Integral of Cot x

Before going to learn what is the integral of cot x, let us recall few facts about the cot (or) cotangent function. In a right angled triangle, if x is one of the acute angles, then cot x is the ratio of the adjacent side of x to the opposite side of x. So it can be written as (cos x)/(sin x) as cos x = (adjacent)/(hypotenuse) and sin x = (opposite)/(hypotenuse). We use these facts to find the integral of cot x.

Let us learn the integral of cot x formula along with its proof and examples.

The integral of cot x dx is equal to ln |sin x| + C. It is represented by ∫ cot x dx and so ∫ cot x dx = ln |sin x| + C, where 'C' is the integration constant. Here,

• '∫' is the symbol of integration.
• cot x is the integrand.
• dx represents that the integration is with respect to x.

We use the integration by substitution method to prove this integration of cot x formula. We will see it in the upcoming section.

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

∫ cot x dx = ∫ (cos x)/(sin x) dx

Substitute sin x = u. Then cos x dx = du. Then the above integral becomes

= ∫ (1/u) du

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

Substitute u = sin x back here,

= ln |sin x| + C

Thus, ∫ cot x dx = ln |sin x| + C.

Hence proved.

The definite integral of cot x dx is an integral with lower and upper bounds. To evaluate it, we substitute the upper and lower limits in the value of the integral cot x dx and subtract them in the same order. We will work on some definite integrals of cot x dx.

Integral of Cot x From 0 to pi/2

∫\(_0^{\pi/2}\) cot x dx = ln |sin x| \(\left. \right|_0^{\pi/2}\)

= ln |sin π/2| - ln |sin 0|

= ln |1| - ln 0

We know that ln 1 = 0 and ln 0 is NOT defined. So

∫\(_0^{\pi/2}\) cot x dx = Diverges

Therefore, the integral of cot x from 0 to π/2 diverges.

Integral of Cot x From pi/4 to pi/2

∫\(_{\pi/4}^{\pi/2}\) cot x dx = ln |sin x| \(\left. \right|_{\pi/4}^{\pi/2}\)

= ln |sin π/2| - ln |sin π/4|

= ln |1| - ln |1/√2|

By one of 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 of cot x from π/4 to π/2 is equal to ln √2.

Important Notes on Integral of Cot x dx:

• ∫ cot x dx = ln |sin x| + C
• Since sin x and csc x are reciprocals of each other,
  ∫ cot x dx = ln |csc x|^-1 + C = - ln |csc x| + C
• ∫ cot²x dx = - csc x + C as d/dx(csc x) = -cot²x + C

Topics Related to Integral of Cot x dx:

• Integration Formulas
• Integral of Sin x
• Integral of Sec x
• Integral Calculator
• Calculus Calculator

What is the Integral of Cot x?

The integral of cot x is ln |sin x| + C. It is mathematically denoted as ∫ cot x dx = ln |sin x| + C.

Is Derivative of Cot x Equal to Integral of Cot x?

No, the derivative of cot x is -csc²x and the integral of cot x is ln |sin x| + C. i.e.,

• d/dx(cot x) = -csc²x
• ∫ cot x dx = ln |sin x| + C

What is the Integral of Cot x From 0 to Pi?

We know that ∫ cot x dx = ln |sin x| + C. Substituting the limits, ln |sin π| - ln |sin 0| = Diverges (as 0 is NOT in the domain of logarithmic function).

What is the Integral of Cot x Sec x?

∫ cot x sec x dx = ∫(cos x)/(sin x) · (1/cos x) dx = ∫ 1/sin x dx = ∫ csc x dx = - ln |csc x + cot x| + C.

How to Derive the Integral of Cot x Formula?

To derive the formula for ∫ cot x dx, we write cot x as (cos x)/(sin x). Then the integral becomes ∫ (cos x)/(sin x) dx. Assume that sin x = u, then cos x dx = du. Then the integral becomes ∫ (1/u) du = ln |u| + C (or) ln |sin x) + C. Therefore, ∫ cot x dx = ln |sin x| + C.

What is the Integral of Cot^2x Equal to?

We know that the derivative of csc x is -cot²x. So the derivative of -csc x is cot²x. Since integration is the reverse process of differentiation, ∫ cot²x dx = -csc x + C.

What is the Integral of Cot x Csc²x dx?

Assume that cot x = u then -csc²x dx = du. Then the given integral becomes ∫ -u du = -u²/2 + C = -cot²x/2 + C.