High Numbers Find the indefinite integral of secx to the third power

i=∫(secx)^3dx=(1/2)×(secxtanx+ln|secx+tanx|)+c

The process is as follows:

secxd(tanx)

secxtanx-∫tanxd(secx)=secxtanx-∫secx(tanx)^2dx=secxtanx-∫(secx)^3dx+∫secxdx=secxtanx-i+ln|secx+tanx|i=(1/2)×(secxtanx+ln|secx+tanx|)+cExtended Materialsy=Properties of secx:

1) Define the domain, 2) Value range, |secx|1 is secx 1 or secx 1;

3) y=secx is an even function, i.e., sec(x)=secx image symmetry on the y-axis;

4) y=secx is a periodic function with a period of 2k (k z, and k ≠0) and a minimum positive period t=2

The secant and cosine are reciprocal to each other, and the cosecant and sine are reciprocal to each other.

5) secθ=1/cos。

The indefinite integral results and derivation process of secx to the third power are shown in the figure above.

The indefinite integral of secx to the power of 3 really can't be calculated now.

The details are as follows:(secx)^3dx

secx(secx)^2dx

secxdtanx

secxtanx-∫tanxdsecx

secxtanx-∫(tanx)^2secxdxsecxtanx-∫(secx)^2-1)secxdxsecxtanx-∫(secx)^3dx+∫secxdxsecxtanx+ln│secx+tanx│--secx)^3dxSo (secx) 3dx=1 2(secxtanx+ln secx+tanx).

Indefinite Integral:According to the Newton-Leibniz formula, the calculation of definite integrals of many functions can be easily performed by finding indefinite integrals. Here we should note the relationship between indefinite integrals and definite integrals:

The Striker Sundefinite integral is a number, and the indefinite integral is an expression.

They are just mathematically a chain of silver and a computational relationship. 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.

An indefinite integral of sec 3(x) is required, which we can solve using the commutation method. First of all, let's do the following element swapping:

Let u = sec(x) +tan(x), then du = sec(x)tan(x) +sec 2(x)) dx

Now, we can express sec 3(x) as a wanton Zen function of u, i.e., sec 3(x) = sec(x)tan(x) +sec 2(x)) sec(x) =u sec(x).

Replacing these basis changes back into the original integral expression, we get:

sec^3(x) dx = u sec(x) dx

Now, we can bring the relationship between du and dx into the integrals and get:

sec^3(x) dx = u sec(x) dx = u du

For u du, we can solve the problem by splitting the front dust and get:

u du = 1/2) u^2 + c

Substituting u = sec(x) +tan(x) back, we end up with:

sec^3(x) dx = 1/2) (sec(x) +tan(x))^2 + c

where c is the integral constant. So, the indefinite integral of sec 3(x) is (1 2) (sec(x) +tan(x)) 2 + c.

The indefinite integrals of the cubic of secx are as follows:(secx)^3dx

secx(secx)^2dx

secxdtanx

secxtanx-∫tanxdsecx

secxtanx- (tanx) 2secxdxsecxtanx- (secx) 2-1)secxdxsecxtanx- (secx) 3dx+ secxdxsecxtanx+ln secx+tanx --secx) 3dx, so (secx) 3dx=1 2(secxtanx+ln secx+tanx).

Formula for indefinite integrals:

1. Bright numbers adx=ax+c, a and c are constants.

2. x adx=[x (a+1)] a+1)+c, where a is a constant rotten rock and a≠-1

3、∫1/xdx=ln|x|+c

4. A xdx = (1 lna) a x+c, where a >0 and a ≠15, e xdx = e x+c

6. Cosxdx=sinx+c

7、∫sinxdx=-cosx+c

8、∫cotxdx=ln|sinx|+c=-ln|cscx|+c

Since there is no simple standard formula for the indefinite integral sec 3(x) dx, dealing with it usually requires some advanced mathematical skills, specifically partial integrals and trigonometric identities.

First, let's decompose sec 3(x) dx into sec x · sec 2(x) dx. Then let u=sec x,dv=sec 2(x) dx. This allows us to calculate using the partial product state division according to the partial integration formula udv = uv - vdu.

We get the following results:

sec(x)·tan(x) -tan^2(x)·sec(x) dx

sec(x)·tan(x) -sec^2(x) -1]·sec(x) dx

sec(x)·tan(x) -sec^3(x) dx + sec(x) dx

Then you have a formula sec 3(x) dx = sec(x)·tan(x) -sec 3(x) dx + sec(x) dx.

Solve this dry equation to get:

2∫sec^3(x) dx = sec(x)·tan(x) +sec(x) dx

sec^3(x) dx = 1/2 [sec(x)·tan(x) +sec(x) dx]

where sec(x) dx is a standard integral solved by ln|sec(x)+tan(x)|。

So the end result is:

sec^3(x) dx = 1/2 [sec(x)·tan(x) +ln|sec(x)+tan(x)|c, where no macro lead c is a constant.

To calculate the indefinite integral of sec 3(x), it can be solved by the partial integral method. The formula for the partial integral method is:

u dv = uv - v du

where you and v are the microscopic numbers. For sec 3(x), we can write it as the product of two functions: u = sec(x) and dv = sec 2(x) dx.

First, find du and v:

du = sec(x) *tan(x) dx

v = sec^2(x) dx = tan(x)

Now substitute the above results into the partial integration formula:

sec^3(x) dx = sec(x) *tan(x) -tan(x) *sec(x) *tan(x) dx

Simplification: rent and acres.

sec^3(x) dx = sec(x) *tan(x) -sec(x) *tan^2(x) dx

Again, the partial integral method is used to calculate sec(x) *tan 2(x) dx:

Let u = tan(x) and dv = sec(x) *tan(x) dx

then du = sec 2(x) dx and v = sec(x).

Substitute the above results into the partial integration formula:

sec(x) *tan^2(x) dx = tan(x) *sec(x) -sec^3(x) dx

Now substitute this result into the previous equation:

sec^3(x) dx = sec(x) *tan(x) -tan(x) *sec(x) -sec^3(x) dx)

Move sec 3(x) dx to the right of the equal sign:

2∫ sec^3(x) dx = sec(x) *tan(x) -tan(x) *sec(x)

Simplification: sec 3(x) dx = 1 2) *sec(x) *tan(x) +c

where c is the integral constant. So, the indefinite integral of sec 3(x) is (1 2) *sec(x) *tan(x) +c.

The indefinite integral of secx to the nth power is calculated as follows:

Functional properties of secant functions1) Define the domain.

x cannot take the equivalent of 90 degrees, 270 degrees, -90 degrees, -270 degrees; That is.

2) Value range. secx 1 or secx 1, i.e. (-1] [1,+3) y=secx is an even function, i.e. sec( =sec The image is symmetrical to the y-axis.

4) y=secx is a periodic function.

The period is 2k (k z, and k ≠ 0), and the minimum positive period t=2.

5) Monotonia.

2k - 2,2k ], 2k + 2k + 3 2), decreasing on k z; Increments over the interval [2k, 2k+2), (2k+2, 2k+kz).

i=∫(secx)^3dx=(1/2)×(secxtanx+ln|secx+tanx|)+c。

The process is as follows:

secxd(tanx)。

secxtanx-∫tanxd(secx)。

secxtanx-∫secx(tanx)^2dx。

secxtanx-∫(secx)^3dx+∫secxdx。

secxtanx-i+ln|secx+tanx|。

i=(1/2)×(secxtanx+ln|secx+tanx|)+c。

1. Partial integration method is an important and basic method for calculating integrals in calculus.

2. The formula of the partial integration method is: u(x)v'(x)dx= u(x)dv(x)=u(x)*v(x)-only excitation- v(x)du(x).

3. The division refers to the common form of the celebration socks.

1. Find the integral of the function containing e x.

x*e^xdx=∫xd(e^x)=x*e^x-∫e^xdx。

2. Find the integral of a function with trigonometric functions.

x*cosxdx=∫x*d(sinx)=x*sinx-∫sinxdx。

3. Find the integral of the function containing arctanx.

x*arctanxdx=1/2∫arctanxd(x^2)=1/2(x^2)*arctanx-1/2∫(x^2)d(arctanx)。