How to find an indefinite integral, how to find an indefinite integral

Indefinite integral concept.

In differential calculus we already know that if the equation for a body to move in a straight line is s=f(t), the instantaneous velocity of the object is known v=f(t), and the law of motion of the object is required to be s=f(t). This is obviously a question of reversing the demand for the "original function" from the derivative of the function, and that's what this section is going to discuss.

Definition 1 knows that f(x) is a function defined over an interval where if there is a function f(x) such that at any point within that interval there is:

In this interval then we call the function f(x) the original function of the function f(x).

Of course, not all functions have primitive functions, and in the next chapter we will prove that continuous functions have primitive functions. If f(x) has the original function f(x), then f(x)+

c is also its original function, where c is an arbitrary constant. Thus, if f(x) is the original function, it has an infinite number of original functions, and f(x)+

c contains all the primitive functions of f(x).

In fact, let g(x) be any of its original functions, then.

According to the corollary of the differential median value theorem, h(x) should be a constant c, so there is.

g(x)=f(x)+

c This means that any two original functions of f(x) differ only one constant.

The whole primitive function that defines the 2 function f(x) is called the indefinite integral of f(x) and is denoted as.

where is called the integral sign, f(x) is called the integrand, and f(x).

dx is called the product expression, and x is called the integral variable.

If f(x) is a primitive function of f(x), then there is a defined one.

where c is an arbitrary constant, which is called the integral constant.

The operation of finding the original function or indefinite integral is called the integral method.

The numerator and denominator are multiplied by sinx at the same time

sinxdx/(sinx)^4=-∫

dcosx/(1-(cosx)^2)^2

Let cosx=t

1/(1-t)+1/(1+t)]^2*dt1/(1-t)^2dt-∫

1/(1+t)^2dt-1/2∫

1/(1-t)dt-1/2∫

1/(1+t)dt

1/(1-t)+1/(1+t)+

Then just restore t to x.

The calculation process is as follows:

indefinite integral of sinx (sinx+cosx).

sinxcosx)/(sinx + cosx) dx

1/2)∫ 2sinxcosx)/(sinx + cosx) dx

1/2)∫ 1 + 2sinxcosx) -1]/(sinx + cosx) dx

1/2)∫ sin²x + 2sinxcosx + cos²x)/(sinx + cosx) dx - 1/2)∫ dx/(sinx + cosx)

1/2)(-cosx + sinx) -1/(2√2)]ln|csc(x + 4) -cot(x + 4)| c

Proof of Indefinite Integral:

If f(x) has a primitive function in the interval i, i.e. there is a function f(x) such that for any x i, there is f'(x)=f(x), then there is obviously [f(x)+c] for any constant'=f(x).That is, for any constant c, the function f(x)+c is also the original function of f(x). This means that if f(x) has a primitive function, then f(x) has an infinite number of primitive functions.

Let g(x) be another primitive function of f(x), i.e., x i,g'(x)=f(x)。So Hui lacks [g(x)-f(x)].'g'(x)-f'(x)=f(x)-f(x)=0。

Since the derivative is constant zero over an interval, the function of the game must be constant, so g(x)-f(x)=c'(c' is a constant).

Solved with fractional credits.

arctanx dx

xarctanx- x d(arctanx)xarctanx- x (1+x 2) dxxarctanx-(1 2) 1 (1+x 2) d(1+x 2)xarctanx-(1 2)ln(1+x 2)+c The relationship between indefinite and definite integrals is determined by the fundamental theorem of calculus. where f is the indefinite integral of f.

A function can have indefinite integrals and no definite integrals, or it can have definite integrals without indefinite integrals. continuous functions, there must be definite integrals and indefinite integrals; If there are only finite discontinuities on the finite interval [a,b] and the function is bounded, then the definite integral exists; If there are jumps, going, and infinite discontinuities, then the original function must not be buried in the original function, that is, the indefinite integral must not exist.

1/2ln[(1+x)/(1-x)]+c

The process of solving the problem is as follows:

1/2∫[1/(1-x)+1/(1+x)]dx1/2[-ln(1-x)+ln(1+x)]+c1/2ln[(1+x)/(1-x)]+c

In calculus, an indefinite integral of a function f, or the original function, or the antiderivative, is a function f where the derivative is equal to f, i.e., f f.

The relationship between indefinite integrals and definite integrals is determined by the basic calculus theory. where f is the indefinite integral of f.

The details are as follows:∫ (cosx)^3 dx

cosx)^2*cosx dx

cosx)^2dsinx

（1-(sinx)^2） dsinx

1 dsinx-∫(sinx)^2 dsinx=sinx-1/3*(sinx)^3+c

Significance of Indefinite Integral:A function can have indefinite integrals and no definite integrals, or it can have definite integrals without indefinite integrals. continuous functions, there must be definite integrals and indefinite integrals;

If there are only finite discontinuities on the finite interval [a,b] and the function is bounded, then the definite integral exists; If there are jump, go, and infinite discontinuities, then the original function must not exist, i.e., the indefinite integral must not exist.

The formula for finding the integral starts with the following:1. Definition of 0dx=c indefinite integral.

2、∫x^udx=(x^(u+1))/u+1)+c3、∫1/xdx=ln|x|+c

4. Swift repentance a xdx=(a x) lna+c5, mu jingzheng e xdx=e x+c

6、∫sinxdx=-cosx+c

7、∫cosxdx=sinx+c

8、∫1/(cosx)^2dx=tanx+c9、∫1/(sinx)^2dx=-cotx+c<>

10、∫1/√（1-x^2)dx=arcsinx+c11、∫1/(1+x^2)dx=arctanx+c12、∫1/(a^2-x^2)dx=(1/2a)ln|(a+x)/(a-x)|+c

13、∫secxdx=ln|secx+tanx|+c14、∫1/(a^2+x^2)dx=1/a*arctan(x/a)+c

15、∫1/√(a^2-x^2)dx=(1/a)*arcsin(x/a)+c

16、∫sec^2xdx=tanx+c

17、∫shx dx=chx+c

18、∫chx dx=shx+c

19、∫thx dx=ln(chx)+c

This problem is the calculation of definite integrals, and the calculation process needs to be separated as follows:

0，1](3/ⅹ^2+1+sinⅹ)dⅹ∫[0，1]3dx/(x^2+1)+∫0，1]sinⅹdx3arctanx[0，1]-cosx[0，1]3*π/4-(cos1-cos0)

3π/4+1-cos1。

In this problem, we use the formulas for finding the derivatives of arctangent functions and sinusoidal trigonometric functions.

3π/4+1-cos1

Here's how, please refer to:

Solution analysis: The sum and difference method of the integral and the basic integral are used to calculate the definite integral.

Separately, they are all a few basic integrals.

The original function of 3 (1+x 2) is 3arctanx

The original function of sinx is -cosx