To solve the integral \(\int x \operatorname{cosec}^2 x \, dx\), we will use integration by parts, which is given by the formula:
[tex]\[
\int u \, dv = uv - \int v \, du
\][/tex]
Here, we need to identify parts of the integrand that fit \(u\) and \(dv\). Let's choose:
[tex]\[
u = x \quad \text{and} \quad dv = \operatorname{cosec}^2 x \, dx
\][/tex]
Next, we need to compute \(du\) and \(v\):
1. Differentiate \(u = x\):
[tex]\[
du = dx
\][/tex]
2. Integrate \(dv = \operatorname{cosec}^2 x \, dx\):
[tex]\[
v = -\cot x
\][/tex]
Using the integration by parts formula, we now have:
[tex]\[
\int x \operatorname{cosec}^2 x \, dx = uv - \int v \, du
\][/tex]
Substitute \(u\), \(v\), \(du\), and \(dv\) into the formula:
[tex]\[
\int x \operatorname{cosec}^2 x \, dx = x(-\cot x) - \int (-\cot x) \, dx
\][/tex]
Simplify the equation:
[tex]\[
\int x \operatorname{cosec}^2 x \, dx = -x \cot x + \int \cot x \, dx
\][/tex]
Next, solve \(\int \cot x \, dx\):
The integral \(\int \cot x \, dx\) is a standard integral, and its result is:
[tex]\[
\int \cot x \, dx = \log |\sin x|
\][/tex]
Substitute this result back into our previous equation:
[tex]\[
\int x \operatorname{cosec}^2 x \, dx = -x \cot x + \log |\sin x|
\][/tex]
Thus, the final solution to the integral \(\int x \operatorname{cosec}^2 x \, dx\) is:
[tex]\[
-x \cot x + \log |\sin x| + C
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
