∫cot²(x)dx
To find the integral of cot²(x) with respect to x, we can use trigonometric identities and integration techniques
To find the integral of cot²(x) with respect to x, we can use trigonometric identities and integration techniques.
First, let’s rewrite cot²(x) using trigonometric identities. The identity we can use is:
cot²(x) = csc²(x) – 1
So, the integral becomes:
∫cot²(x)dx = ∫(csc²(x) – 1)dx
Now, let’s integrate both terms separately.
∫csc²(x)dx:
To integrate csc²(x), we can use the identity:
csc²(x) = cot²(x) + 1
So, the integral becomes:
∫csc²(x)dx = ∫(cot²(x) + 1)dx
Since we already know that ∫cot²(x)dx = ∫(csc²(x) – 1)dx, we can rewrite the integral as:
∫csc²(x)dx = ∫(cot²(x) + 1)dx = ∫cot²(x)dx + ∫1 dx
Now, let’s integrate the second term:
∫1 dx:
Integrating 1 with respect to x gives us x. So,
∫1 dx = x
Now, let’s go back to the expression:
∫cot²(x)dx + ∫1 dx
We can substitute the integral of cot²(x) with ∫(csc²(x) – 1)dx:
∫(csc²(x) – 1)dx + ∫1 dx
Simplifying, we get:
∫cot²(x)dx = ∫csc²(x)dx – ∫1 dx + ∫1 dx
∫cot²(x)dx = ∫csc²(x)dx
Now, we can integrate ∫csc²(x)dx:
∫csc²(x)dx:
To integrate csc²(x), we can use the substitution method. Let’s substitute u = cot(x) and see what happens.
Differentiating both sides with respect to x:
du/dx = -csc²(x)
dx = -du/csc²(x)
Substituting into the integral:
∫csc²(x)dx = ∫csc²(x) (-du/csc²(x))
∫csc²(x)dx = -∫du
∫csc²(x)dx = -u + C
Substituting back u = cot(x), we get:
∫csc²(x)dx = -cot(x) + C
Hence, the integral of cot²(x)dx is:
∫cot²(x)dx = -cot(x) + C
Where C represents the constant of integration.
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