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Integral of Cot x is ln |sin x| + C. Cot x is among one of the trigonometric functions that is the ratio of cosine and sine. The integral of cot x is mathematically represented as ∫cot x dx = ln |sinx| + C. In this article, we will explore the integral of cot x, the integral of cot x formula, the derivation of the integral of cot x, definite integral of cot x along with some examples based on 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. The integral of cot x means finding the antiderivative of the cot x. The process of finding the anti-derivative of a function is called the integration. The result of the integration is called as integral. Hence, the antiderivative of the cot x is ln |sin x| + C.
Read in Detail:
The integral of cot x formula is given by:
∫cot x dx = ln |sin x| + C
Integral of Cot x in terms of cosec x is given as follows:
∫cot x dx = - ln |cosec x| + C
We can derive the integral of cot x by using the Substitution Method in integration.
To prove integral of cot x we will use the integration by substitution method which is described below:
We know that,
cot x = cos x / sin x
Integrating both sides we get,
∫cot x dx = ∫ [cos x / sin x] dx ----(1)
Let t = sin x
Differentiating both sides w.r.t t, we get
dt = cos x dx
Putting the above values in equation (1)
∫cot x dx = ∫ [1 / t] dt
∫cot x dx = ln |t| + C
Putting value of t
∫cot x dx = ln |sin x| + C
The integral of cot x is ln |sin x| + C.
Integral of cot x with the upper and lower limit is called as the definite integral of cot x. In this we apply the limits and evaluate the resultant value for the integral. The value of definite integral of cot x is given below:
The value of integral of cot x with lower limit 0 and upper limit π/2 is given below:
We know that,
∫cot x dx = ln |sin x| + C
Applying lower limit = 0 and upper limit = π/2, we get
∫0π/2cot x dx = [ln |sin x| ]0π/2
∫0π/2cot x dx = ln |sin(π/2) | - |ln sin (0) |
∫0π/2cot x dx = ln |sin(π/2) | - |ln 0|
Since, the ln 0 is not defined, the definite integral ∫0π/2cot x dx diverges.
The value of integral of cot x with lower limit π/4 and upper limit π/2 is given below:
We know that,
∫cot x dx = ln |sin x| + C
Applying lower limit = π/4 and upper limit = π/2
∫π/4π/2cot x dx = [ln |sin x| ]π/4π/2
⇒ ∫π/4π/2cot x dx = ln |sin(π/2) | - |ln sin(π/4) |
⇒ ∫π/4π/2cot x dx = ln 1 - ln (1/√2)
⇒ ∫π/4π/2cot x dx = ln 1 - [ln 1 - ln √2]
⇒ ∫π/4π/2cot x dx = ln (√2)
Integral of Cot x from pi/4 to pi/2 is ln (√2).
Some important points related to integral of cot x are:
Read More:
Example 1: Find ∫cot 6x dx
Solution:
We have ∫cot 6x dx ------(1)
Let t = 6x
Differentiating w.r.t t
dt = 6 dx
⇒ dx = dt / 6
Putting in (1)
∫cot 6x dx = ∫cot t (dt / 6)
⇒ ∫cot 6x dx = (1 / 6) ∫cot t dt
⇒ ∫cot 6x dx = (1 / 6) [ln |sin t| + C]
⇒ ∫cot 6x dx = (1 / 6) [ln |sin (6x) | + C]
Example 2: Evaluate: ∫cot x cosec2x dx
Solution:
Let I = ∫cot x cosec2x dx -----(1)
Take t = cot x
Differentiating w.r.t t
dt = - cosec2x dx
putting in (1)
I = -∫t dt
⇒ I = -t2 / 2 + C (putting values)
⇒ I = - cot2x / 2 + C
⇒ ∫cot x cosec2x dx = - cot2x / 2 + C
Example 3: Solve ∫cot x. sec x dx
Solution:
I = ∫cot x. sec x dx
We know that,
cot x = cos x / sin x and sec x = 1 / cos x
Putting in I
I = ∫ [cos x / sin x]. [1 / cos x] dx
⇒ I = ∫ [1 / sin x] dx
⇒ I = ∫ cosec x dx
⇒ I = - ln | cosec x + cot x| + C
Example 4: Evaluate ∫cot2 x dx
Solution:
I = ∫cot2 x dx
We know that,
[d / dx] (cosec x) = - cot2 x
cot2 x = - [d / dx] (cosec x)
Putting in I
I = ∫ - [d / dx] (cosec x) dx
By the property ∫[d / dx] f(x) dx = f(x) + C
I = - cosec x + C
Q1. Solve ∫cot x. cos x dx.
Q2. Evaluate the integral ∫ [cot x / √ (6 + 16cot2x)] dx.
Q3. Find ∫ cot (4x) dx.
Q4. Evaluate ∫ (1 + cot x) / (1 - cot x) dx
Q5. Evaluate ∫(sin x)/(cos2x) dx.
Q6. Find ∫(sec2x)/(sec x + tan x) dx.
Q7. Evaluate ∫cot2x dx.
Q8. Find ∫(cos2x)/(sin x)dx.
Q9. Evaluate ∫sin3x cos x dx.
Q10. Find ∫(1 - cos2 x) / (sin2x)dx.
In practice, this integral provides a useful tool for solving problems involving trigonometric functions, particularly in calculus. By understanding the derivation and applying the integral in definite form, one can effectively handle a range of problems in both theoretical and applied contexts.
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