不定积分换元法例题.docx

不定积分的第一类换元法】已知 J f (u )du = F (u) + C凑微分】求 J g (x)dx = J f (p (x))p'(x)dx = J f (p (x))dp (x)=J f (u )du = F (u) + C【做变换，令u =申(x)，再积分】=F ((P (x)) + C【变量还原，u =申(x)】【求不定积分J g(x)dx的第一换元法的具体步骤如下:】1)变换被积函数的积分形式: Jg(x)dx=Jf(p(x))p'(x)dx2)凑微分: Jg(x)dx=J f(p(x))p'(x)dx=J f(p(x))dp(x)3)作变量代换u =p ( x)得： Jg(x)dx=J f(p(x))p'(x)dx=J f(p(x))dp(x) =Jf(u)du4)利用基本积分公式J f (u)du = F(u) + C求出原函数:J g (x)dx = J f (p (x))p'(x)dx = J f (p (x))dp (x) = J f (u )du = F (u) + C5)将u =申(x)代入上面的结果，回到原来的积分变量x得:J g (x)dx = J f (p(x))p'(x)dx = J f (p(x))dp(x) = J f (u)du = F(u) + C = F(p (x)) + C【注】熟悉上述步骤后，也可以不引入中间变量U =申(x)，省略(3)(4)步骤，这与复合函数的求导法则类似。

第一换元法例题】J (5 x+7)9 dx=J (5 x+7)9 必=J (5 x+7)9 - 5 d (5 x+7) = 5J (5 x+7)9 -d (5 x+7) =1J (5 x + 7)9 d (5 x + 7) =1 - — (5 x + 7)io + C =丄(5 x + 7)10 + C5 5 10 50【注】(5x + 7)' = 5, n d(5x + 7) = 5dx, n dx = 5d(5x + 7)1、J i^dx = J In x 丄 dx = J In x - d In x xx=J ln x -d ln x=1(ln x)2+C=2(ln x)2+C【注】(Inx)' = , n d(lnx) = dx, n dx = d(lnx) x x x2、sin x sin xdx -d cos x d cos x3(1)tan xdx = dx = = = cos x cos x cos x cos xcos x=-fd cos x =- ln I cos x I+C = -ln I cos x I+C cos x注】(cosx)' = -sin x, n d(cos x) = -sin xdx, n sin xdx = -d(cos x)3（2）cos x cos xdx d sin x cot xdx = dx = =sin xsin xsin xd sin xJ = ln I sin x I+C = ln I sin x I+Csin x注】(sin x)' = cos x, n d (sin x) = cos xdx, n cos xdx = d (sin x)4（1）1 1 1J dx = J - dx = J - d (a + x)a+x a+x a+x1J - d(a + x) = ln I a + x I+C = ln I a + x I+Ca+x注】(a + x)' = 1, n d (a + x) = dx, n dx = d (a + x)4（2）1 1 1J dx = J - dx = J - d (x - a)x - a x - a x - a1J - d (x - a) = ln I x - a I +C = ln I x - a I+Cx - a注】(x - a)' = 1, n d(x - a) = dx, n dx = d(x - a)3）J 1 dx = J 1 dx =丄 Jx2 - a2 x2 - a2 2adx=12aJ-^ dx —I x — adx、 x + a 丿2aI x — a I — In I x + a l)+ C =丄In2a1）secx(secx+tanx) seci x+secxtanxsec xdx = dx = - dxsec x + tan xsec x + tan x2）J d (tan x + sec x) = J d (tan x + sec x)=山〔sec x + 仙 x i +csecx+tanx=J cos x = J cos x - dx = J d sin x— dx — —cos2 x cos2 x 1 - sin2 xsecx+tanxJ sec xdx = J —-—dx cos xcos2 xJ d sin x1 - sin2 xsin x -1 sin x +1 丿1sin x — 11.1 — sin x—ln+ C = — ln2sin x +121 + sin x-d sin x =+Ccscx(cscx+cotx) csci x+cscxcotxcsc x + cot xcsc x + cot xcsc xdx = dx = - dxJ d (- cot x - csc x) = J d (csc x + cot x) = [nIcsc x + cot x I +c cscx+cotx cscx+cotxcscx(cscx-cotx) csc2x-cscxcotx(2) J csc xdx = dx = • dxcscx-cotx cscx-cotxCSC x - cot x=J d (- cot x + csc x) = J d(csc x - cot x)=山 | csc x cot x | +c csc x - cot x1)J亠\'1 - x2dx =dxdx dx ==arcsin x + C2)dxa 2 - x 2 a 2 - x 22J a丿x \=arcsin - + Ca1)J 亠x = A = arctan x + C1 + x 2 1 + x 22)1 dxdx =a2 + x2 a2 + x2=Jdxa22axarctan + C , (a > 0 )a1)Ja丿Ja丿J sin3 x cos5 xdx = J sin2 x cos5 x - sin xdx = -J sin2 x cos5 x - d cos xJ (1-cos2 x) • cos5 x • d cos x = J (cos7 x - cos5 x) • d cos x =沁-沁 + C8 6(2) Jsin3 xcos5 xdx = Jsin3 xcos4 x- cosxdx = Jsin3 xcos4 x- d sin xsin 4 x sin 6 x=J sm3 x(1-sm2 x)2 - d sin x = J (sm3 x - 2sm5 x + sm7 x) - d sin x = -4sin 8 x+ C810( 1)10( 2)J仝=J丄丄dx = J x ln x ln x xJ出」丄x ln2 x ln2 x1 1一 • d ln x = J • d ln x = lnlln x + Cln x ln x—• dx = J — d ln x = J xln2 x11 d ln x = - + C ln2 x ln x111)J 2 xdxx 4 + 2 x 2 + 2 x 4 + 2 x 2 + 2=J2 xdx Jdx 2x 4 + 2 x 2 + 2=Mx 2 + 1L = arctan( x 2 +1) + C1 + (x 2 + 1)2J xdx = 1Jx 4 + 2 x 2 + 5 2 x 4 + 2 x 2 + 5 2 x 4 + 2 x 2 + 5 2 4 + (x 2 +1)22xdx=1Jdx2=1 J d (x 2 +1)12、d=1 f d (x 2 +1) = 1 J = =—<81+4—+2 = 4arctan( 2J 血丄“ dx = J sin、x •-^ •dx = 2J sin Jx • I •dx = 2J sin £ x d Jx x x 2、；'x=2J sin Jx -d\ x = -2cos x + C = -2cos、；'x + C13、111e2 xdx = e 2 xd 2 x = e 2 xd 2 x = e 2 x + C2 2 214、J sin3 xcos xdx = J sin3 x - cos xdx = J sin3 x - d sin x = J sin3 x - d sin x =迎 + C 4J (2 x+5)10如=J (2 x+5)100 dx=J (2 x+5)100 - 2d (2 x+5)=2J (2 x+5)100 -d (2 x+5) =1 • J (2 x + 5)100 d (2 x + 5) =1 •丄(2 x + 5)101 + C =丄(2 x + 5)101 + C2 2 101 20215、16、J x sin x2dx = J sin x2 - xdx = — J sin x2 - dx2 = — J sin x2 - dx2 = - — cos x2 + C2 2 217、ln x ln x 1 ln x (1+ ln x) - 1dx = • dx = • d ln x = ^ • d ln xx、1 + ln x 1 + ln x x 1 + ln x '1 + ln x=J <1 + ln x - d ln x - J ]——-d ln x'1 + ln x= J、1 + ln x - d (1+ ln x) - J 1 - d (1+ ln x)'1 + ln x=3(1+ Inx)3 -2(1+ Inx)2 + C18、earctan x 1dx = J earctan x • dx earctanx •darctan x = e arctan x •darctan x = earctanx +C1 + x 2 1 + x 2J . dx = J . • xdx = J • dx2 = —。