Mathematics

# Evaluate the following integral:$\displaystyle\int{\sqrt{\dfrac{1-\sin{2x}}{1+\sin{2x}}}dx}$

##### SOLUTION
$\displaystyle\int{\sqrt{\dfrac{1-\sin{2x}}{1+\sin{2x}}}dx}$

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

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

$=\displaystyle\int{\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}$
Let $t=\sin{x}+\cos{x}\Rightarrow\,dt=\left(\cos{x}-\sin{x}\right)dx=-\left(\sin{x}-\cos{x}\right)dx$

$=-\displaystyle\int{\dfrac{dt}{t}}$

$=-\log{\left|t\right|}+c$

$=-\log{\left|\sin{x}+\cos{x}\right|}+c$ where $t=\sin{x}+\cos{x}$

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

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