Solving the Integral of csc^2(x) dx | Using Trigonometric Identities and Integration by Parts

∫ csc^2(x) dx

To solve the integral of csc^2(x) dx, we can use a trigonometric identity to rewrite the integral in terms of another trigonometric function

To solve the integral of csc^2(x) dx, we can use a trigonometric identity to rewrite the integral in terms of another trigonometric function.

Recall the identity: csc^2(x) = 1 + cot^2(x)

Using this identity, we can rewrite the integral as:

∫ csc^2(x) dx = ∫ (1 + cot^2(x)) dx

Now, let’s integrate each term separately.

∫ 1 dx = x + C (where C is the constant of integration)

∫ cot^2(x) dx: This is a bit more tricky, as it requires a technique called “integration by parts”.

Integration by parts states:

∫ u dv = uv – ∫ v du

In this case, let’s choose u = cot(x) and dv = cot(x) dx.
Differentiating u, we get du = -csc^2(x) dx (since the derivative of cot(x) is -csc^2(x)),
and integrating dv, we get v = ln|sin(x)| (since the integral of cot(x) is ln|sin(x)|).

Using the formula for integration by parts, we have:

∫ cot^2(x) dx = ∫ u dv = uv – ∫ v du
= cot(x) ln|sin(x)| – ∫ ln|sin(x)| (-csc^2(x)) dx
= cot(x) ln|sin(x)| + ∫ ln|sin(x)| csc^2(x) dx

The integral on the right side, ∫ ln|sin(x)| csc^2(x) dx, is the same as the original integral we started with, ∫ csc^2(x) dx.

So, let’s call this integral I. We can rewrite it as:

I = ∫ csc^2(x) dx

Now, let’s substitute this back into the equation for ∫ cot^2(x) dx:

∫ cot^2(x) dx = cot(x) ln|sin(x)| + I

Therefore, we have:

I = ∫ csc^2(x) dx = cot(x) ln|sin(x)| + I

To find I, we can subtract I from both sides:

0 = cot(x) ln|sin(x)|

This equation is true for certain values of x. For example, when x is any multiple of π (e.g., x = π, 2π, etc.), both cot(x) and sin(x) become 0, making I = 0 for those values of x.

Hence, the integral ∫ csc^2(x) dx does not have a simple closed-form solution.

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