wordpress文章点赞,丈哥seo博客,贵阳市网站开发,在wordpress上添加播放视频文章目录1 第一类换元法1.1 定理11.2 例题1.2 常见凑微分形式1.2.1常见基本的导数公式的逆运算1.2.2被积函数含有三角函数2 第二类换元法2.1 定理22.2 常见第二换元代换方法2.2.1 三角代换-弦代换2.2.2 三角代换-切代换2.2.3 三角代换-割代换2.2.4 三角代换汇总2.2.5 倒代换2.2…
文章目录1 第一类换元法1.1 定理11.2 例题1.2 常见凑微分形式1.2.1常见基本的导数公式的逆运算1.2.2被积函数含有三角函数2 第二类换元法2.1 定理22.2 常见第二换元代换方法2.2.1 三角代换-弦代换2.2.2 三角代换-切代换2.2.3 三角代换-割代换2.2.4 三角代换汇总2.2.5 倒代换2.2.6 根式代换3 积分推导公式后记把复合函数的微分法反过来用于求不定积分利用中间变量的代换得到复合函数的积分法称为换元积分法。换元法分为两类。1 第一类换元法
1.1 定理1
设f(u)f(u)f(u)具有原函数F(u)F(u)F(u),即
F′(u)f(u),∫f(u)duF(u)CF^{}(u)f(u),\int{f(u)du}F(u)CF′(u)f(u),∫f(u)duF(u)C
如果u是中间变量:uϕ(x)且设ϕ(x)u\phi(x)且设\phi(x)uϕ(x)且设ϕ(x)可微那么根据复合函数微分法有
dF[ϕ(x)]f[ϕ(x)]ϕ′(x)dxdF[\phi(x)]f[\phi(x)]\phi^{}(x)dxdF[ϕ(x)]f[ϕ(x)]ϕ′(x)dx,
根据不定积分的定义就得
∫f[ϕ(x)]ϕ′(x)dxF[ϕ(x)]C[∫f(u)du]uϕ(x)\int{f[\phi(x)]\phi^{}(x)dx}F[\phi(x)]C[\int{f(u)du}]_{u\phi(x)}∫f[ϕ(x)]ϕ′(x)dxF[ϕ(x)]C[∫f(u)du]uϕ(x) 定理1 设f(u)f(u)f(u)具有原函数uϕ(x)u\phi(x)uϕ(x)可导则有换元公式 ∫f[ϕ(x)]ϕ′(x)dx[∫f(u)du]uϕ(x)\int{f[\phi(x)]\phi^{}(x)dx}[\int{f(u)du}]_{u\phi(x)}∫f[ϕ(x)]ϕ′(x)dx[∫f(u)du]uϕ(x) 1.2 例题
例1 求∫132xdx\int{\frac{1}{32x}dx}∫32x1dx 解∫132xdx12∫132xd(32x)12ln∣32x∣C解\\ \int{\frac{1}{32x}dx}\frac{1}{2}\int{\frac{1}{32x}d(32x)}\frac{1}{2}\ln|32x|C 解∫32x1dx21∫32x1d(32x)21ln∣32x∣C 例2 求∫x2(x2)3dx\int{\frac{x^2}{(x2)^3}dx}∫(x2)3x2dx
注分式积分分母越简单越易积 解令ux2,xu−2∫x2(x2)3dx∫(u−2)2u3du∫1udu−∫4u2du∫4u3duln∣x2∣4x2−12(x2)2C解\\ 令ux2,xu-2 \\ \int{\frac{x^2}{(x2)^3}dx}\int{\frac{(u-2)^2}{u^3}du}\\ \int{\frac{1}{u}du}-\int{\frac{4}{u^2}du}\int{\frac{4}{u^3}du}\\ \ln|x2|\frac{4}{x2}-\frac{1}{2(x2)^2}C 解令ux2,xu−2∫(x2)3x2dx∫u3(u−2)2du∫u1du−∫u24du∫u34duln∣x2∣x24−2(x2)21C
例3 求∫1a2x2dx\int{\frac{1}{a^2x^2}dx}∫a2x21dx
解析基本积分公式∫11x2dxarctanxC\int{\frac{1}{1x^2}dx}\arctan xC∫1x21dxarctanxC 解:∫1a2x2dx1a2∫11(xa)2dx1a∫11(xa)2d(xa)1aarctanxaC解:\\ \int{\frac{1}{a^2x^2}dx}\frac{1}{a^2}\int{\frac{1}{1(\frac{x}{a})^2}dx}\\ \frac{1}{a}\int{\frac{1}{1(\frac{x}{a})^2}d(\frac{x}{a})}\frac{1}{a}\arctan\frac{x}{a}C 解:∫a2x21dxa21∫1(ax)21dxa1∫1(ax)21d(ax)a1arctanaxC
例4 求∫1x2−a2dx\int{\frac{1}{x^2-a^2}dx}∫x2−a21dx
注裂项公式分母为两个一次项乘积形式1(xa)(xb)1b−a(1xa−1xb)\frac{1}{(xa)(xb)}\frac{1}{b-a}(\frac{1}{xa}-\frac{1}{xb})(xa)(xb)1b−a1(xa1−xb1) 解:∫1x2−a2dx12a[∫1x−ad(x−a)−∫1xad(xa)]12a(ln∣x−a∣−ln∣xa∣)C12aln∣x−axa∣C解:\\ \int{\frac{1}{x^2-a^2}dx}\frac{1}{2a}[\int{\frac{1}{x-a}d(x-a)}-\int{\frac{1}{xa}d(xa)}]\\ \frac{1}{2a}(\ln|x-a|-\ln|xa|)C\frac{1}{2a}\ln|\frac{x-a}{xa}|C 解:∫x2−a21dx2a1[∫x−a1d(x−a)−∫xa1d(xa)]2a1(ln∣x−a∣−ln∣xa∣)C2a1ln∣xax−a∣C
例5 求∫dxx(12lnx)\int{\frac{dx}{x(12\ln x)}}∫x(12lnx)dx
解析(lnx)′1x(\ln x)^{}\frac{1}{x}(lnx)′x1 解:∫dxx(12lnx)12∫d(12lnx)12lnx12ln∣12lnx∣C解:\\ \int{\frac{dx}{x(12\ln x)}}\frac{1}{2}\int{\frac{d(12\ln x)}{12\ln x}}\\ \frac{1}{2}\ln|12\ln x|C 解:∫x(12lnx)dx21∫12lnxd(12lnx)21ln∣12lnx∣C
例6 求∫e3xxdx\int{\frac{e^{3\sqrt{x}}}{\sqrt{x}}dx}∫xe3xdx
解析(x)′12⋅1x(\sqrt{x})^{}\frac{1}{2}\cdot\frac{1}{\sqrt{x}}(x)′21⋅x1 解:∫e3xxdx23∫e3xd(3x)23e3xC解:\\ \int{\frac{e^{3\sqrt{x}}}{\sqrt{x}}dx}\frac{2}{3}\int{e^{3\sqrt{x}}d(3\sqrt{x})}\\ \frac{2}{3}e^{3\sqrt{x}}C 解:∫xe3xdx32∫e3xd(3x)32e3xC
例7 求∫sin2xcos5xdx\int{sin^2x\cos^5xdx}∫sin2xcos5xdx
解析(sinx)′cosx(cosx)′−sinxsin2xcos2x1(\sin x)^{}\cos x\quad (\cos x)^{}-\sin x\quad \sin^2x\cos^2x1(sinx)′cosx(cosx)′−sinxsin2xcos2x1 解∫sin2xcos5xdx∫sin2xcos4xd(sinx)∫sin2x(1−sin2x)2d(sinx)∫(sin6x−2sin4xsin2x)d(sinx)17sin7x−25sin5x13sin3xC解\int{sin^2x\cos^5xdx}\int{sin^2x\cos^4xd(\sin x)}\\ \int{sin^2x(1-\sin^2x)^2d(\sin x)}\int{(\sin^6x-2\sin^4x\sin^2x)d(\sin x)}\\ \frac{1}{7}\sin^7x-\frac{2}{5}\sin^5x\frac{1}{3}\sin^3xC 解∫sin2xcos5xdx∫sin2xcos4xd(sinx)∫sin2x(1−sin2x)2d(sinx)∫(sin6x−2sin4xsin2x)d(sinx)71sin7x−52sin5x31sin3xC 一般地对于sin2k1xcosnx或sinnxcos2k1x(其中k∈N)\sin^{2k1}x\cos^nx或\sin^nx\cos^{2k1}x(其中k\in N)sin2k1xcosnx或sinnxcos2k1x(其中k∈N)型函数的积分总可依次做变换ucosx或usinxu\cos x或u\sin xucosx或usinx,求得结果。 例8 求∫tanxdx\int{\tan xdx}∫tanxdx 解∫tanxdx∫sinxcosxdx−∫1cosxd(cosx)−ln∣cosx∣C解\int{\tan xdx}\int{\frac{\sin x}{\cos x}dx}-\int{\frac{1}{\cos x}d(\cos x)}\\ -\ln|\cos x|C 解∫tanxdx∫cosxsinxdx−∫cosx1d(cosx)−ln∣cosx∣C 类似可得∫cotxdxln∣sinx∣C\int{\cot xdx}\ln|\sin x|C∫cotxdxln∣sinx∣C 例9 求∫x3(x2−2x2)\int{\frac{x^3}{(x^2-2x2)}}∫(x2−2x2)x3
1.2 常见凑微分形式
1.2.1常见基本的导数公式的逆运算
∫f(1x)⋅1x2dx−∫f(1x)d1x\int{f(\frac{1}{x})\cdot\frac{1}{x^2}dx}-\int{f(\frac{1}{x})d\frac{1}{x}}∫f(x1)⋅x21dx−∫f(x1)dx1∫f(x)1xdx2∫f(x)dx\int{f(\sqrt{x})\frac{1}{\sqrt{x}}dx}2\int{f(\sqrt{x})d\sqrt{x}}∫f(x)x1dx2∫f(x)dx∫f(sinx)cosxdx∫f(sinx)dsinx\int{f(\sin x)\cos xdx}\int{f(\sin x)d\sin x}∫f(sinx)cosxdx∫f(sinx)dsinx∫f(cosx)sinxdx−∫f(cosx)dcosx\int{f(\cos x)\sin xdx}-\int{f(\cos x)d\cos x}∫f(cosx)sinxdx−∫f(cosx)dcosx∫f(lnx)1xdx∫f(lnx)dlnx\int{f(\ln x)\frac{1}{x}dx}\int{f(\ln x)d\ln x}∫f(lnx)x1dx∫f(lnx)dlnx∫f(xlnx)(1lnx)dx∫f(xlnx)d(xlnx)\int{f(x\ln x)(1\ln x)dx}\int{f(x\ln x)d(x\ln x)}∫f(xlnx)(1lnx)dx∫f(xlnx)d(xlnx)∫f(secx)secxtanxdx∫f(secx)dsecx\int{f(\sec x)\sec x\tan xdx}\int{f(\sec x)d\sec x}∫f(secx)secxtanxdx∫f(secx)dsecx∫f(cscx)cscxcotxdx−∫f(cscx)dcscx\int{f(\csc x)\csc x\cot xdx}-\int{f(\csc x)d\csc x}∫f(cscx)cscxcotxdx−∫f(cscx)dcscx∫f(tanx)sec2xdx∫f(tanx)dtanx\int{f(\tan x)\sec^2xdx}\int{f(\tan x)d\tan x}∫f(tanx)sec2xdx∫f(tanx)dtanx∫f(cotx)csc2xdx−∫f(cotx)dcotx\int{f(\cot x)\csc^2xdx}-\int{f(\cot x)d\cot x}∫f(cotx)csc2xdx−∫f(cotx)dcotx∫f(arcsinx)11−x2dx∫f(arcsinx)darcsinx\int{f(\arcsin x)\frac{1}{\sqrt{1-x^2}}dx}\int{f(\arcsin x)d\arcsin x}∫f(arcsinx)1−x21dx∫f(arcsinx)darcsinx∫f(arctanx)11x2dx∫f(arcsinx)darctanx\int{f(\arctan x)\frac{1}{1x^2}dx}\int{f(\arcsin x)d\arctan x}∫f(arctanx)1x21dx∫f(arcsinx)darctanx∫f(ex)exdx∫f(ex)dex\int{f(e^x)e^xdx}\int{f(e^x)de^x}∫f(ex)exdx∫f(ex)dex
1.2.2被积函数含有三角函数
常用的三角恒等式
sin2xcos2x1\sin^2x\cos^2x 1sin2xcos2x11tan2xsec2x1\tan^2x\sec^2x1tan2xsec2x1cot2xcsc2x1\cot^2x\csc^2 x1cot2xcsc2x两角和差的三角公式 cos(α±β)cosαcosβ±sinαsinβ\cos(\alpha\pm\beta)\cos\alpha\cos\beta\pm\sin\alpha\sin\betacos(α±β)cosαcosβ±sinαsinβsin(α±β)sinαcosβ±cosαsinβ\sin(\alpha\pm\beta)\sin\alpha\cos\beta\pm\cos\alpha\sin\betasin(α±β)sinαcosβ±cosαsinβ 2倍角公式降幂公式 cos2αcos2α−sin2α2cos2α−11−2sin2α\cos2\alpha\cos^2\alpha-\sin^2\alpha2\cos^2\alpha-11-2\sin^2\alphacos2αcos2α−sin2α2cos2α−11−2sin2αsin2α2sinαcosα\sin2\alpha2\sin\alpha\cos\alphasin2α2sinαcosαcos2α12(1cos2α)\cos^2\alpha\frac{1}{2}(1\cos2\alpha)cos2α21(1cos2α)sin2α12(1−cos2α)\sin^2\alpha\frac{1}{2}(1-\cos2\alpha)sin2α21(1−cos2α) 积化和差和差化积 一般地对于sin2k1xcosnx或sinnxcos2k1x(其中k∈N)\sin^{2k1}x\cos^nx或\sin^nx\cos^{2k1}x(其中k\in N)sin2k1xcosnx或sinnxcos2k1x(其中k∈N)型函数的积分总可依次做变换ucosx或usinxu\cos x或u\sin xucosx或usinx,求得结果。 tannxsec2kx或者tan2k−1xsecnx(n,k∈N)\tan^nx\sec^{2k}x或者\tan^{2k-1}x\sec^nx(n,k\in N_)tannxsec2kx或者tan2k−1xsecnx(n,k∈N)型积分可依次做变换utanx或usecxu\tan x或u\sec xutanx或usecx 2 第二类换元法
2.1 定理2 设xϕ(t)x\phi(t)xϕ(t)是单调可导的函数并且ϕ′(t)̸0.又设f[ϕ(t)]ϕ′(t)\phi^{}(t)\not0.又设f[\phi(t)]\phi^{}(t)ϕ′(t)0.又设f[ϕ(t)]ϕ′(t)具有原函数则有换元公式 ∫f(x)dx[∫f[ϕ(t)]ϕ′(t)dt]tϕ−1(x)\int{f(x)dx}[\int{f[\phi(t)]\phi^{}(t)dt}]_{t\phi^{-1}(x)}∫f(x)dx[∫f[ϕ(t)]ϕ′(t)dt]tϕ−1(x) 其中ϕ−1(x)是xϕ(t)\phi^{-1}(x)是x\phi(t)ϕ−1(x)是xϕ(t)的反函数。 证明设f[ϕ(t)]ϕ′(t)的原函数为Φ(t),记Φ(t)Φ[ϕ−1(x)]F(x)利用复合函数和反函数的求导法则有F′(x)dΦdt⋅dtdxf[ϕ(t)]ϕ′(t)⋅1ϕ′(t)f(x)即F(x)是f(x)的原函数所以有∫f(x)dxF(x)cΦ[ϕ−1(x)]C[∫f[ϕ(t)]ϕ′(t)dt]tϕ−1(x)证明\\ 设f[\phi(t)]\phi^{}(t)的原函数为\Phi(t),记\Phi(t)\Phi[\phi^{-1}(x)]F(x)\\ 利用复合函数和反函数的求导法则有\\ F^{}(x)\frac{d\Phi}{dt}\cdot\frac{dt}{dx}f[\phi(t)]\phi^{}(t)\cdot\frac{1}{\phi^{}(t)}f(x) \\ 即F(x)是f(x)的原函数所以有 \\ \int{f(x)dx}F(x)c\Phi[\phi^{-1}(x)]C[\int{f[\phi(t)]\phi^{}(t)dt}]_{t\phi^{-1}(x)} 证明设f[ϕ(t)]ϕ′(t)的原函数为Φ(t),记Φ(t)Φ[ϕ−1(x)]F(x)利用复合函数和反函数的求导法则有F′(x)dtdΦ⋅dxdtf[ϕ(t)]ϕ′(t)⋅ϕ′(t)1f(x)即F(x)是f(x)的原函数所以有∫f(x)dxF(x)cΦ[ϕ−1(x)]C[∫f[ϕ(t)]ϕ′(t)dt]tϕ−1(x)
注
必须写出变量代换xϕ(t)x\phi(t)xϕ(t)必须替换微分dxϕ′(t)dtdx\phi^{}(t)dtdxϕ′(t)dt最后结果必须还原成原来积分变量tϕ−1(x)t\phi^{-1}(x)tϕ−1(x)
2.2 常见第二换元代换方法
三角代换倒代换根式代换
2.2.1 三角代换-弦代换
正弦或者余弦代换
例1 求∫a2−x2dx(a0)\int{\sqrt{a^2-x^2}dx}(a\gt0)∫a2−x2dx(a0) 解令xasint,−π2tπ2a2−x2acost,dxacostdt∫a2−x2dx∫acost⋅acostdta22ta24sin2tCa22ta22sintcostC其中tarcsinxa,sintxa,costa2−x2a∫a2−x2dxa22arcsinxaxa2−x22C解令xa\sin t,-\frac{\pi}{2}\lt t\lt\frac{\pi}{2}\\ \sqrt{a^2-x^2}a\cos t,dxa\cos tdt \\ \int{\sqrt{a^2-x^2}dx}\int{a\cos t\cdot a\cos tdt}\frac{a^2}{2}t\frac{a^2}{4}\sin2tC\\ \frac{a^2}{2}t\frac{a^2}{2}\sin t\cos tC\\ 其中t\arcsin\frac{x}{a},\sin t\frac{x}{a},\cos t\frac{\sqrt{a^2-x^2}}{a}\\ \int{\sqrt{a^2-x^2}dx}\frac{a^2}{2}\arcsin\frac{x}{a}\frac{x\sqrt{a^2-x^2}}{2}C 解令xasint,−2πt2πa2−x2acost,dxacostdt∫a2−x2dx∫acost⋅acostdt2a2t4a2sin2tC2a2t2a2sintcostC其中tarcsinax,sintax,costaa2−x2∫a2−x2dx2a2arcsinax2xa2−x2C
2.2.2 三角代换-切代换
正切或者余切
例2 求∫1x2a2dx(a0)\int{\frac{1}{\sqrt{x^2a^2}}dx}(a0)∫x2a21dx(a0) 解令xatant,−π2tπ2x2a2asect,dxasec2t∫1x2a2dx∫1asect⋅asec2tdt∫sectdtln∣secttant∣C其中tantxa,sectx2a2a∫1x2a2dxln∣x2a2axa∣C1ln(x2a2x)C解令xa\tan t,-\frac{\pi}{2}\lt t\lt\frac{\pi}{2} \\ \sqrt{x^2a^2}a\sec t,dxasec^2t \\ \int{\frac{1}{\sqrt{x^2a^2}}dx}\int{\frac{1}{a\sec t}\cdot a\sec^2tdt}\\ \int{\sec tdt}\ln|\sec t\tan t|C \\ 其中\tan t\frac{x}{a},\sec t\frac{\sqrt{x^2a^2}}{a} \\ \int{\frac{1}{\sqrt{x^2a^2}}dx}\ln|\frac{\sqrt{x^2a^2}}{a}\frac{x}{a}|C_1\\ ln(\sqrt{x^2a^2}x)C 解令xatant,−2πt2πx2a2asect,dxasec2t∫x2a21dx∫asect1⋅asec2tdt∫sectdtln∣secttant∣C其中tantax,sectax2a2∫x2a21dxln∣ax2a2ax∣C1ln(x2a2x)C
2.2.3 三角代换-割代换
正割或者余割
例3 求∫1x2−a2dx(a0)\int{\frac{1}{\sqrt{x^2-a^2}}dx}(a\gt0)∫x2−a21dx(a0) 解定义域xa或者x−a(1)当xa时令xasect,0tπ2x2−a2atant,dxasecttantdt∫1x2−a2dx∫1atantasecttantdt∫sectdtln∣secttant∣C1其中sectxa,tantx2−a2a∫1x2−a2dxln(xx2−a2)C(2)当x−a时,令u−x,则ua∫1x2−a2dx−∫1u2−a2du−ln(uu2−a2)C−ln(−xx2−a2)Cln(−x−x2−a2)C1综上当xa时∫1x2−a2dxln(xx−a2)Cx−a时∫1x2−a2dxln(−x−x2−a2)C所以∫1x2−a2dxln∣xx2−a2∣C解定义域x\gt a或者x\lt -a \\ (1)当x\gt a时令xa\sec t,0\lt t\lt\frac{\pi}{2} \\ \sqrt{x^2-a^2}a\tan t,dxa\sec t\tan tdt \\ \int{\frac{1}{\sqrt{x^2-a^2}}dx}\int{\frac{1}{a\tan t}a\sec t\tan tdt}\\ \int{\sec tdt}\ln|\sec t\tan t|C_1 \\ 其中\sec t\frac{x}{a},\tan t\frac{\sqrt{x^2-a^2}}{a} \\ \int{\frac{1}{\sqrt{x^2-a^2}}dx}\ln(x\sqrt{x^2-a^2})C \\ (2)当x\lt -a时,令u-x,则u\gt a \\ \int{\frac{1}{\sqrt{x^2-a^2}}dx}-\int{\frac{1}{\sqrt{u^2-a^2}}du}\\ -\ln(u\sqrt{u^2-a^2})C-\ln(-x\sqrt{x^2-a^2})C\ln(-x-\sqrt{x^2-a^2})C_1\\ 综上当x\gt a时\int{\frac{1}{\sqrt{x^2-a^2}}dx}\ln(x\sqrt{x^-a^2})C \\ x\lt -a时\int{\frac{1}{\sqrt{x^2-a^2}}dx}\ln(-x-\sqrt{x^2-a^2})C\\ 所以\int{\frac{1}{\sqrt{x^2-a^2}}dx}\ln|x\sqrt{x^2-a^2}|C 解定义域xa或者x−a(1)当xa时令xasect,0t2πx2−a2atant,dxasecttantdt∫x2−a21dx∫atant1asecttantdt∫sectdtln∣secttant∣C1其中sectax,tantax2−a2∫x2−a21dxln(xx2−a2)C(2)当x−a时,令u−x,则ua∫x2−a21dx−∫u2−a21du−ln(uu2−a2)C−ln(−xx2−a2)Cln(−x−x2−a2)C1综上当xa时∫x2−a21dxln(xx−a2)Cx−a时∫x2−a21dxln(−x−x2−a2)C所以∫x2−a21dxln∣xx2−a2∣C
2.2.4 三角代换汇总
被积函数中函数含有三角代换a2−x2\sqrt{a^2-x^2}a2−x2xasint,−π2tπ2xa\sin t,-\frac{\pi}{2}\lt t\lt \frac{\pi}{2}xasint,−2πt2πa2x2\sqrt{a^2x^2}a2x2xatant,−π2tπ2xa\tan t,-\frac{\pi}{2}\lt t\lt \frac{\pi}{2}xatant,−2πt2πx2−a2\sqrt{x^2-a^2}x2−a2xa,xasect,0tπ2x\gt a,xa\sec t,0\lt t\lt \frac{\pi}{2}xa,xasect,0t2π
2.2.5 倒代换 适用分母次数分子次数 例4 求∫a2−x2x4dx(a̸0)\int{\frac{\sqrt{a^2-x^2}}{x^4}dx}(a\not0)∫x4a2−x2dx(a0) 解(1)利用上面的三角代换自己做(1)倒代换,令x1tdx−1t2,a2−x2a2t2−1∣t∣∫a2−x2x4dx∫a2t2−1∣t∣⋅t4⋅(−1t2)dt−∫a2t2−1⋅∣t∣dt当x0时t1x0∫a2−x2x4dx−12a2∫(a2t2−1)12d(a2t2−1)dt−(a2−x2)323a2x3C当x0时t1x0∫a2−x2x4dx−(a2−x2)323a2x3C综上∫a2−x2x4dx−(a2−x2)323a2x3C解(1)利用上面的三角代换自己做\\ (1)倒代换,令x\frac{1}{t} \\ dx-\frac{1}{t^2},\sqrt{a^2-x^2}\frac{\sqrt{a^2t^2-1}}{|t|} \\ \int{\frac{\sqrt{a^2-x^2}}{x^4}dx}\int{\frac{\sqrt{a^2t^2-1}}{|t|}\cdot t^4\cdot(-\frac{1}{t^2})dt} \\ -\int{\sqrt{a^2t^2-1}\cdot|t|dt} \\ 当x\gt0时t\frac{1}{x}\gt0\\ \int{\frac{\sqrt{a^2-x^2}}{x^4}dx}-\frac{1}{2a^2}\int{(a^2t^2-1)^{\frac{1}{2}}d(a^2t^2-1)dt}-\frac{(a^2-x^2)^{\frac{3}{2}}}{3a^2x^3}C\\ 当x\lt0时t\frac{1}{x}\lt0\\ \int{\frac{\sqrt{a^2-x^2}}{x^4}dx}-\frac{(a^2-x^2)^{\frac{3}{2}}}{3a^2x^3}C \\ 综上 \int{\frac{\sqrt{a^2-x^2}}{x^4}dx}-\frac{(a^2-x^2)^{\frac{3}{2}}}{3a^2x^3}C 解(1)利用上面的三角代换自己做(1)倒代换,令xt1dx−t21,a2−x2∣t∣a2t2−1∫x4a2−x2dx∫∣t∣a2t2−1⋅t4⋅(−t21)dt−∫a2t2−1⋅∣t∣dt当x0时tx10∫x4a2−x2dx−2a21∫(a2t2−1)21d(a2t2−1)dt−3a2x3(a2−x2)23C当x0时tx10∫x4a2−x2dx−3a2x3(a2−x2)23C综上∫x4a2−x2dx−3a2x3(a2−x2)23C
2.2.6 根式代换
例5 求∫112xdx\int{\frac{1}{1\sqrt{2x}}dx}∫12x1dx 解令2xt,xt22,dxt∫112xdx∫t1tdtt−ln∣t1∣C2x−ln∣sqrt2x1∣C解令\sqrt{2x}t,x\frac{t^2}{2},dxt \\ \int{\frac{1}{1\sqrt{2x}}dx}\int{\frac{t}{1t}dt}t-\ln|t1|C\\ \sqrt{2x}-\ln|sqrt{2x}1|C 解令2xt,x2t2,dxt∫12x1dx∫1ttdtt−ln∣t1∣C2x−ln∣sqrt2x1∣C
3 积分推导公式
常用积分公式除了基本积分表中在添加下面几个前面推导的公式
∫tanxdx−ln∣cosx∣C\int{\tan xdx}-\ln|\cos x|C∫tanxdx−ln∣cosx∣C∫cotxdxln∣sinx∣C\int{\cot xdx}\ln|\sin x|C∫cotxdxln∣sinx∣C∫secxdxln∣secxtanx∣C\int{\sec xdx}\ln|\sec x\tan x|C∫secxdxln∣secxtanx∣C∫cscxdxln∣cscx−cotx∣C\int{\csc xdx}\ln|\csc x-\cot x|C∫cscxdxln∣cscx−cotx∣C∫1a2x2dx1aarctanxaC\int{\frac{1}{a^2x^2}dx}\frac{1}{a}\arctan\frac{x}{a}C∫a2x21dxa1arctanaxC∫dxx2−a212aln∣x−axa∣C\int{\frac{dx}{x^2-a^2}}\frac{1}{2a}\ln|\frac{x-a}{xa}|C∫x2−a2dx2a1ln∣xax−a∣C∫dxa2−x2arcsinxaC\int{\frac{dx}{\sqrt{a^2-x^2}}}\arcsin\frac{x}{a}C∫a2−x2dxarcsinaxC∫dxx2a2ln(xx2a2)C\int{\frac{dx}{\sqrt{x^2a^2}}}\ln(x\sqrt{x^2a^2})C∫x2a2dxln(xx2a2)C∫dxx2−a2ln∣xx2−a2∣C\int{\frac{dx}{\sqrt{x^2-a^2}}}\ln|x\sqrt{x^2-a^2}|C∫x2−a2dxln∣xx2−a2∣C
例6 求 ∫x3(x2−2x2)2dx\int{\frac{x^3}{(x^2-2x2)^2}dx}∫(x2−2x2)2x3dx 解x2−2x2(x−1)21,令x−1tant,(−π2tπ2)∫x3(x2−2x2)2dx∫(tant1)3sec4tsec2tdt∫(sin3tcos−1t3sin2t3sintcostcos2t)dt−lncost−cos2t2t−sintcostC按tantx−1做辅助三角形cost1x2−2x2,sintx−1x2−2x2∫x3(x2−2x2)2dx12ln(x2−2x2)2arctan(x−1)−xx2−2x2C解\\ x^2-2x2(x-1)^21 ,令x-1\tan t ,(-\frac{\pi}{2}\lt t\lt\frac{\pi}{2})\\ \int{\frac{x^3}{(x^2-2x2)^2}dx}\int{\frac{(\tan t1)^3}{\sec^4t}\sec^2tdt}\\ \int{(\sin^3t\cos^{-1}t3\sin^2t3\sin t\cos t\cos^2t)dt}\\ -\ln\cos t-\cos^2t2t-\sin t\cos tC\\ 按\tan tx-1做辅助三角形\\ \cos t\frac{1}{\sqrt{x^2-2x2}},\sin t\frac{x-1}{\sqrt{x^2-2x2}} \\ \int{\frac{x^3}{(x^2-2x2)^2}dx}\frac{1}{2}\ln(x^2-2x2)2\arctan(x-1)-\frac{x}{x^2-2x2}C 解x2−2x2(x−1)21,令x−1tant,(−2πt2π)∫(x2−2x2)2x3dx∫sec4t(tant1)3sec2tdt∫(sin3tcos−1t3sin2t3sintcostcos2t)dt−lncost−cos2t2t−sintcostC按tantx−1做辅助三角形costx2−2x21,sintx2−2x2x−1∫(x2−2x2)2x3dx21ln(x2−2x2)2arctan(x−1)−x2−2x2xC 辅助三角形图示
后记 ❓QQ:806797785 ⭐️文档笔记地址https://gitee.com/gaogzhen/math 参考
[1]同济大学数学系.高等数学 第七版 上册[M].北京:高等教育出版社,2014.7.P193~p207.
[2]【梨米特】同济七版《高等数学》全程教学视频纯干货知识点解析应该是全网最细微积分 | 高数[CP/OL].2020-04-16.p28.
三角函数公式
