Mathematics

# Evaluate the following integral$\int { \cfrac { 1 }{ \sqrt { 1-\cos { 2x } } } } dx\quad$

##### SOLUTION
$I=\displaystyle\int{\dfrac{dx}{\sqrt{1-\cos{2x}}}}$

$I=\displaystyle\int{\dfrac{dx}{\sqrt{1-1+2{\sin}^{2}{x}}}}$

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

$I=\dfrac{1}{\sqrt{2}}\displaystyle\int{\csc{x}dx}$

$I=\dfrac{1}{\sqrt{2}}\displaystyle\int{\dfrac{\csc{x}\left(\csc{x}-\cot{x}\right)}{\left(\csc{x}-\cot{x}\right)}dx}$

Let $t=\csc{x}-\cot{x}\Rightarrow\,\left(-\csc{x}\cot{x}+{\csc}^{2}{x}\right)dx=dt$

$\Rightarrow\,\csc{x}\left(-\cot{x}+\csc{x}\right)dx=dt$

$I=\dfrac{1}{\sqrt{2}}\displaystyle\int{\dfrac{dt}{t}}$

$I=\dfrac{1}{\sqrt{2}}\log{\left|t\right|}+c$

$I=\dfrac{1}{\sqrt{2}}\log{\left|\csc{x}-\cot{x}\right|}+c$

$I=\dfrac{1}{\sqrt{2}}\log{\left|\dfrac{1-\cos{x}}{\sin{x}}\right|}+c$

$I=\dfrac{1}{\sqrt{2}}\log{\left|\dfrac{2{\sin}^{2}{\dfrac{x}{2}}}{2\sin{\dfrac{x}{2}\cos{\dfrac{x}{2}}}}\right|}+c$

$I=\dfrac{1}{\sqrt{2}}\log{\left|\dfrac{\sin{\dfrac{x}{2}}}{\cos{\dfrac{x}{2}}}\right|}+c$

$\therefore\,I=\dfrac{1}{\sqrt{2}}\log{\left|\tan{\dfrac{x}{2}}\right|}+c$

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

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