Mathematics

# Evaluate:$\int { \cfrac { \cos { 2x } +2\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } } dx$

##### SOLUTION
Given $\displaystyle\int{\dfrac{\cos{2x}+2{\sin}^{2}{x}}{{\sin}^{2}{x}}dx}$

$=\displaystyle\int{\dfrac{2{\cos}^{2}{x}-1+2{\sin}^{2}{x}}{{\sin}^{2}{x}}dx}$    $[\because \cos2x=2\cos^2-1]$

$=\displaystyle\int{\dfrac{2\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)-1}{{\sin}^{2}{x}}dx}$

$=\displaystyle\int{\dfrac{2-1}{{\sin}^{2}{x}}dx}$ since $\left({\sin}^{2}{x}+{\cos}^{2}{x}=1\right)$

$=\displaystyle\int{\dfrac{1}{{\sin}^{2}{x}}dx}$

$=\displaystyle\int{{\csc}^{2}{x}dx}$

$=-\cot{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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