∫cot²(x)dx
To find the integral of cot²(x), we can use a trigonometric identity and a basic integration technique
To find the integral of cot²(x), we can use a trigonometric identity and a basic integration technique.
First, let’s recall the trigonometric identity for cotangent squared:
cot²(x) = 1 + csc²(x)
By substituting this identity into the integral, we get:
∫cot²(x)dx = ∫(1 + csc²(x))dx
Next, we can split this integral into two separate integrals:
∫(1 + csc²(x))dx = ∫1dx + ∫csc²(x)dx
The integral of 1dx is simply x:
∫1dx = x + C1, where C1 is the constant of integration for the first integral.
Now, let’s focus on the second integral, which is the integral of csc²(x). To solve this integral, we can use a u-substitution.
Let u = cot(x), then du = -csc²(x)dx
Rearrange the equation to solve for dx:
dx = -du/csc²(x)
Substituting du and -du/csc²(x) for dx in the integral, we get:
∫csc²(x)dx = ∫csc²(x)(-du/csc²(x))
The csc²(x) terms cancel out, leaving us with:
∫csc²(x)(-du/csc²(x)) = -∫du
The integral of -du is simply -u:
-∫du = -u + C2, where C2 is the constant of integration for the second integral.
Now, let’s put everything together:
∫cot²(x)dx = x – cot(x) + C, where C = C1 + C2 is the constant of integration for the entire expression.
Therefore, the integral of cot²(x) is x – cot(x) + C, where C is a constant.
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