Before walk to discover what is the integral the cot x, let united state recall couple of facts about the cot (or) cotangent function. In a right angled triangle, if x is among the acute angles, climate cot x is the proportion of the surrounding side the x to the opposite next of x. Therefore it can be written as (cos x)/(sin x) together cos x = (adjacent)/(hypotenuse) and sin x = (opposite)/(hypotenuse). We usage these facts to uncover the integral of cot x.

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Let us find out the integral of cot x formula together with its proof and also examples.

1.What is the Integral that Cot x dx?
2.Integral the Cot x proof by Substitution
3.Definite Integral that Cot x dx
4.FAQs ~ above Integral that Cot x

What is the Integral that Cot x dx?

The integral of cot x dx is same to ln |sin x| + C. It is stood for by ∫ cot x dx and also so ∫ cot x dx = ln |sin x| + C, wherein 'C' is the integration constant. Here,

'∫' is the price of integration.dx represents the the integration is with respect to x.


We usage the integration through substitution technique to prove this integration that cot x formula. Us will check out it in the upcoming section.

Integral of Cot x proof by Substitution

Here is the source of the formula the integral of cot x by using integration by substitution method. Because that this, let us recall the cot x = cos x/sin x. Climate ∫ cot x dx becomes

∫ cot x dx = ∫ (cos x)/(sin x) dx

Substitute sin x = u. Climate cos x dx = du. Climate the over integral becomes

= ∫ (1/u) du

= ln |u| + C (Because ∫ 1/x dx = ln|x| + C)

Substitute u = sin x earlier here,

= ln |sin x| + C

Thus, ∫ cot x dx = ln |sin x| + C.

Hence proved.

Definite Integral that Cot x dx

The identify integral that cot x dx is an integral through lower and upper bounds. To evaluate it, us substitute the upper and lower borders in the worth of the integral cot x dx and also subtract them in the same order. Us will work on some definite integrals that cot x dx.

Integral of Cot x indigenous 0 to pi/2

∫(_0^pi/2) cot x dx = ln |sin x| (left. ight|_0^pi/2)

= ln |sin π/2| - ln |sin 0|

= ln |1| - ln 0

We recognize that ln 1 = 0 and also ln 0 is not defined. So

∫(_0^pi/2) cot x dx = Diverges

Therefore, the integral the cot x indigenous 0 come π/2 diverges.

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Integral that Cot x indigenous pi/4 come pi/2

∫(_pi/4^pi/2) cot x dx = ln |sin x| (left. ight|_pi/4^pi/2)

= ln |sin π/2| - ln |sin π/4|

= ln |1| - ln |1/√2|

By among the properties of logarithms, ln (m/n) = ln m - ln n. So

= ln 1 - ln 1 + ln √2

= ln √2

∫(_pi/4^pi/2) cot x dx = ln √2

Therefore, the integral that cot x indigenous π/4 to π/2 is same to ln √2.

Important notes on Integral the Cot x dx:

∫ cot x dx = ln |sin x| + C∫ cot2x dx = - csc x + C as d/dx(csc x) = -cot2x + C

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