∫(csc²x)dx
To find the integral of the function ∫(csc²x)dx, we can make use of trigonometric identities and integration techniques
To find the integral of the function ∫(csc²x)dx, we can make use of trigonometric identities and integration techniques.
First, let’s rewrite the given function using trigonometric identities. The reciprocal of sin(x) is known as cosec(x), so we can rewrite csc²(x) as (1/sin(x))².
Now, we have the integral of (1/sin(x))² dx. To simplify this further, we can use another trigonometric identity, namely the Pythagorean identity: 1 + cot²(x) = csc²(x). Rearranging this identity, we get cot²(x) = csc²(x) – 1.
Our initial integral can now be written as ∫(1/sin(x))² dx = ∫(csc²(x))dx = ∫(cot²(x) – 1)dx.
Now, let’s split the integral into two parts:
∫(cot²(x) – 1)dx = ∫cot²(x)dx – ∫1 dx.
The integral of cot²(x) can be approached by using integration by parts. Let’s define u = cot(x) and dv = cot(x) dx. Therefore, du = -csc²(x) dx and v = ∫cot(x) dx.
Integrating v (i.e., ∫cot(x) dx) requires using the natural logarithm (ln) function. Evaluating ∫cot(x) dx would give us ln|sin(x)|.
Now, we can apply integration by parts:
∫cot²(x) dx = ∫u dv
= uv – ∫v du
= cot(x)∫cot(x) dx – ∫ln|sin(x)|(-csc²(x)) dx
= cot(x)ln|sin(x)| + ∫ln|sin(x)| csc²(x) dx.
Notice how the last term on the right-hand side is similar to our original integral. Let’s denote it as I:
I = ∫ln|sin(x)| csc²(x) dx.
Now, we can substitute I back into the equation:
∫cot²(x) dx = cot(x)ln|sin(x)| + I.
Now, let’s solve the second part of the integral: ∫1 dx. The integral of a constant is equal to the constant multiplied by the variable of integration. In this case, the integral of 1 dx is simply x.
Therefore, ∫1 dx = x.
Putting everything together:
∫(csc²x)dx = ∫cot²(x)dx – ∫1 dx
= (cot(x)ln|sin(x)| + I) – x
= cot(x)ln|sin(x)| + C,
where C is the constant of integration.
Thus, the integral of ∫(csc²x)dx is equal to cot(x)ln|sin(x)| + C, where C is the constant of integration.
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