**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.