Question

Integrate :-


`:\implies \: \boxed{ \displaystyle \int \sf{ \frac{1}{sin {}^(2) \bigg((x - 2)/(3) \bigg) } \: dx }} \: \red\bigstar`
​

Answer 1

`\large\displaystyle\text{$\begin{gathered}\sf \bf{Your \ exercise \ is \to\int\limits (1)/(sin^2\left((x-2)/(3)\right) )dx } \end{gathered}$}`

`\boldsymbol{Replaces \ u=(x-2)/(3) \longmapsto (du)/(dx)=(1)/(3) \longmapsto \ dx=3 \ du }`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=\int\limits (3)/(Sin^2(u) )du } \end{gathered}$}`

`\boldsymbol{\sf{Simplify }}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=3\int\limits csc^2(u)du } \end{gathered}$}`

`\boldsymbol{\sf{Solving \ now }}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\int\limits csc^2(u)du } \end{gathered}$}`

`\boldsymbol{\sf{This\:is\:a\:standard\:integral. }}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=-cot(u)} \end{gathered}$}`

`\boldsymbol{\sf{We\:replace\:the\:integrals\:already\:solved. }}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=\int\limits (3)/(Sin^2(u) )du } \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=-cot(u)} \end{gathered}$}`

`\boldsymbol{The\:substitution\:is\:undone\: u=(x-2)/(3) ;}`

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=-3cot\left((x-2)/(3)\right) } \end{gathered}$}`

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf Answer \longmapsto \bf{\int\limits (1)/(sin^2\left((x-2)/(3)\right) )dx }=\boldsymbol{ -3cot\left((x-2)/(3)\right)+C } \end{gathered}$}}`

`\large\displaystyle\text{$\begin{gathered}\sf \underline{\bf{\green{I\:hope \ it \ helps \ you..... Regards }}} \end{gathered}$}`

Answer 2

Recall that

`(d)/(dx)\left[\cot(x)\right] = -\csc^2(x)`

so that

`\displaystyle \int \frac{dx}{\sin^2\left(\frac{x-2}3\right)} = \int \csc^2\left(\frac{x-2}3\right) \, dx = \boxed{-3 \cot\left(\frac{x-2}3\right) + C}`