天津工业大学数学系：《高等数学》课程电子教案（PPT课件）第四章 不定积分（4.3）分部积分法

⑩串紫学 Teaching Plan on Advanced Mathematics 第三节分部积分法 设函数u=u(x)和v=v(x)具有连续导数, uv=u'v+uv uv=(uv)-u'v uv'cx=uv-Ju'vdx 「uby=uv

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 设函数 和 具有连续导数， uv dx uv u vdx,    = −  udv uv vdu.   = − u = u(x) v = v( x) (uv)'= u'v + uv' uv'= (uv)'−u'v 第三节 分部积分法

⑩串紫学 Teaching Plan on Advanced Mathematics 注 解 ①显然,L,v选择不当,积分更难进行 ②若被积函数是幂函数和正(余)弦函数或幂函数和指数函 数的乘积,就考虑设幂函数为l,,使其降幂一次(假定幂 指数是正整数) ③若被积函数是幂函数和对数函数或幂函数和反三角函 数的乘积,就考虑设对数函数或反三角函数为

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 显然， u,v 选择不当，积分更难进行. 若被积函数是幂函数和正(余)弦函数或幂函数和指数函 数的乘积,就考虑设幂函数为 , 使其降幂一次(假定幂 指数是正整数) u, ① ② 若被积函数是幂函数和对数函数或幂函数和反三角函 数的乘积，就考虑设对数函数或反三角函数为 . ③ u 注 解

⑩串紫学 Teaching Plan on Advanced Mathematics 例1求积分xdk 解(一)令u=cosx,xt=a2=h xcos xdx =cosx+ SInd dx (二)令u=x, sxd= d sin=b xcos xdx=lxdsinx=xsinx-sinxdx =sinx+cosx+C 例2求积分「x2ex 解=x,eI=le=加h, xe dx=xte xe dx (再次使用分部积分法)=X,eb= =x2e2-2(xe-e2)+C

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics cos .  x xdx 解 令 u = cos x, xdx = dx = dv 2 2 1  xcos xdx  = + xdx x x x sin 2 cos 2 2 2 （二）令 u = x, cos xdx = d sin x = dv  xcos xdx  = xd sin x  = xsin x − sin xdx = xsin x +cos x +C. 例1 求积分 . 2  x e dx x 解 , 2 u = x e dx de dv, x x = =  x e dx 2 x  = x e − xe dx x x 2 2 2( ) . 2 x e xe e C x x x = − − + （再次使用分部积分法） u = x, e dx dv x = 例2 求积分 （一）

⑩串紫学 Teaching Plan on Advanced Mathematics 例3求积分 xarctanxd 解令u= arctan,xr=dx 2 2 xarctanxdx =— arctan d(arctan) 2 2 2 arctan 2 21+x arctan-9·(1 2 21+x =arctanx-(x-arctan x)+C. 2

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例3 求积分 arctan .  x xdx 解 令 u = arctan x , dv x xdx = d = 2 2  xarctan xdx (arctan ) 2 arctan 2 2 2 d x x x x  = − dx x x x x 2 2 2 1 1 2 arctan 2 + = −   dx x x x ) 1 1 (1 2 1 arctan 2 2 2 + = −  −  ( arctan ) . 2 1 arctan 2 2 x x x C x = − − +

⑩串紫学 Teaching Plan on Advanced Mathematics 例4求积分x3lnxx 解 u=nx xdx= d ∫x3 In xdx =-xtlnx--x+ AC =-x Inx 16

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例4 求积分 ln . 3  x xdx 解 u = ln x, , 4 4 3 dv x x dx = d =  x ln xdx 3  = x x − x dx 4 3 4 1 ln 4 1 . 16 1 ln 4 1 4 4 = x x − x + C

⑩串紫学 Teaching Plan on Advanced Mathematics 例5求积分sn(nx)hx 解∫smnx)x=xsi(mx)- jxd(sin(In x) =xsin(x)-「 x cos(Inx)t xsin( x)-xcos( x)+ xd cos(In x)I x(sin( x)-cos(In x)1-sin(x)dx jsin(n x) dx= Isin(In x)-cos(Inx)+C

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例5 求积分 sin(ln ) .  x dx 解  sin(ln x)dx= −  xsin(ln x) xd[sin(ln x)]  = −  dx x x x x x 1 sin(ln ) cos(ln )  = xsin(ln x) − xcos(ln x) + xd[cos(ln x)] = − −  x[sin(ln x) cos(ln x)] sin(ln x)dx  sin(ln x)dx [sin(ln ) cos(ln )] . 2 x x C x = − +

⑩串紫学 Teaching Plan on Advanced Mathematics 例6求积分 e sin xdx 解 e sin xdx=sin xde e" d(sin x) =e"- e cos xdx=e sin x-cos xde -e sinx-e cosx-e d cosX e (sin x-cos x)=e sinxdx 注意循环形式 ∫ e sin xx=(ix-cosx)+C

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例6 求积分 sin .  e xdx x 解  e xdx x sin  = x sin xde = −  e sin x e d(sin x) x x  = e x − e xdx x x sin cos = −  x x e sin x cos xde  = e sin x − (e cos x − e d cos x) x x x = − −  e x x e xdx x x (sin cos ) sin  e xdx x sin (sin cos ) . 2 x x C e x = − + 注意循环形式

⑩串紫学 Teaching Plan on Advanced Mathematics arctan 例7求积 √1+x 解 1+x2)= 2 1+x x arctan dx= arctanxd√1+x2 √1+x W1+xarctanx-v1+xd(arctan) =√1+x2 arctan-|√1+x 2 1+x

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例7 求积分  + . 1 arctan 2 dx x x x 解 ( ) , 1 1 2 2 x x x + =   +  +  dx x x x 2 1 arctan  = + 2 arctan xd 1 x 1 arctan 1 (arctan ) 2 2 x x x d x  = + − + dx x x x x 2 2 2 1 1 1 arctan 1 + = + − +  

⑩串紫学 Teaching Plan on Advanced Mathematics =√1+x2 arctan 2x令x=tant √1+x sec tdt= sectdt 1+x 2 √1+tan2t =In(sect+tant)+C=In(x+ 1+x)+C arctan 2 1+x =√+x2 arctan-ln(x+√1+x2)+C

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics dx x x x  + = + − 2 2 1 1 1 arctan 令 x = tant dx x  + 2 1 1  + = tdt t 2 2 sec 1 tan 1  = sectdt = ln(sec t + tant) +C= ln( x + 1+ x ) + C 2  +  dx x x x 2 1 arctan 1 x arctan x 2 = + ln( 1 ) . 2 − x + + x + C

⑩串紫学 Teaching Plan on Advanced Mathematics 例8已知f(x)的一个原函数是eX,求「xf(x)x 解∫y(x)=xd(x)=(x)-J/(x f(x)x)=∫(x) ∫(x)x=e+C, 两边同时对x求导,得f(x)=-2xe, lxf(x)dx=xf(x)-If(x)dx 2. r e e +c

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例 8 已知 f (x)的一个原函数是 2 x e − , 求 xf (x)dx. 解  xf (x)dx  = xdf (x) ( ) ( ) ,  = xf x − f x dx ( ) , 2   = + − f x dx e C x ( f (x)dx) = f (x),    两边同时对 x 求导, 得 ( ) 2 , 2 x f x xe− = −   =  xf (x)dx  xf (x) − f (x)dx 2 2 2 x x e − = − . 2 e C x − + −