2xx2x2x20ln(x )dxxln(x )0xdln(x 0ln(

2a2) dxln(x2x2

x22ax20ln(2a2)2a2|a| 设f(x)在[ab]连续.证明af(x)dxba)0f(ab 证令xaba)t,则0a,1bdxba af(x)dx(ba)0f(a(ba)t)dt(ba)0f(a(ba 1证明0xf(x)dx20xf证令x2t,则x0时t0xa时ta2a 1a 1 10xf(x)dx20xf(x)dx20tf(t)dt20xf 1xm(1x)ndx1xn(1 证令x1t,则x0时x1时t0.dxdt,1xm(1x)ndx0(1t)mtndt1(1t)mtndt1xn(1 利用分部积分公式证明,若f(x)连续x 00f(x)dxdt0f(t)(x证xtf(x)dxdtttf(x)dxxxttf(x)dxa 0xf(x)dx0tf(t)dt0xf(t)dt0 x0f(t)(x 利用换元积分法证明 f(sin 证xtx0时tx时dxdt 0xf(sinx)dx(t)f(sin( 0(t)f(sint)dt0f(sint)dt0tf(sin 0f(sint)dt0xf(sin 20xf(sinx)dx0f(sinxf(sinx)dx1f(sin 1/2f(sint)dt1 f(sin /令ut,则t2时u2,t时u0du //2f(sint)dt/2f(sin(u))du f(sin 0xf(sinx)dx f(sin xsinx xsinx解 xsinxdx/ sin dx/2dcos01cos2 1cos2 1cos2arctancosx|/2 设函数f(x)在(,)上连续,以T为周期,证明函数F(xxTf(x)dxxf(t)dt也以TT lim1xf(t)dt1Tfxx 证(1)F(xTxTTf(x)dxxTf xTf(x)dxTf(x)dx xf(t)dtxTf(t)dtT0 xTf(x)dxTf(x)dx xf(t)dtTfT xTf(x)dxxf(t)dtFT (2)1xf(t)dt1Tf

Tf(x)dx

xf(t)dtF(x)xT xf(t)dt Tf(x)dxlimF(x)xx T x lim1xf(t)dt1Tfxx TT设f(x)是以T为周期的连续函数,f(x0T0,且

f(x)dx0,中间值定理,f在(x0x0T)恒正设其最小值为m.则m0,x0

f(x)dx

x0

mdxmT0.由周期性和假设

x0

f(x)dx

f(x)dxT T.故f在(x0x0T)至少有一个根x1.若f在(x0x0T)再无其它根,由 f(x)dx f(x)dx f(x)dx0,.故f在(x,x)或(x,xT)至 sin4xcos4其中m为正整数解被积函数以2为周期,

dxm sin4xcos4 0sin4xcos4 dx 0sin4xcos4 0sin2xcos2