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\[\int{x.{{\csc }^{2}}x.dx}\]

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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.

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.

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