∫cot(x)dx
-ln|csc(x)| +C
To integrate ∫cot(x)dx, we can use the method of substitution. Let’s consider the following substitution:
u = sin(x)
Then, we can express cot(x) in terms of u:
cot(x) = cos(x) / sin(x) = √(1 – sin^2(x)) / sin(x) = √(1 – u^2) / u
We can also express dx in terms of u:
du / dx = cos(x)
dx = du / cos(x)
Substituting these expressions into the integral, we obtain:
∫cot(x)dx = ∫(√(1 – u^2) / u) (du / cos(x))
Simplifying this expression, we get:
∫cot(x)dx = ∫(√(1 – u^2) / u) sec(x) du
Now, let’s recall a trigonometric identity:
sec(x) = 1 / cos(x)
Substituting this into the integral, we obtain:
∫cot(x)dx = ∫(√(1 – u^2) / u) (1 / cos(x)) cos(x) du
Simplifying this expression, we get:
∫cot(x)dx = ∫√(1 – u^2) du
This integral can be evaluated using the substitution u = sin(t):
u = sin(t)
du = cos(t) dt
Substituting these expressions into the integral, we obtain:
∫cot(x)dx = ∫√(1 – u^2) du = ∫cos(t) cos(t) dt
∫cot(x)dx = ∫cos^2(t) dt
Using the trigonometric identity:
cos^2(t) = (1 + cos(2t)) / 2
We can express the integral as:
∫cot(x)dx = (1/2) ∫(1 + cos(2t)) dt
∫cot(x)dx = (1/2)(t + (1/2) sin(2t)) + C
Substituting back u = sin(x), we obtain:
∫cot(x)dx = (1/2)(arcsin(u) + (1/2) sin(2arcsin(u))) + C
Therefore, the antiderivative of cot(x) is:
∫cot(x)dx = (1/2)(arcsin(sin(x)) + (1/2) sin(2arcsin(sin(x)))) + C
Simplified as:
∫cot(x)dx = (1/2)(x + sin(x) cos(x)) + C
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