Question

Show that $\int{\csc xdx}=\ln \left| \tan \left( \dfrac{x}{2} \right) \right|+c$?

Answer 1

We first change the given expression of $\csc x$ to $\dfrac{\csc x\left( \csc x-\cot x \right)}{\left( \csc x-\cot x \right)}$ by multiplying $\left( \csc x-\cot x \right)$ to both its numerator and denominator. We change the variable from the assumption of $\left( \csc x-\cot x \right)=z$. The differential gives $\left( {{\csc }^{2}}x-\cot x\csc x \right)dx=dz$. We change the function and find the solution of the integral. We then sue the formulas of submultiple $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ to get the final solution.

Complete step by step answer:
We first change the given function of $\csc x$ to $\dfrac{\csc x\left( \csc x-\cot x \right)}{\left( \csc x-\cot x \right)}$ by multiplying $\left( \csc x-\cot x \right)$ to both its numerator and denominator.
We get $\csc x=\dfrac{\csc x\left( \csc x-\cot x \right)}{\left( \csc x-\cot x \right)}=\dfrac{{{\csc }^{2}}x-\csc x\cot x}{\left( \csc x-\cot x \right)}$.
We take $\left( \csc x-\cot x \right)=z$.
Differentiating we get
$\begin{aligned} & \left( \csc x-\cot x \right)=z \\\ & \Rightarrow \left( {{\csc }^{2}}x-\cot x\csc x \right)dx=dz \\\ \end{aligned}$
So, we get

& \int{\csc xdx} \\\ & =\int{\dfrac{{{\csc }^{2}}x-\csc x\cot x}{\left( \csc x-\cot x \right)}dx} \\\ & =\int{\dfrac{dz}{z}} \\\ & =\log \left| z \right|+c \\\ \end{aligned}$$ We replace the values and get $\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+c$. Now we simplify the function $\csc x-\cot x$. $\csc x-\cot x=\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}=\dfrac{1-\cos x}{\sin x}$ We use the formulas of submultiple and get $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$. $\dfrac{1-\cos x}{\sin x}=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}=\tan \dfrac{x}{2}$. Therefore, $\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+c=\ln \left| \tan \dfrac{x}{2} \right|+c$ **Note:** Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi +{{\left( -1 \right)}^{n}}a$ for $\sin \left( x \right)=\sin a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$. For our given problem $\sin \dfrac{x}{2}\ne 0$, the primary solution is $x\ne 0$. The condition will be $x\ne n\pi $. Here $n\in \mathbb{Z}$.