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
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