Question

Evaluate the given integral
\[\int{x.{{\csc }^{2}}x.dx}\]

Answer 1

Hint: Solve the integral by doing integration by parts. Take \[u=x\] and \[v={{\csc }^{2}}x\]. Hence apply the value of u and v in the formula and simplify it. Find the integral of v and the differentiation of u, so you can apply these directly in the formula.

Complete step by step answer:
We have been given an integral, which we need to evaluate. Let us take the integral as I.
\[I=\int{x.{{\csc }^{2}}x.dx}\]
We can solve the expression by doing integration by parts. It is a special method of integration that is often useful when two functions are multiplied together. If u and v are two functions, the integration by parts is given as,
\[\int{uv.dx=u\int{v.dx}-\int{u'\left( \int{v.dx} \right).dx}}-(1)\]
Now, let us take, \[u=x\] and \[v={{\csc }^{2}}x\].
\[\therefore u'=\dfrac{du}{dx}=1\] and \[\int{v.dx}=-\cot x\].
Thus we can substitute these values in equation (1).
\[\begin{align}
  & \int{x.{{\csc }^{2}}x.dx}=x\left( -\cot x \right)-\int{1\times \left( -\cot x \right)dx} \\
 & \int{x.{{\csc }^{2}}x.dx}=-x\cot x+\int{\cot x.dx} \\
\end{align}\]
We know that, \[\int{\cot x.dx}=\ln \left| \sin x \right|+C\].
\[\therefore \int{x.{{\csc }^{2}}x.dx}=-x\cot x+\ln \left| \sin x \right|+C\]
Thus, we have integrated \[\int{x.{{\csc }^{2}}x}\] by integration by parts.

\[\therefore \int{x.{{\csc }^{2}}x.dx}=-x\cot x+\ln \left| \sin x \right|+C\]

Note: In integration by parts we consider the functions as u and v. If you take \[u={{\csc }^{2}}x\] and \[v=x\], the entire integration becomes complex. Thus always take \[u=x\] in such kind of integration questions. And remember the basic trigonometric and integral formulas we have used here.