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).

& \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{aligned}$$ We know that, $$\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.