求下列各定積分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| 設(shè)f(x)在[ab]連續(xù).證明af(x)dxba)0f(ab 證令xaba)t,則0a,1bdxba af(x)dx(ba)0f(a(ba)t)dt(ba)0f(a(ba 1證明0xf(x)dx20xf證令x2t,則x0時t0xa時ta2a 1a 1 10xf(x)dx20xf(x)dx20tf(t)dt20xf 1xm(1x)ndx1xn(1 證令x1t,則x0時x1時t0.dxdt,1xm(1x)ndx0(1t)mtndt1(1t)mtndt1xn(1 利用分部積分公式證明,若f(x)連續(xù)x 00f(x)dxdt0f(t)(x證xtf(x)dxdtttf(x)dxxxttf(x)dxa 0xf(x)dx0tf(t)dt0xf(t)dt0 x0f(t)(x 利用換元積分法證明 f(sin 證xtx0時tx時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,則t2時u2,t時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 設(shè)函數(shù)f(x)在(,)上連續(xù),以T為周期,證明函數(shù)F(xxTf(x)dxxf(t)dt也以TT lim1xf(t)dt1Tfxx 證(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設(shè)f(x)是以T為周期的連續(xù)函數(shù),f(x0T0,且

f(x)dx0,中間值定理,f在(x0x0T)恒正設(shè)其最小值為m.則m0,x0

f(x)dx

x0

mdxmT0.由周期性和假設(shè)

x0

f(x)dx

f(x)dxT T.故f在(x0x0T)至少有一個根x1.若f在(x0x0T)再無其它根,由 f(x)dx f(x)dx f(x)dx0,.故f在(x,x)或(x,xT)至 sin4xcos4其中m為正整數(shù)解被積函數(shù)以2為周期,

dxm sin4xcos4 0sin4xcos4 dx 0sin4xcos4 0sin2xcos2