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
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