What is the integral of 1 minus x squared under the root number

(1/2)[arcsinx + x√(1 - x²)]c

The process of solving the problem is as follows:

Let x = sin, then dx = cos d

1 - x²) dx = ∫ 1 - sin²θ)cosθ dθ) = ∫ cos²θ dθ

Using the descending formula, the original formula = 1 + cos2 ) 2 d = 2 + sin2 ) 4 + c

Because =arcsinx, 2 + sin2 ) 4 + c

arcsinx)/2 + x√(1 - x²))/2 + c

1/2)[arcsinx + x√(1 - x²)]c

The commutation integral method can be divided into the first type of commutation method and the second type of commutation method.

1. The first type of commutation method (i.e., the method of differential calculation).

By making up the differentiation, it finally relies on a certain integral formula. Then the original indefinite integral is obtained. For example.

2. Note: The transformation formula of the second type of commutation method must be reversible.

The second type of commutation method is often used to eliminate the radical in the integrand. When the integrand is a binomial with a high degree, in order to avoid cumbersome formulas, it is sometimes possible to use the second type of commutation method to solve the problem.

Using trigonometry, the root number 1+x 2 can make x=tant, the root number x 2-1 can make x=sect, and the root number 1-x 2 can make x=sint

Original = tant sect*secttantdt

tan^2tdt

(sec^2t-1)dt

tant-t)+c

tan(artsecx)-arcsecx+c

The root sign is a symbol used to represent the opening operation of a number or an algebraic formula. If a = b, then a is the nth root of b to the nth power or a is the 1 nth power of b. The open nth power handwriting and typography are denoted by the number or algebraic form written on the left side of the symbol and the area below the horizontal part above the symbol, and cannot go out of bounds.

Writing specifications

The writing of the root number is exactly the same in the printed and handwritten fonts, and only the handwriting specifications are introduced here.

1. Write the root number:

First draw a short diagonal line to the upper right corner in the middle of the grid, and then continue to draw the lower right middle diagonal line with the strokes, and then draw a horizontal line of moderate length according to your needs near the top of the grid, and then make up for it if it is not enough. (Here only focus on stroke order and writing, according to the printing body refer to this article to imitate the writing, not hard requirements).

2. Write the number or formula of the square to be opened:

The number or algebraic formula to be opened is written in the area enclosed by the right side of the V-shaped part on the left side of the symbol and the lower part of the horizontal part above the symbol, and cannot go out of bounds, if the number or algebraic formula of the square is too long, the upper horizontal must be extended to ensure that the open square or algebraic formula below is covered.

3. Write the square number or formula:

N to the nth power is written on the left side of the symbol, and n can be ignored when n=2 (square root), but it must be written if it is a cubic root (cubic root), a fourth square root, etc.

∫xdx/√(1 - x²)(1/2)∫2xdx/√(1 - x²)

1/2)∫dx²/√1 - x²)

1 2) d(-x) 1 - x )-1 2) d(1-x) brilliant (1 - x) (1 2)[1 (1- 1 2)] 1 - x )-1 - x ) c

Interpretation. According to the Newton-Leibniz formula, the definite integral of many functions.

The computation can be easily performed by finding the indefinite integral. Here we should pay attention to the relationship between indefinite integrals and definite integrals: a definite integral is a number, and an indefinite integral is an expression.

They are simply mathematically computationally related. A function can have indefinite integrals and no definite integrals, or it can have definite integrals without indefinite integrals.

The integral process of the root number x squared plus one half:

(x^2+1) dx

Let x=tanz, dx=sec 2z dz

Original = sec 3z dz

1/2)tanzsecz+(1/2)∫secz dz=(1/2)tanzsecz+(1/2)ln(secz+tanz)+c

1/2)x√(x^2+1)+(1/2)ln[x+√(x^2+1)]+c

The simple integral is the known derivative to find the original function, and if the derivative of f(x) is f(x), then the derivative of f(x) + c (c is a constant) is also f(x), that is, the integration of f(x) may not necessarily get f(x), because the derivative of f(x) + c is also f(x), c is an arbitrary constant, so the result of f(x) integration is infinite, it is uncertain, we always use f(x) + c instead, this is called an indefinite integral.

Here's how to solve the problem:

Let x tan , then: (1 x 2).

1＋（tanα）^2］＝1/cosα,dx＝［1/（cosα）^2］dα.

sinα＝√sinα）^2/［（sinα）^2＋（cosα）^2］｝

（tanα）^2/［1＋（tanα）^2｝

x (1 x 2),原式 {1 cos ) 1 (cos ) 2 }d

cosα/（cosα）^4］dα

1/［1－（sinα）^2］^2｝d（sinα）.

Then let sin u, then:

Original 1 (1 U 2) 2 du

1/4）∫［1＋u＋1－u）^2/（1－u^2）^2］du

1/4）∫［1＋u）^2/（1－u^2）^2］du＋（1/2）∫［1－u^2）/（1－u^2）^2］du

1/4）∫［1－u）^2/（1－u^2）^2］du

1/4）∫［1/（1－u）^2］du＋（1/2）∫［1/（1－u^2）］du＋（1/4）∫［1/（1＋u）^2］du

1/4）∫［1/（1－u）^2］d（1－u）＋（1/4）∫［1＋u＋1－u）/（1－u^2）］du

1/4）∫［1/（1＋u）^2］d（1＋u）

1/4）［1/（1－u）］－1/4）［1/（1＋u）］＋1/4）∫［1/（1－u）］du

1/4）∫［1/（1＋u）］du

1/4）［1/（1－sinα）］1/4）［1/（1＋sinα）］

1/4）∫［1/（1－u）］d（1－u）＋（1/4）∫［1/（1＋u）］d（1＋u）

1/4）｛1/［1－x/√（1＋x^2）］｝1/4）｛1/［1＋x/√（1＋x^2）］｝

1/4）ln｜1－u｜＋（1/4）ln｜1＋u｜＋c

1/4）［1＋x/√（1＋x^2）－1＋x/√（1＋x^2）］/1－x^2/（1＋x^2）］

1 4) ln 1 sin 1 4) ln 1 sin fan xiao c

1/4）［2x/√（1＋x^2）］/1＋x^2－x^2）/（1＋x^2）］

First Chop 1 4) LN 1 x (1 x 2) 1 x (1 x 2) c

1/2）x√（1＋x^2）＋（1/4）ln｜［√1＋x^2）＋x］/［1＋x^2）－x］｜＋c

1 2) x (1 x 2) (1 4) ln 1 x 2) x 2 (1 x 2 x 2) c

1/2）x√（1＋x^2）＋（1/2）ln｜x＋√（1＋x^2）｜＋c

Replace x=sec t, tan t= root number (sec 2 t-1) = root number (x 2-1).

dx=sec t tan t

integral = integral sect * root number (sec 2 t-1) sect tan t dt

Integral sect * root number (tan 2 t) sect tan t dt integral sect * tan t sect tan t dt integral sec 2 t * tan 2 t dt integral tan 2 t d (tan t).

1/3*tan^3 t +c

1 3 * [Genling shirt scrambling manuscript ruler Li cavity (x 2-1)] 3 +c

Replace x=sec t, tan t= root number (sec 2 t-1) = root number (x 2-1).

dx=sec t tan t

Integral = integral sect * root scrambling manuscript number (sec 2 t-1) sect tan t dt

Integral sect * root number (tan 2 t) sect tan t dt integral sect * tan t sect tan t dt integral sec 2 t * tan 2 t dt integral mausoleum shirt tan 2 t d (tan t).

1/3*tan^3 t +c

1 3 * [Root ruler Li cavity number (x 2-1)] 3 +c