Integral Of Cscx

(cscx) Find the integral part of 4 dx?

I wrote for csc 4x (csc x) 4 and other functions.

It is useful to know or table that csc 2 x dx = an integral part of cot x. (This is common knowledge, because when you study the derivative you calculate all the derivatives of sin. The derivative of cot x is csc 2x.

It is also useful to know or table what is 1 + cot 2x = csc 2x. (This sin comes from dividing both sides of both 2x + cos 2x = 1 sin x 2x.) So you can apply this problem in the following way:

Integral csc 4 x dx = required csc 2 x csc 2 x dx

= csc 2 x (1 + cot 2 x) Integral of dx.

[... with ID only]

= Mandatory csc 2 x dx + Mandatory csc 2 x cot 2 x dx.

[... spreads uct and distributes mandatory on money]

= cot x + csc 2 x cot 2 x is an integral part of dx.

[... because I remember that the essential part of csc 2 x dx is only kno and cot x]

Try adding u = cot x for the rest of the integration. You have du = csc 2 x dx then 1 / (csc 2 x) dx = you and so on

csc 2 x cot 2 x dx = integral part of (csc 2 x) u 2 [1 / (csc 2 x)] of

Integral of = u 2 du = u 3/3 = (cot 3x) / 3.

So when we connect it to ve, we see.

An integral part of csc 4 x dx = cot x (cot x) 3) / 3 + C.

The difficulty of this requirement is not so much in the technique used, but in the memory of the identification which is satisfied with csc and cot.

For related issues you may be familiar with, you can use integer seconds 4 x dx in the same way. One theory (derived from the division of cos 2x by sin 2 x + cos 2x = 1) x) = (s 2 x) (tan 2 x + 1) divided by a combination of integrals of seconds 2 x dx and integrals of seconds 2 x tan 2 x. The first can do this by remembering that the essential part of the tan is dry 2, and the second can do this by adding u = tan (since you are dry 2).

In fact, I knew W would do the necessary thing because I remember that W was embedded in سیک 4x and the memory was similar to CSC 4x. So it's just a matter of memorizing / minimizing relevant ideas so that the same strategy works.

Integral Of Cscx

(cscx) Find the integral of 4 dx? 3
I wrote csc 4x for (csc x) 4 and the same for other functions.
It is useful to know that the integration of csc 2 x dx = cot x is useful. (This is known because when you study derivatives, you calculate all the derivatives of sin / cos / tan / csc / sec / tan and therefore d / dx (cot x) = csc 2 x find.) or equivalent is a derivative from which cradle x csc 2x.
It is also useful to know that 1 + cot 2x = csc 2x. (This is obtained by dividing both sides of sin 2x + cos 2x = 1 by sin 2x.) So you can apply this problem as follows:
Integral CSC 4 x DX = Integral CSC 2 x CSC 2 x DX
= csc 2 x (1 + cot 2 x) Integral of dx
[... with ID only]
= Integral CSC 2 x DX + Integral CSC 2 x Quote 2 x DX
[... spreads uct and distributes integrals over aggregates]
= Integral of Cot x + csc 2 x cot 2 x dx
[... because I remember that the integral of csc 2 x dx is only kno and cot x]
Try adding u = cot x for the rest of the integrals. You have du = csc 2 x dx, then 1 / (csc 2x) dx = you etc.
Integral of csc 2 x cot 2 x dx = (csc 2 x) u 2 [1 / (csc 2 x)]
= Integral of u 2 du = u 3/3 = (cot 3x) / 3
So when we connect it to ve, we see.
Integral of csc 4 x dx = cot x (cot x) 3) / 3 + C.
The difficulty of this integral is not in the technique used but in the identity memory filled by csc and cradle.
For related issues you may already be familiar with, you can use the same integer second 4 x dx. Returns an iodine (divided by sin 2 x + cos 2 x = 1 by cos 2 x) tan 2 x + 1 = sec 2 x, which is an integral of sec 4 x = (sec 2). x) (sec 2 x) = (s 2 x) (tan 2 x + 1) sec تقسیم 2 x dx integrals and seconds divided by 2 x tan 2 x integrals. The first can do this by remembering that the integral of tan is dry 2, and the second can do this by adding u = tan (since you are dry 2).
In fact, I knew that w would do its job because I remember that w was built-in sec 4x and its memory was like csc 4x. So it's just a matter of remembering / lowering the relevant ideal so that the same strategy works.