Question

integral 1+cos squarex/1-cos2x dxxnotequaltonpie,nbelongs to z​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{1 + {cos}^{2}x}{1 - cos2x} \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ 1 - cos2x = {2sin}^{2}x}}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{1 + {cos}^{2}x }{ {2sin}^{2}x} \: dx$

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int\rm \bigg[\dfrac{1}{ {sin}^{2} x} + \dfrac{ {cos}^{2}x}{ {sin}^{2}x}\bigg] \: dx$

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int\rm \bigg[ {cosec}^{2}x + {cot}^{2}x\bigg] \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1}}}$

So, using this, we get

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int\rm \bigg[ {cosec}^{2}x + {cosec}^{2}x - 1\bigg] \: dx$

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int\rm \bigg[ 2{cosec}^{2}x - 1\bigg] \: dx$

$\rm \:  =  \: \dfrac{1}{2}[ - 2cotx - x] + c$

$\rm \:  =  \: - \: cotx \: - \: \dfrac{x}{2} \: + \: c$

Hence,

$\red{\boxed{\tt{ \rm \: \displaystyle\int\rm \frac{1 + {cos}^{2} x}{1 - cos2x}dx =  \: - \: cotx \: - \: \dfrac{x}{2} \: + \: c}}}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$