不定積分換元法例的題目1

1、實(shí)用標(biāo)準(zhǔn)文案精彩文檔【第一換元法例題】1(5x7)9d(5x 7)1 1(5x 7)10C (5x 7)1055 10501【注】(5x 7)5, d(5x 7) 5dx, dx -d(5x 7)In x12、dx In x dx In x d In xxx1212In x d Inx - (In x) C (Inx) C2 2,cosxcos xdx dsinxcot xdxdxsin xsin x sin xIn |sin x | C In | sin x| C【注】(sin x) cosx, d (s inx) cos xdx, cos xdx d (s inx)【注】(a x)1,d

2、(ax)dx,dxd (ax)1114(2)dxdxd(xa)xax ax a1d(xa) In |xa | Cln| x a |Cx a【注】(x a)1,d(xa)dx,dxd(xa)d(a(5x 7)9dx(5x7)9dx(5x 7)9 1d(5x 7)1(5x 7)955d(5x 7)【注】（Inx）1xd (In x) dx, xdx d(In x) xtan xdxd cosxcosxsin x sin xdx d cosx dxcosxcosx cosxIn | cosx | C In |cosx | Cd cosxcosx【注】（cosx）sinx, d (cosx)sin x

3、dx, sin xdx d(cos x)d sin xsin xdxdxd (ax)實(shí)用標(biāo)準(zhǔn)文案精彩文檔x) In |ax | C In |a x | C實(shí)用標(biāo)準(zhǔn)文案精彩文檔111111 11(3)- x22dxdx - dx- dxx axa2ax axa2a x ax a1 , ,1 ,x aln | x a |ln |x a |C lnC2a2ax a2, secx(secx tan x) sec x secxtanx ,(1)secxdx -dx - dxsecx tan xsecx tan xcscx(cscx cot x), dxcscx cot xcsc2x cscx cot x

4、 , dxcsc x cot xcscx(cscx cot x), dxcscx cot xcsc2x cscxcot x , dxcscx cot x(2)1dxdxdx122dx222455667788(2)d(tanx secx) d(tanx secx) ,，小In | secx tanx| Csecx tan xsecx tan x1 ,cosx ,dx2dxcosxcos xsecxdxcos x dxd sin xd sin x2cos x1 sin2x1 sin2x2 sin x 1 sin xdsin x1In2sin x 1sin x 111 sin xln -21 sin

5、 xd( cotx cscx)cscx cotxd(cscx cotx)cscx cot xln | cscx cot x| C(1)d( cotx cscx) cscxcot xd(cscx cotx)ln | cscx cot x | Ccscx cot x1dxdx1 x2arcs in x C1 x21(2)=22dxf 22- a xdx.a2x2x arcsi n Ca(1)x1 x2dx1 x2arcta n x C(1)cscxdx(2)cscxdx實(shí)用標(biāo)準(zhǔn)文案精彩文檔a xax2彳x a 1一adxa1dxa1,x小22arctanC, ( a u)丿xa丿xaa1 -1aa

6、實(shí)用標(biāo)準(zhǔn)文案精彩文檔9(1)sin3xcos5xdxsin2xcos5x sinxdxsin2xcos5x d cosx(1 cos2x) cos5x d cos x(cos7xcos5x) d cosxcos8x6cos xC6 35.9(2)sin xcos xdxsin3xcos4x cosxdxsin3xcos xdsinxsin3x(1 sin2x)2d sin x(sinx 2sin5x4sin7x) dsinxsin xsin6x10(1)10(2)dxxln xdxxln2x1ln2x1dxx11(1)2 xdxx42x2212、13、11-d l n x-d ln x lnl

7、n xCln xln x1dxln x x丁d ln xln2x亍d l nx ln2x1ln x2xdxx42x22dx2x42x22d(x21)1 (x21)22arctan(x 1) Cxdx12xdx1dx21d(x21)x42x252x42x252x42x2524 (x21)211(2)d Jd(x21)sin x2 2x21dxx21冷心）C1xdxsinxdx2 sin、x d x2 sin x d /x2cos x C2cos、x Ce2xdx12x12xe d2x e d2x221e2x14、sin3x cos xdxsinx cos xdxsin3x d sin x 43s

8、in xsin x d si n xC4實(shí)用標(biāo)準(zhǔn)文案精彩文檔實(shí)用標(biāo)準(zhǔn)文案精彩文檔15、(2x100 .5) dx(2x100 .5) dx(2x5)1001 d(2x25)12(2x 5)100d(2x 5)1(2x1005) d(2x5)1 1(2x5)101C丄(2x5)101C22 10120221、20、sin x2 exdx1sin xdxcos x3cos2xd cosx12cos2x Cexdxdexex) In(2 ex) C216、xsin x dx212 2sin x xdx sin x dx21.2. 2sin x dx2-cosx2C217、In xdxx1 In xI

9、n x 1dx.1 In x xIn x,d I n x.1 In x(1JnxL1dInx、1 In x、1 In x d In xd I n x.1 Inx18、19、1 Inx d(1 Inx)d(1 In x)2(13In x)2arcta nxevdxdx12(1 In x)2Carcta nxe-2dx1 x2xdxarcta nxe d arctan xarcta nxarcta nxe d arctan xe C_1_2 Jx2dx2_1_2、1 x2d(1x2)x1實(shí)用標(biāo)準(zhǔn)文案精彩文檔實(shí)用標(biāo)準(zhǔn)文案精彩文檔22、ln 2xdxln2xdx2ln x d ln x2ln3xln

10、x d ln xCxx3【不定積分的第二類換元法】sin x cosxdx12 2 2 22d(a sin x b cos x) b2)2(a2【解答】-sin xcosxdxsin xcosxdx1d(a22sin xb2cos x)2 2 2 2a sin x b cosx2 2 2 2a sin x b cos xa2b22、a2 2sin xb2cos2x所以：d (a2sin2x b2cos2x)23、dx.1 2x x2dx2 (1 x)2d(1 x)2 (1 x)2d(1 x).(2)2(1 x)2arcsin1 xCV224、dxx2x 2(xdx2)2d(x 2)二12v 7

11、2(x2)2()2arcta nd(x；)(xyx 12飛飛2d(x(x 1)2(27)2arctan2xJC25、計(jì)算sin xcosxa2sin2x b22cos xdx,a2【分析】因?yàn)?2 2 2 2(a sin x b cos x)2 2 2a 2sin xcosx b 2cos x( sin x) 2(ab2)sin xcosx2 22(a b )sin xcosxdx實(shí)用標(biāo)準(zhǔn)文案精彩文檔已知f (t)dtF(t) C求g(x)dxg( (t)d (t)g( (t) (t)dt【做變換，令x(t)，再求微分】f (t)dt F(t) C【求積分】1F( (x) C1【變量還原，t(

12、x)】變量還原_2cost C2cos x Ct x令 x t2(1)1護(hù) xt2dt21 t2tdt1 t2 dt 21 t變量還原2 t ln|1 t| Ct x2 xln|1、x|C1 dt1 t2(2)令 1+ x t2x (t 1)12td(t 1)21 2(t 1)dt2 dtdtx t2迥dt2Sint2tdt2si ntdt【第二換元法例題】令 x t實(shí)用標(biāo)準(zhǔn)文案精彩文檔變量還原2 t ln|t|C2 1 ,x ln|1 ,x| Ct i伐dx令314x tx (t31)41t d(t31)4ut31)4I332E4(t 1)3tdt12 (t6t3)dt1dx-x(1 x)2

13、arcta ntxdx1 ex t2In |t| In |1dx(1：x)sx12t7變量還原12J(1冏J(1 &t314x74變量還原令 extx In tt| Cx t66(tarcta nt)12t(1 t)dtt(1t2)2tdt2arctan、x丄dlnt1 tIn1(1 t2)t3t(1 t)dtdt變量還原t exdt6Inxex1 eL 6t5dt(1 t2)t3t21 t2dt11 t2dt變量還原Ct6xarctan , x) C【注】被積函數(shù)中出現(xiàn)了兩個(gè)根式、x,X時(shí)，可令，其中k為m, n的最小公倍數(shù)。dx1：x 2令3廠2 tt2xt323r?t3匸t ln

14、|12t| C實(shí)用標(biāo)準(zhǔn)文案精彩文檔【注】變量還原3t6xJ(X 2)27(2)變量還原被積函數(shù)中含有簡單根式dx令1tx7T2 ln|1tAdt1X o令:ax1|2t 2ln|t 1| ln|t21xln| 1| Cxnax b或cx時(shí)，可令這個(gè)簡單根式為1|t，即可消去根式。8x(1x2)t7t5t3+t7531 In x8(1)arcta nt1xt(x In x)2dx變量還原tint211 11 -in- x x1t8(1d1t111變量還原CInlnt)dt：dtt241 t8上dt1 t2t6t4t2n dt t2tIn -tx In x17x715x513x31arcta n

15、CxdtIn1tIn1tt12dt叫dttint1 tint2d(1tint)1 tint【注】當(dāng)被積函數(shù)中分母的次數(shù)較高時(shí)，可以試一試倒變換。實(shí)用標(biāo)準(zhǔn)文案精彩文檔sinx(11 sinx , dxcosx)x 2arctanti JLi t2二(ii t2d2arctaJ22t2dt1 t2-t 21dt - t -ln|t| C2t 422x變量還原tan % ix2tan -ln|tan-| Ct142 22x【注】對(duì)三角函數(shù)有理式的被積函數(shù)，可以用萬能公式變換，化為有理分式函數(shù)的積分問題。令 x asint,|t|10(1). a2x2dxt arcsinF a a2a2s in2t2

16、 2dasint a cos tdt_ dx- a2asint, |t| 2das int.x arcsin_aa2sin2tdt變量還原Ct.x arcsin_aarcs屛C a變量還原1 cos2t ,dt22a2(1cos2t)dtsin2txt arcs in aaxarcsin2x2Cdxx令 x atart,|t| 2datantx arctanaa2a2tan tsecdtIn | sec tant|變量還原xt arctanaln|-a2 2a x .| C In |xa.a2x21因?yàn)?(xa2x2)2 a2x22aa2所以:(x a2x2)dx2、a2x2dx-2dxa x

17、即:a2x2dx2(x a2 x2)dx a2dx2x實(shí)用標(biāo)準(zhǔn)文案精彩文檔10(3)Tx2dx令 x aseC , t _2變量還原x asect因?yàn)?所以:即:a2ln|-adasect.a2seC tC In | x(x、. x2a2)2 x2a2(x . x2a2)dx 2 . x2、Cdx1a2dxx2sectdtIn | sect tant |_ dx2 2x a(x x2a2fxJx27a2x2| C【注】當(dāng)被積函數(shù)中出現(xiàn).a2x2a2x2,.x2a2因子時(shí)，可以用三角變換，化為三角函數(shù)的積分問題。實(shí)用標(biāo)準(zhǔn)文案精彩文檔【附加】【應(yīng)用題】已知生產(chǎn)x單位的某種產(chǎn)品，邊際單位成本是C(x)102，又知邊際收益為R(x)120.1x，且R(0)0，求：(1)利潤函數(shù)L(x)；(2)利潤最大時(shí)的產(chǎn)量；(3)利潤最大時(shí)的平均價(jià)格。(3)利潤最大時(shí)的平均價(jià)格為：P迥竺7【解答】_C(x)100(1)因?yàn)?/p>