∫(cotx)dx
To find the integral of cot(x) dx, we can use a technique called substitution
To find the integral of cot(x) dx, we can use a technique called substitution.
Let’s start by carefully considering the integral:
∫(cot(x))dx
We can rewrite cot(x) as cos(x)/sin(x):
∫(cos(x)/sin(x))dx
Now, let’s substitute u = sin(x):
du/dx = cos(x)
So, we have:
∫(du/u)dx
Now, substitute back in x for u using x = arcsin(u):
∫(dx/asin(x))
To simplify this further, we can rewrite the denominator in terms of x:
∫(dx/(x/sqrt(1 – x^2)))
Now, we can rewrite this as:
∫(sqrt(1 – x^2)/x)dx
This integral requires another technique called trigonometric substitution. Consider substituting x = sin(θ):
dx = cos(θ)dθ
sqrt(1 – x^2) = sqrt(1 – sin^2(θ)) = sqrt(cos^2(θ)) = cos(θ)
Now, we have:
∫(cos(θ)/sin(θ))cos(θ)dθ
Simplify this as:
∫(cos^2(θ)/sin(θ))dθ
Using the trig identity cos^2(θ) = 1 – sin^2(θ), we can substitute to get:
∫((1 – sin^2(θ))/sin(θ))dθ
Simplify further:
∫(1/sin(θ) – sin(θ)/sin(θ))dθ
∫(cosec(θ) – 1)dθ
Finally, we have:
∫cosec(θ)dθ – ∫dθ
Now, let’s integrate each term separately:
∫cosec(θ)dθ = ln|cosec(θ) + cot(θ)| + C1
∫dθ = θ + C2
So, the overall integral becomes:
ln|cosec(θ) + cot(θ)| – θ + C
Now, substitute back x for θ:
ln|cosec(x) + cot(x)| – x + C
Therefore, the integral of the cot(x) function is:
∫(cot(x))dx = ln|cosec(x) + cot(x)| – x + C, where C is the constant of integration.
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