sinx 1 x

The original function of x(sin2x-sinx) is -1 2xcos2x+1 4sin2x+xcosx-sinx+c, where c is a constant. Analysis: Partial integrals are used to solve the original function of x (sin2x-sinx).

Solution: x(sin2x-sinx)dx= xsin2xdx- xsinxdx=-1 2 xdcos2x+ xdcosx=-1 2xcos2x+1 2 cos2xdx+xcosx- cosxdx=-1 2xcos2x+1 4sin2x+xcosx-sinx+cFractional Credits:

An important and fundamental class of methods for calculating integrals in calculus. Its main principle is to reverse the differential formula of two multiplicative functions and convert the required integration into the integration of another simpler function. According to the basic function types that make up the integrand, the order of the partial integrals is organized into a formula:

to the inverse power of three fingers". They refer to the integral order of five basic functions: logarithmic functions, inverse trigonometric functions, power functions, trigonometric functions, and exponential functions.

In this case, it is the penultimate type of trigonometric function.

2x points are calculated as (2 3) x 4 3

sinx/[1+√(1-x^2)]

The denominator is a constant and lower bound even function, and the numerator is a bounded odd function, so the fraction is a bounded odd function with an integral = 0 in the symmetric interval

sin x is equal to (1-cos2x) 2.

sin x can be obtained according to the formula sin = sinx*sinx = [1-cos(2)]2, and sin x = (1-cos2x) 2.

sin is a sinusoidal function that refers to the ratio of the opposite side of any acute angle a to the hypotenuse of any right triangle.

There are several explanations for this question:

sin 2x represents the product of 2 sinx, mathematically expressed as:

sin^2x=sinx*sinx.

Since the value of x in this problem is not determined, the product of its sine cannot be determined.

Sin(x) 2 and (sinx) 2 are both equivalent to x when x=0.

In the case of higher mathematics equivalent infinitesimal substitution, sinx x, then (sinx) 2 can be replaced by x 2 (squared).

When x 0, the Taylor formula of sinx is sinx x o(x)o(x) refers to the higher-order infinitesimal of x, so when x 0.

Yes (sinx) x when x 0 (sinx) x o(x ) so when x 0 can (sinx) x.

Equivalent infinitesimal :1、e^x-1～x (x→0)

2、 e^(x^2)-1～x^2 (x
-cosx～1/2x^2 (x
-cos(x^2)～1/2x^4 (x→0)5、sinx~x (x→0)

6、tanx~x (x→0)

7、arcsinx~x (x→0)

8、arctanx~x (x→0)

cosx~1/2x^2 (x→0)

This can be done using the partial integration method, x 2 as u, sinx as v, and then the original function can be calculated according to the calculation principle. The solution is as follows:

It can be found using the two-part integral formula.

y=sin(x) is the odd function Wang Rang.

So sin(-x) = -sin(x).

sin (-x)=(sinx) trap = sin debate x<>

Since sin(-x)=-sin(x), sin(-x)=(sin(x)) sin(x). The specific derivation process is as follows:

Let x be any angle, then there is sin(-x)=-sin(x).

The square of the left and right sides of the above equation is obtained by sin (-x)=(sin(x)) The lead excavation sedan is scattered in (-a)(-b)=ab, so there is (-sin(x)) sin (x).

i.e. sin (x) sin x ) is proven.

According to the periodic referentiality of the sinusoidal function, sin(x) = sin(x + 2), therefore:

sin²(-x) =sin²(-x + 2π) sin²(2π -x) =sin²(x - 2π) sin²x

Therefore, sin is only mu pei (-x) = sin x.

The original function of the function sinx (without the constant rolling mask term) can be obtained by integral. First, we split sinx into sinx multiplied by sinx before doing the integrals.

sinx dx = sinx * sinx dx Next, we can use the commutation method to make a big deal about integrals, so that u = sinx and du = 2sinx*cosx dx.

Substituting du into the above equation gives :

sinx³ dx = 1/2)u du

Integrate u and get:

1 2) u du = 1 4) u + c Finally, replace u back with sinx to get:

1/4)sinx^4 + c

where c is a constant term, which represents any constant of an indefinite integral. Therefore, the original function of sinx is (1 4) sinx 4 + c.

d∫2x²sin(x+1)dx

I need to solve $ int 2x 2 sin(x+1) dx$. According to the derivative of the integral, we have $$ frac left( int f(x) dxight) =f(x)$$ so, $$d int 2x 2 sin(x+1) dx = 2x 2 sin(x+1)dx$$ and now we can do indefinite integrals for $2x 2 sin(x+1)dx$. First of all, we can use the slow return to correct which stool meta-method, so that $u = x+1$, and get:

int 2x 2 sin(x+1) dx = int 2(u-1) 2 sin u du$$ and simplify, we get: $$begin int 2 (u 2 - 4u + 4) sin u du &=2 int (u 2 sin u - 4u sin u + 4 sin u) du 2 left(-u 2 cos u + 4u cos u - int (-2u cos u + 4 cos u) du ight) 2 left(-u 2 cos u + 4u cos u + 2u s

The original function of 1 sin x is: -cotx+ is a constant.

The process is as follows:

Finding the original function of 1 sin x is an indefinite integral of 1 sin x.

1/sin²xdx

csc²xdx

cotx+c

∫1/sin²xdx

csc²xdx

cotx+c

This is the basic integral formula.

Keep it in mind.