# Textbook fourth-indefinite integral exercise arrangement

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Trigonometric Correlation Types

1. Using triangular changes and trigonometric identities, such as $\sin^2 x + \cos^2 x=1$ and twice-fold formula, and differential product, accumulation and difference.

For example: 4-1 of 1-(5), so that $1=\sin^2 x+ \cos^2 x$ can;

4-1 of 1-(6), twice-angle formula

4-1 of 1-(7), twice-angle formula $\cos 2x =\cos^2 x-\sin^2 x$

4-2 of 1-(12), integrable and differential formulas

4-3 (5), $\tan^2 x=\sec^2 x-1$

2. The type of $\sin^m x \cos^n x$, if the $m, n$ two items are even, the use of twice-angle formula to reduce the order; one is odd, then one comes out of the differential, and the remaining even items use $\sin^2 x+ \cos^2 x=1$ into integrals of the same function, if higher than 2 The twice-angle formula is used to reduce the order.

such as 4-1 of 1-(6);

For example, 4-2 of 1-(13):
$\int \cos^3 x dx = \int \cos^2 x D (\sin x) =\int (1-\sin^2 x) d (\sin x) =\sin x-\frac13 \sin^3 x +c.$

3. $\frac{\sin^m x}{\cos^n x}$ or $\frac{\cos^m x}{\sin^n x}$ type, if $m, n$ are even, use $\frac{1}{\cos^2 x} dx= D (\tan x)$ or $\frac{1}{\sin^2 X} dx= D (-\cot x)$, the remaining even times trigonometric functions, using $\sin^2 x+\cos^2 x=1$ and $\tan^2 x = \sec^2 x-1$ to be consistent; if $m, n$ has One is odd, a differential is eliminated, and then a triangular identity is used.

For example 4-4 of (13): Because the odd item is in the denominator, unlike the previous type, we do not propose a $\cos x$, and the numerator denominator is multiplied by a $\cos x$
\begin{aligned} \int \frac{1}{\sin^3 x \cos x} dx=\int \frac{\cos x}{\sin^3 x \cos^2 x} DX = \int \frac{1}{\sin^3x (1-\sin^2 x)} d (\sin x) = \int \frac{(1-\sin^2 x) +\sin^2 x}{\sin^3x (1-\sin^2 x)} d (\sin x) \\= \int \frac{1}{\sin^3 x} d (\sin x) + \int \frac{1}{\sin x (1-\sin^2 x)} d (\sin x) =-\FRAC12 \csc^2 x + \int \frac{1-\sin^2 x +\sin^2 x}{\sin x (1-\sin^2 x)} d (\sin x) \\=-\FRAC12 \csc^2 x + \int (\frac1 {\sin x}-+ \frac{\sin x}{1-\sin^2 x}) d (\sin x) =-\FRAC12 \csc^2 X+\ln|\sin x| -\ln|\cos X|+c. \end{aligned}

For example, it's an even number of times.
\begin{aligned} \int \tan^2 x dx = \int \frac{\sin^2 x}{\cos^2 x} dx = \int \sin^2 x D (\tan x) =\int (1-\cos^2 x) d (\tan x) \\= \tan x-\int \frac{1}{\tan^2 +1} d (\tan x) =\tan x-x +c. \end{aligned}
In fact, this is just an example of how to show an even number of times, and this integral is very easy to do because
$\int \tan^2 x DX =\int (\sec^2 x-1) dx.$

(Of course, in some cases, partial integrals are helpful in this type)

4. $\tan x \sec x$ type, if two functions appear simultaneously, take advantage of $d (\sec x) = \tan x \sec x dx$ and $\sec^2 x =\tan^2 x +1$. (In fact, do not control this technique, put $\tan x =\sin x/\cos x$ and $\sec x =1/\cos x$ generation, use case 2 direct calculation can also)

Like 4-1 of 1-(4)

Like 4-2 of 1-(14)
$\int \tan^3 x \sec x dx= \int \tan^2 x D (\sec x) = \int (\sec^2 x-1) d (\sec x) = \frac13 \sec^3 x-\sec x +c.$

5. If the trigonometric functions are not thought out, it is always helpful to write it in the form of $\sin x$ and $\cos x$.

Like 4-2 of 1-(22)
$\int\frac{\cot x}{1+\sin x} dx = \int \frac{\cos x}{\sin x (1+\sin x)} dx = \int \frac{1}{\sin x (1+\sin x)} d (\sin x)$

Rational Function

1. True fraction, simple decomposition can

For example: Exercise 4-1 of 1-(8), Exercise 4-4 (1), (2), (3)

4-4 (5): Because the $x ^{10}-2$ differs from the $x$ constant by 2, so
\begin{aligned} \int \frac{1}{x (x^{10}-2)}dx =\frac12\int \frac{2}{x (x^{10}-2)}dx= \frac12 \int \frac{2-x^{10}+x^{10}}{x (x^{10}-2)} Dx \\= \frac12 \int \frac{x^9}{x^{10}-2}dx-\frac12 \int \frac 1x Dx=\frac 1{20} \ln|x^{10}-2| -\FRAC12 \ln|x|+c. \end{aligned}
(Description: Observe the denominator factor before the constant difference, and then propose a constant re-formulation is the basic method)

2. Is not a true fraction, that is, the use of polynomial division. For example: Exercise 4-4 of 1-(4)

Universal Transform

More suitable for the universal transformation is, $\sin x$ and $\cos x$ Plus and minus operations, and factors in front of the factor; because other aspects of dealing with this problem are often particularly troublesome, such as review questions 43, 4-4 (11), (12), (15)

For example 4-4 of (12): Make $u =\tan \frac x2$,
\begin{aligned} \int \frac{1}{3+5\cos x} dx = \int \frac{1}{3+5 \frac{1-u^2}{1+u^2}} \cdot \frac{2}{1+u^2} du= \int \frac{1}{4-u^2} du \\=-\frac14 \ln|u-2| +\frac14 \ln |u+2|+c = \frac14 \ln |\tan \frac x2 +2| -\frac14 \ln |\tan \frac x2-2|+c. \end{aligned}

Basic triangular transformations, $\sqrt{x^2-a^2}$, $\sqrt{a^2-x^2}$ and $\sqrt{x^2+a^2}$

1. Simple transformations, such as the denominator of the rational, the molecules have physical and chemical

For example, 4-2 of 3-(2):
$\int \frac{\sqrt{x^2-1}}{x} dx = \int \frac{x^2-1}{x\sqrt{x^2-1}} dx= \int\frac{x}{\sqrt{x^2-1}} DX-\int \FRAC{1}{X\SQR T{X^2-1}} dx= \sqrt{x^2-1}-\arcsin|\frac 1x| +c.$
(using the results of (1))

4-2 of 3-(3):
$\int \frac{dx}{1+\sqrt{1-x^2}} = \int (\frac 1{x^2}-\frac{\sqrt{1-x^2}}{x^2}) dx =-\frac 1x-\int \frac{\sqrt{1-x^ 2}}{x^2} DX,$
which
$\int \frac{\sqrt{1-x^2}}{x^2} dx = \int \sqrt{1-x^2} d (\frac 1x) = \frac {\sqrt{1-x^2}} {x}-\int \frac 1x \cdot \frac{- X}{\SQRT{1-X^2}} DX =\frac {\sqrt{1-x^2}} {x}+ \arcsin x+c,$
So
$\int \frac{dx}{1+\sqrt{1-x^2}} = \arcsin x +\frac{\sqrt{1-x^2}}{x}-\frac 1x +c.$

2. Formulations written in standard form

Like 4-2 of 3-(9)
$\int \frac{dx}{\sqrt{x^2-2x-3}} =\int \frac{dx}{\sqrt{(x-1) ^2-4}}$
Next, directly using the p.190 formula

3. Types of $\sqrt{x^2-a^2}$ (pay particular attention to this type of discussion)

For example 4-2 of 3-(1): When $x >1$, set $x =\sec t \ (0<t<\frac \PI 2)$, then $dx = \sec t \tan t dt$, and
$\int \frac{dx}{x\sqrt{x^2-1}} =\int \frac{1}{\sec t \tan t} \sec t \tan t dt = t+c.$
Because $x =\sec t$, so $t =\arccos \frac1x$, so
$\int \frac{dx}{x\sqrt{x^2-1}} =\arccos \frac1x +c=\frac \pi 2-\arcsin \frac 1x +c =-\arcsin \frac 1x +C.$
When $x <-1$, $t =-x$ $t >1$ and
$\int \frac{dx}{x\sqrt{x^2-1}} = \int \frac{1}{(-t) \sqrt{t^2-1}} d (-t) = \int \frac{1}{t \sqrt{t^2-1}} dt = -\arcsin \frac 1t +c=-\arcsin \frac{-1}{x}+c,$
So
$\int \frac{dx}{x\sqrt{x^2-1}} =-\arcsin |\frac 1x|+c.$
(note, consistent with the use of inverted transform conclusions)

For example 4-2 of 3-(2): When $x >1$, set $x =\sec t \ (0<t<\frac \PI 2)$, then $dx = \sec t \tan t dt$, and
$\int \frac{\sqrt{x^2-1}}{x} dx = \int \tan^2 t dt = \int (\sec^2 t-1) DT =\tan t-t+c.$
And because $\tan T =\sqrt{\sec^2 t-1}$, so
$\int \frac{\sqrt{x^2-1}}{x} dx=\sqrt{x^2-1}-\arccos \frac 1x +c,$
When $x <-1$, set $x =-t$, then $t >-1$ and
$\int \frac{\sqrt{x^2-1}}{x} dx = \frac{\sqrt{t^2-1}}{t}dt = \sqrt{t^2-1}-\arccos \frac1t +C = \sqrt{x^2-1}-\arccos (-\fra C 1x) +c.$
Note that $\arccos (-\frac 1x) = \pi-\arccos (\frac 1x)$, when $x <-1$,
$\int \frac{\sqrt{x^2-1}}{x} DX =\sqrt{x^2-1} +\arccos (\frac 1x) +c.$
(Note: It is obviously very simple and straightforward to get the answer, actually the basic triangulation is one of the best methods)

4. Types of $\sqrt{a^2-x^2}$

4-2 of 3-(3): Make $x =\sin T, \ (-\frac \pi 2< T < \frac \PI 2)$, then
$\int \frac{dx}{1+\sqrt{1-x^2}} = \int \frac{\cos t}{1+\cos t} dt = \int (1-\frac 1{1+\cos t}) DT$
and notice the
$\int \frac{1}{1+\cos t} DT =\int \frac{1-\cos t}{1-\cos^2 t}dt = \int (\frac 1{\sin^2 T}-\frac{\cos t}{\sin^2 t}) DT =-\cot +\frac 1{\sin T} +c,$
So
$\int \frac{dx}{1+\sqrt{1-x^2}} = t +\cot t-\frac{1}{\sin t}+c,$
Because
$\cot t = \frac{\cos t}{\sin t}=\frac{\sqrt{1-\sin^2 t}}{\sin t}= \frac{\sqrt{1-x^2}}{x}$
So
$\int \frac{dx}{1+\sqrt{1-x^2}} = \arcsin x + \frac{\sqrt{1-x^2}}{x}-\frac1x +c.$
(Note: It is not more troublesome than physical and chemical)

4-2 of 3-(4): Make $x =\sin T, \ (-\frac \pi 2 < T < \frac \PI 2)$, then
$\int \frac{1}{x+\sqrt{1-x^2}} dx = \int \frac{\cos t}{\sin t +\cos t} DT,$
Using the method of undetermined coefficients,
$\cos T =a (\sin t +\cos t) ' + B (\sin t +\cos t)$
Compare the coefficients of $\sin t$ and $\cos t$ on both sides.
\begin{aligned} \int \frac{\cos t}{\sin t +\cos t} dt= \frac12 \int \frac{(\sin t +\cos t) '}{\sin t +\cos t} DT +\frac \int \frac{\sin t+\cos t}{\sin t +\cos t}dt = \frac12 \ln|\sin t+\cos t|+\frac t +c \\= \FRAC12 \ln|x+\sqrt{1-x^2}| +\FRAC12 \arcsin x +c. \end{aligned}

5. Types of $\sqrt{x^2+a^2}$

4-2 of 3-(5): Make $x =\tan T, \ (-\frac \pi 2 < T < \frac \PI 2)$, then
$\int \frac{1}{(x^2+1) ^{3/2}} DX = \int \frac{1}{\sec t} dt = \sin t +c.$
Notice that the
$|\sin t| = \sqrt{\frac{1}{1+\cos^2 T}}=\frac{1}{1+\frac 1{x^2}} = \frac{|x|} {\sqrt{1+x^2}},$
When $0<t<\frac \pi 2$, $x =\tan t >0$, $\sin t >0$, so $\sin t =\frac{x}{\sqrt{1+x^2}}$. And when $-\frac \pi 2<t<0$,
$x =\tan t <0$, $\sin t<0$, so $-\sin t =-\frac{x}{\sqrt{x^2+1}}$, i.e. $\sin t =\frac{x}{\sqrt{x^2+1}}$. So
$\int \frac{1}{(x^2+1) ^{3/2}} dx = \sin T +c =\frac{x}{\sqrt{x^2+1}}+c.$

4-2 of 3-(10): Make $x =\tan t,\ (-\frac \pi 2 < T<\frac \PI 2)$, then
$\int \frac{1}{x^2\sqrt{1+x^2}} dx =\int \frac{\cos t}{\sin^2 t} +c.$
Notice that the
$\frac 1{\sin T} =\sqrt{1+\cos^2 t} =\frac{\sqrt{1+x^2}}{x},$
So
$\int \frac{1}{x^2\sqrt{1+x^2}} DX =-\frac{\sqrt{1+x^2}}{x}+c.$

Note: Although the use of triangular transformation will be more complex, such as the above example, a few problems can be solved by other methods more simple, but triangular transformation is still a basic means, theoretically these three types of problems must be solvable, so if there is no way to master too much skill, then honestly use the three kinds of transformation. In fact, it is better than many methods, such as 4-2 of 3-(10) than the inverted transformation of the simple and direct.

inverted transformation, the applicable range is the denominator is higher than the number of molecules more than two orders, if the denominator is multiplied by the factor, it is generally recommended

For example, 4-2 of 3-(1): The front with the basic transformation is troublesome, if the inverted transformation will be much easier. Assuming that the $x >1$, $t =\frac1x$
$\int \frac{dx}{x \sqrt{x^2-1}} =\int \frac{1}{\frac 1t \frac{\sqrt{1-t^2}}{|t|}} \cdot \frac{-1}{t^2} dt =-\int \frac{1}{\sqrt{1-t^2}} dt=-\arcsin \frac 1x +c,$
And when the $x <-1$
$\int \frac{dx}{x \sqrt{x^2-1}} =\int \frac{1}{\frac 1t \frac{\sqrt{1-t^2}}{|t|}} \cdot \frac{-1}{t^2} dt = \int \frac{1}{\sqrt{1-t^2}} dt=\arcsin \frac 1x +c,$
That
$\mbox{original}=-\arcsin |\frac1x|+c.$

For example 4-2 of 3-(5): When $x >0$
$\int \frac{dx}{(x^2+1) ^{3/2}} =-\int \frac{|t|} {(1+t^2) ^{3/2}} dt = (1+t^2) ^{-1/2}+c =\frac{\sqrt{1+x^2}}{x} +c,$
And when $x <0$,
$\int \frac{dx}{(x^2+1) ^{3/2}} =-\int \frac{|t|} {(1+t^2) ^{3/2}} dt =-(1+t^2) ^{-1/2}+c =-\frac{\sqrt{1+x^2}}{|x|} +c = \frac{\sqrt{1+x^2}}{x} +c.$

For example 4-2 of 3-(10): Make $t =\frac 1x$ when $x >0$
$\int \frac{dx}{x^2 \sqrt{1+x^2}} =-\int \frac{|t|} {\sqrt{1+t^2}} dt =-\sqrt{1+t^2} =-\frac{\sqrt{1+x^2}}{|x|} +c=-\frac{\sqrt{1+x^2}}{x} +c.$
And when $x <0$,
$\int \frac{dx}{x^2 \sqrt{1+x^2}} =-\int \frac{|t|} {\sqrt{1+t^2}} dt = \sqrt{1+t^2} = \frac{\sqrt{1+x^2}}{|x|} +c=-\frac{\sqrt{1+x^2}}{x} +c.$
(When using inverted transformations, it is important to note the absolute value as it relates to the root)

irrational transformations: such as radical transformations, exponential transformations, logarithmic transformations

1. $\sqrt{\frac{a+x}{b+x}}$ type

Select one: Make $t =\sqrt{\frac{a+x}{a-x}}$, then $x =a-\frac{2a}{t^2+1}$
\begin{aligned} \int \sqrt{\frac{a+x}{a-x}} dx = \int t \cdot \frac{4at} {(t^2+1) ^2} dt = 4a \int (\frac{1}{t^2+1}-\frac{1}{(t^2+1) ^2}) Dt \\= 2a \arctan T-\frac{2at}{t^2+1}+c =2A \arctan \sqrt{\frac{a+x}{a-x}}-\sqrt{a^2-x^2}+c. \end{aligned}
(where a triangular transformation of the second integral, or the use of $\int \frac{1}{(x^2+1) ^2} dx$ known conclusions, see the workbook or the following sub-integral discussion)

Option two: If 4-2 of 3-(6): Make $t =\sqrt{a+x}$, then
\begin{aligned} \int \sqrt{\frac{a+x}{a-x}} dx = \int \frac{2t^2}{\sqrt{2a-t^2}} dt=-\int 2t D (\sqrt{2a-t^2}) = -2t\sqrt{2a-t^2} + 2\int \sqrt{2a-t^2} DT \\= -2t\sqrt{2a-t^2} + 2\int \frac{2a-t^2}{\sqrt{2a-t^2}} DT = -2t\sqrt{2a-t^2} + \int \frac{4a}{\sqrt{2a-t^2}} dt-\int \frac{2t^2}{\sqrt{2a-t^2}} DT \end{aligned}
Notice that the same item appears, so the equation above
$\int \frac{2t^2}{\sqrt{2a-t^2}} dt = \int \frac{2a}{\sqrt{2a-t^2}} dt-t\sqrt{2a-t^2} =2a \arcsin \frac t{\sqrt {2a}}-t\ SQRT{2A-T^2},$
So
$\int \sqrt{\frac{a+x}{a-x}} DX = 2a \arcsin \sqrt{\frac{x+a}{2a}}-\sqrt{a^2-x^2} +c.$

Select three: Physical and chemical. Assuming that the $a >0$, then $-a<x<a$, i.e.
$\int \sqrt{\frac{a+x}{a-x}} dx= \int \frac{a+x}{\sqrt{a^2-x^2}} DX =\int (\frac{a}{\sqrt{a^2-x^2}} + \frac{x}{\sqrt{a^2-x^2}}) dx =a \arcsin \frac x a-\sqrt{a^2-x^2}+c.$

such as 4-4 (8): To make $u =\sqrt{\frac{x}{2-x}}$, then
$\int \frac 1{x^3} \sqrt{\frac{x}{2-x}} dx = \int (\frac{1+u^2}{2u^2}) ^3 u \cdot \frac{4 u}{(1+u^2) ^2} dt= \int \frac{1+u^ 2}{2 U^4}du =-\frac16 u^{-3}-\frac12 U^{-1}+c,$
That
$\int \frac 1{x^3} \sqrt{\frac{x}{2-x}} dx =-\FRAC16 (\frac{2-x}{x}) ^{3/2}-\frac12 (\frac{2-x}{x}) ^{1/2} +C.$

2. Type of $\sqrt[a]{x}$: least common multiple transformation of the same pattern

4-2 of 3-(7): of which radicals are $x ^{1/2}$ as well as $x ^{3/4}$, so $4$ and $2$ least common multiple, even if $t =x^{1/4}$,
$\int \frac{\sqrt x}{1+ \sqrt[4]{x^3}} dx = 4 \int \frac{t^5}{1+t^3} DT = \frac43 t^3-\frac \ln |1+t^3| +c= \frac43 \sqrt[4]{x^3}-\frac43 \ln |1+\sqrt[4]{x^3}| +c.$

such as 4-4 (9), (10)

3. Other types that involve complex functions, as transformations that remove complex structures ~

For example, 4-2 of 3-(8): Make $t =\sqrt{e^x +1}$
$\int \frac1 {\sqrt{e^x +1}} dx = \int \frac{2t}{t (t^2-1)} dt=2\ln (\sqrt{e^x+1}-1)-x+c.$

Exercise 4-2 1-(1) to 1-(11) is a simple combination of basic integration tables and integral addition operations

Exercise 4-2 1-(15) to 1-(16) direct differential,

Exercise 4-2 1-(17) to 1-(18) numerator denominator is multiplied directly by a function, making it possible to differentiate

Partial integrals: The case of polynomial multiplication $e ^x$ or trigonometric functions

The partial differential of the exponential function or the triangular function is used to reduce the order of polynomial by using the partial integral. such as 4-3 of 1-(1), (2), (6), (7); For example, 4-3 of 1-(11), (12), is the exchange of elements and the combination of differential and partial integrals

Partial integrals: Contains inverse trigonometric function and logarithmic functions

It is important to note that inverse trigonometric function and logarithmic functions do not correspond to the basic integral tables, so their existence is difficult for us, and the partial integration method is used to remove them first!! such as 4-3 of 1-(3), (4), (9), (13)

For example 4-3 of 1-(8): Notice that there is a logarithmic function, so first remove it
\begin{aligned} \int \sin x \ln \tan x DX =-\int \ln \tan x D (\cos x) =-\ln \tan x \cdot \cos x + \int \frac{\sec X}{\tan x} DX \\=-\ln \tan x \cdot \cos x + \int \frac{1}{\sin x} dx=-\ln \tan x \cos x +\ln| \CSC x-\cot x |+c. \end{aligned}
It is particularly important to note that if the trigonometric functions are not clearly related, they are all written in sine cosine, such as $\frac{\sec X}{\tan x}$.

For example Review Title 44: Note not be here $\frac{1}{1+x^2}$ can together differential for $\arctan x$ confused, always remember inverse trigonometric function, unless can directly together differential out, otherwise cannot solve; The first task is to remove the inverse trigonometric function.
It is observed that the function is not able to get the original function directly, then the rational function of the denominator should be decomposed.
$\int \frac{\arctan x}{x^2 (1+x^2)} DX =\int \arctan x (\frac{1}{x^2}-\frac{1}{1+x^2}) dx = \int \frac{\arctan x}{x^2} dx-\int \frac{\arctan x}{1+x^2} DX$
Note that the second item can be directly differential, i.e.
$-\frac \sin 2x E^{-3x} = \frac12 \arctan x.$
The first item cannot be directly differential, then the use of partial integrals is
$\int \frac{\arctan x}{x^2} DX =-\int \arctan x D (1/x) =-\frac{\arctan x}{x}+ \int \frac{1}{x (1+x^2)}DX =-\frac{\arctan x} {X}++\ln |x| -\FRAC12 \ln|1+x^2|+c.$

Partial integrals: Simultaneous trigonometric and exponential functions

It is noted that the trigonometric function is a trigonometric function after derivation, and the exponential functions are exponential, so only recursion can occur and then it is solved. It is important to note that if you choose to differentiate the trigonometric functions at the outset, then each step will differentiate the trigonometric functions and vice versa.

For example, 4-3 of 1-(10):
\begin{aligned} \int e^{-3x} \sin 2x DX =-\frac13 \sin 2x D (e^{-3x}) =-\frac \sin 2x e^{-3x} +\frac23 \int e^{-3x} \cos 2x DX \\=-\frac \sin 2x e^{-3x}-\frac29 \int \cos 2x D (e^{-3x}) \-\frac13 \sin 2x e^{-3x}-\frac29 \cos 2x e^{-3x}-\frac \int e^{-3x} \sin 2x DX, \end{aligned}
If recursion is present, the original indefinite integral can be solved.

Partial integrals: Converting complex forms to processed forms

Like what
\begin{aligned} \int \frac{1}{(x^2+1) ^2} dx =-\int \frac 1{2x} d (\frac{1}{x^2+1}) =-\frac{1}{2x (x^2+1)}-\int \frac{1}{2x^2 (x^2+1)}DX \\=-\frac{1}{2x (x^2+1)}+\frac1{2x} +\frac12 \arctan X+c = \frac{x}{2 (x^2+1)} +\frac12 \arctan x+c. \end{aligned}
(Note: Using the classic triangulation is also very easy, the main idea here is that through such a partial integration, the structure becomes simple, we can directly decompose the rational function.) )

Textbook fourth-indefinite integral exercise arrangement

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