Definite integral sinx x0 to infinity pai 2

This is a dirchlet integral, which is more cumbersome to calculate, and has to be calculated using the integral with parametric variables.

Using generalized integrals with parameter variables, (0,+ denotes the integral from 0 to +.

Consider: 1 x= (0,+ e (-xt) dt

So sinx x= (0,+ e (-xt)*sinx dt

0,+∞sinx/xdx=∫(0,+∞0,+∞e^(-xt) sinxdt]dx

Changing the order of points will obtain:

0,+∞sinx/xdx=∫(0,+∞0,+∞e^(-xt) sinxdx]dt

Using the partial integration method, we can find:

i1=∫e^(-tx)sinxdx

e^(-tx)d(cosx)

e^(-tx)cosx-t∫e^(-tx)cosxdx

e^(-tx)cosx-ti2

i2=∫e^(-tx)cosxdx

e^(-tx)d(sinx)

e^(-tx)sinx+t∫e^(-tx)sinxdx

e^(-tx)sinx+ti1

The two-formula simultaneous solution is as follows: i1=-e (-tx)*(cosx+tsinx) (1+t 2).

So: i1(0)=-1 (1+t 2).

i1 (+ = 0 (at x + , e (-tx) 0 and cosx + tsinx bounded).

0,+∞sinx/xdx=∫(0,+∞1/(1+t^2)dt=arctant┃(0,+∞=π/2.

Summary. All the original functions of the function f(x) + c (c is an arbitrary constant) of the function f(x) are called indefinite integrals of the function f(x) and denoted as, i.e., f(x)dx=f(x)+cwhere is called the integral sign, f(x) is called the integrand, x is called the integral variable, f(x)dx is called the integrand, c is called the integral constant, and the process of finding the indefinite integral of a known function is called integrating this function.

Note: f(x)dx+c1= f(x)dx+c2, c1=c2sin x indefinite integral cannot be introduced.

Hello, I have seen your question and am sorting out the answer, please wait a while for a while sin x one-tenth indefinite integral.

All the original functions of the function f(x) + c (c is an arbitrary constant) of the function f(x) are called indefinite integrals of the function f(x) and denoted as, i.e., f(x)dx=f(x)+cwhere is called the integral sign, f(x) is called the integrand, x is called the integral variable, f(x)dx is called the integrand, c is called the integral constant, and the process of finding the indefinite integral of a known function is called integrating this function. Note:

f(x)dx+c1= f(x)dx+c2, c1=c2 cannot be pushed

Note: The upper and lower limits of this question are wrong, it should be the upper and lower limits of the integral (- 4, 4)!

Solution: Primitive = (4, 4)(sinx) 2 [1+e (-x)]dx (-4, 4) means from -4 to 4 integral) (4,0)(sinx) 2 [1+e (-x)]dx+ (0, 4)(sinx) 2 [1+e (-x)]dx

(4,0)(sinx)^2/(1+e^x)dx+∫(0,π/4)(sinx)^2/[1+e^(-x)]dx

The first point is obtained by substituting -x for x).

0,π/4)(sinx)^2/(1+e^x)dx+∫(0,π/4)e^x(sinx)^2/(1+e^x)dx

The second integral numerator denominator.

Multiply e x with you).

0,π/4)(1+e^x)(sinx)^2/(1+e^x)dx∫(0,π/4)(sinx)^2dx

1/2∫(0,π/4)[1-cos(2x)]dx1/2[x-1/2sin(2x)]|0,π/4)

Summary. Find the indefinite integral of sinx x from 1 to infinity.

My topic is this.

Here's how to do it, please make the first to check the ginseng test:

If it helps, the reeds are destroyed.

Problem : Find the indefinite integral of [x (sinx) 2] dx.

Antiderivative. In calculus, the indefinite indefinite infusion integral of a function f, or the original function of the pico, or the antiderivative, is a function f whose derivative is equal to f, i.e., f f.

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

Examples of indefinite integrals.

Example 1: DX = X+C

Example 2: sinx dx = -cosx + cExample 3 x 2 dx = 1 3) x 3 + c[x (sinx) 2] dx

1/(sinx)^2 = cscx)^2

x(cscx)^2 dx

dcotx = cscx)^2 dx

x dcotx

Division Points. -xcotx + cotx dx-xcotx + ln|sinx| +c

x (sinx) 2] dx withered socks =-xcotx + ln|sinx| +c