Sinx squared integral of e to the power of x

Ask noDefinite integrals∫(e^x)sin²xdx

Solution: Original formula = (1 2) (e x)(1-cos2x)dx

1/2)[(e^x)-∫e^x)cos2xdx]

1/2)[e^x-∫cos2xd(e^x)]

1/2)(1-cos2x)(e^x)-[sin2x)(e^x)-2∫(e^x)cos2xdx]

1/2)(1-cos2x)(e^x)-(sin2x)(e^x)+2∫(e^x)cos2xdx

The move yields (5 2) (e x)cos2xdx = (1 2)e x-(1 2)(1-cos2x)(e x)+(sin2x)(e x)=(1 2)(cos2x+2sin2x)(e x).

Therefore (e x)cos2xdx=(1 5)(cos2x+2sin2x)(e x)

Therefore, the original formula = (1 2)e x-(1 5)(cos2x+2sin2x)(e x)+c=[(1 2)-(1 5)(cos2x+2sin2x)]e x+c

Interpretation. Note the relationship between indefinite integrals and definite integrals: a definite integral is a number, while 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. Continuous functions.

There must be definite and indefinite integrals; If there are only finite breaks on the finite interval [a,b] and the function is bounded.

then a definite integral exists; If there are jump, go, infinite breaks, then the original function.

must not exist, i.e., indefinite integrals must not exist.

Solution: Original formula = (1 2) (e x)(1-cos2x)dx

1/2)[(e^x)-∫e^x)cos2xdx]

1/2)[e^x-∫cos2xd(e^x)]

1/2)e^x-(1/2)[(e^x)cos2x+2∫(e^x)sin2xdx]

1/2)(1-cos2x)(e^x)-∫e^x)sin2xdx

1/2)(1-cos2x)(e^x)-∫sin2xd(e^x)

1/2)(1-cos2x)(e^x)-[sin2x)(e^x)-2∫(e^x)cos2xdx]

1/2)(1-cos2x)(e^x)-(sin2x)(e^x)+2∫(e^x)cos2xdx

The move yields (5 2) (e x)cos2xdx = (1 2)e x-(1 2)(1-cos2x)(e x)+(sin2x)(e x)=(1 2)(cos2x+2sin2x)(e x).

Therefore (e x)cos2xdx=(1 5)(cos2x+2sin2x)(e x)

Therefore, the original formula = ( 1 2)[e x-(1 5)(cos2x+2sin2x)(e x)]+c=[(1 2)-(1 10)(cos2x+2sin2x)]e x+c

∫(e^x)sinxdx=(e^x)[sinx-cosx]/2+c。

e^x)sinxdx

sinxd(e^x)

sinx(e^x)-∫e^x)dsinx

sinx(e^x)-∫e^x)cosxdx

sinx(e^x)-∫cosxd(e^x)

sinx(e^x)-(e^x)cosx+∫e^xdcosx

sinx(e^x)-(e^x)cosx-∫e^xsinxd

So (e x)sinxdx=(e x)[sinx-cosx] 2+c

Properties: Integration is the inverse of differentiation, that is, knowing the derivative of the function, the original function is reversed. In terms of application, the integral effect is not only that, it is widely used in summation, in layman's terms, to find the area of the curved triangle, this ingenious solution method is determined by the special properties of the integral.

It is mainly divided into definite integrals, indefinite integrals, and other integrals. The properties of integration mainly include linearity, number preservation, maximum minimum, absolute continuity, absolute value integration, etc.

The partial integral method, which is indeed derived using multiplicative derivatives.

e^xsinxdx

sinxde^x

sinxe^x-∫e^xdsinx

sinxe^x-∫cosxe^xdx

sinxe^x-∫cosxde^x

sinxe^x-(cosxe^x-∫e^xdcosx)=sinxe^x-cosxe^x-∫sinxe^xdx2∫e^xsinxdx=sinxe^x-cosxe^x∫e^xsinxdx=e^x(sinx-cosx)/2

The indefinite integral of e to the power of x multiplied by the square of sinx is (1 2)e x-(1 5)(cos2x+2sin2x)(e x)+c=[(1 2)-(1 5)(cos2x+2sin2x)]e x+c.

A function can have indefinite integrals without definite integrals, or it can have indefinite integrals without indefinite integrals. For continuous functions, there must be definite and indefinite integrals.

Tips for calculating indefinite integrals:When some of the integrands are more complex, we can observe that some functions are placed behind d (the function behind d will change), so that the functions behind d d have a similar structure to the integrand of the complex intellectual hall in the front, and finally use the basic integral formula to find them (if not, we will further use other methods to find them).

i = virtual hand e x(sinx) 2dx = 1 2) e x(1-cos2x)dx = 1 Danyu shirt 2)e x - 1 2) e xcos2xdx

i1 = e^xcos2xdx = cos2xde^x = e^xcos2x + 2∫e^xsin2xdx

e^xcos2x + 2∫sin2xde^x = e^x(cos2x+2sin2x) -4i1

The cavity i1 = 1 5)e x(cos2x+2sin2x) is obtained

i = 1/2)e^x - 1/10)e^x(cos2x+2sin2x) +c

E to the power of x multiplied by sinx squared noDefinite integralsYes (1 2)e x-(1 5)(cos2x+2sin2x)(e x)+c=[(1 2)-(1 5)(cos2x+2sin2x)]e x+c.

A letter containing a number can have indefinite integrals without definite integrals, or it can exist with definite integrals without indefinite integrals. Continuous functions.

There must be definite and indefinite integrals;

Tips for calculating indefinite integrals:When some of the integrands are more complex, we can observe that some functions are placed behind d (the functions placed after d will change), so that the functions behind d d have a similar structure to the complex integrand on the front surface, and finally use the basic integral formula.

Find it (if you can't find it, use other methods to find it).

As follows:

Elementary functions. It can't be accumulated, double integral.

The method can be obtained:

exp(x^2)dx]^2

exp(y^2)dy∫exp(x^2)dx∫∫exp(x^2+y^2)dxdy

Substitution with polar coordinates:

rexp(r^2)drdθ

Suppose the radius of the circle is r:

2π[(1/2)exp(r^2)]

exp(a^2)-1]

Therefore, exp(x2)dx=[exp(a2)-1] under the root number.

Representation of points.

The integral of a function represents the overall nature of the function in a certain region, and changing the value of a point of the function does not change its integral value. For functions with Riemann integrables, change the value of a finite number of points, and its integral remains unchanged. For Lebegus integrable functions, a change in the value of a function on a set with a measure of 0 does not affect its integral value.

If two functions are the same almost everywhere, then their integrals are the same.

∫(e^x)²dx∫(e^x)d(e^x)

e^x)²/2+c

e^(2x)]/2+c

Define integralsThere is more than one approach, and the definitions are not exactly equivalent to each other. The main argument is that some special functions are defined in the definition of certain special functions: these functions are not integrable under the definition of some integrals, but their integrals exist under other definite space collisions.

However, there are sometimes differences in definitions due to teaching reasons, and the most common definitions of integral are Riemann integral and Lebegus integral.